text: 3 little bit mistakes edit approved Oct 13 '16 at 12:59 L.K. 16011 gold badge22 silver badges1212 bronze badges So I'll talk about polarization of classical electromagnetic waves just because you've already seen it. Imagine a wave travelling in the $$z$$ direction with the electric field always pointing in the same direction, say $$\pm x$$. This is called a linearly polarized wave. Same if the wave traveled in the $$z$$ direction and the electric field was in the plus or minus y direction. If those two waves were in phase and had the same magnitude, then their superposition would be a wave that oscillates at the same frequency/wavelength as the previous waves, and is still linearly polarized but this time not in the $$x$$ or $$y$$ direction but instead in the direction $$45$$ degrees (halfway) between them. Basically if the electric field always points in plus ofor minus the same direction, then that's linear polarization, and it could in theory be in any direction by adjusting the relative magnitude of an $$x$$ polarized one and a $$y$$ polarized one (that are in phase with each other). Now we're going to steal some math and some terminology. Think of multiplying by $$i$$ as changing the phase of the wave by a quarter period, then we built up a circular polarization by $$X+iY$$ and the other circular polarization by $$X+iiiY=X-iY$$$$X+iii Y=X-iY$$ so given two circular polarization you see that we can add them to get a linearly polarized state $$|+\rangle + |-\rangle$$ gives one of the linearly polarized states and $$-i(|+\rangle - i|-\rangle)$$$$-i(|+\rangle - |-\rangle)$$ gives a linearly polarized state orthogonal to the other one. We can borrow all the math and terminology from the classical waves, and associate the spin states of the photon with the right and left circularly polarized waves. We are stealing the math and stealing the terminology, but the fact is that we have two vectors $$|+\rangle$$ and $$|-\rangle$$ and they span a (complex) two space of possibilities and the basis $$\left\{(|+\rangle + |-\rangle), -i(|+\rangle - |-\rangle) \right\}$$ would work just as well. We could also use $$\left\{((|+\rangle + |-\rangle) - i(|+\rangle - |-\rangle)),((|+\rangle + |-\rangle) +i(|+\rangle - |-\rangle))\right\}$$ which are two more linearly polarized states. Mathematically the spin states are like the left and right circularly polarized waves, so their sum and difference are like the x$$x$$ and y$$y$$ polarized waves but one of them shifted by a phase, and the $$45$$ degreedegrees tilted ones really are literal sums and differences of the $$x$$ and $$y$$ (in phase) waves. So I'll talk about polarization of classical electromagnetic waves just because you've already seen it. Imagine a wave travelling in the $$z$$ direction with the electric field always pointing in the same direction, say $$\pm x$$. This is called a linearly polarized wave. Same if the wave traveled in the $$z$$ direction and the electric field was in the plus or minus y direction. If those two waves were in phase and had the same magnitude, then their superposition would be a wave that oscillates at the same frequency/wavelength as the previous waves, and is still linearly polarized but this time not in the $$x$$ or $$y$$ direction but instead in the direction $$45$$ degrees (halfway) between them. Basically if the electric field always points in plus of minus the same direction, then that's linear polarization, and it could in theory be in any direction by adjusting the relative magnitude of an $$x$$ polarized one and a $$y$$ polarized one (that are in phase with each other). Now we're going to steal some math and some terminology. Think of multiplying by $$i$$ as changing the phase of the wave by a quarter period, then we built up a circular polarization by $$X+iY$$ and the other circular polarization by $$X+iiiY=X-iY$$ so given two circular polarization you see that we can add them to get a linearly polarized state $$|+\rangle + |-\rangle$$ gives one of the linearly polarized states and $$-i(|+\rangle - i|-\rangle)$$ gives a linearly polarized state orthogonal to the other one. We can borrow all the math and terminology from the classical waves, and associate the spin states of the photon with the right and left circularly polarized waves. We are stealing the math and stealing the terminology, but the fact is that we have two vectors $$|+\rangle$$ and $$|-\rangle$$ and they span a (complex) two space of possibilities and the basis $$\left\{(|+\rangle + |-\rangle), -i(|+\rangle - |-\rangle) \right\}$$ would work just as well. We could also use $$\left\{((|+\rangle + |-\rangle) - i(|+\rangle - |-\rangle)),((|+\rangle + |-\rangle) +i(|+\rangle - |-\rangle))\right\}$$ which are two more linearly polarized states. Mathematically the spin states are like the left and right circularly polarized waves, so their sum and difference are like the x and y polarized waves but one of them shifted by a phase, and the $$45$$ degree tilted ones really are literal sums and differences of the $$x$$ and $$y$$ (in phase) waves. So I'll talk about polarization of classical electromagnetic waves just because you've already seen it. Imagine a wave travelling in the $$z$$ direction with the electric field always pointing in the same direction, say $$\pm x$$. This is called a linearly polarized wave. Same if the wave traveled in the $$z$$ direction and the electric field was in the plus or minus y direction. If those two waves were in phase and had the same magnitude, then their superposition would be a wave that oscillates at the same frequency/wavelength as the previous waves, and is still linearly polarized but this time not in the $$x$$ or $$y$$ direction but instead in the direction $$45$$ degrees (halfway) between them. Basically if the electric field always points in plus or minus the same direction, then that's linear polarization, and it could in theory be in any direction by adjusting the relative magnitude of an $$x$$ polarized one and a $$y$$ polarized one (that are in phase with each other). Now we're going to steal some math and some terminology. Think of multiplying by $$i$$ as changing the phase of the wave by a quarter period, then we built up a circular polarization by $$X+iY$$ and the other circular polarization by $$X+iii Y=X-iY$$ so given two circular polarization you see that we can add them to get a linearly polarized state $$|+\rangle + |-\rangle$$ gives one of the linearly polarized states and $$-i(|+\rangle - |-\rangle)$$ gives a linearly polarized state orthogonal to the other one. We can borrow all the math and terminology from the classical waves, and associate the spin states of the photon with the right and left circularly polarized waves. We are stealing the math and stealing the terminology, but the fact is that we have two vectors $$|+\rangle$$ and $$|-\rangle$$ and they span a (complex) two space of possibilities and the basis $$\left\{(|+\rangle + |-\rangle), -i(|+\rangle - |-\rangle) \right\}$$ would work just as well. We could also use $$\left\{((|+\rangle + |-\rangle) - i(|+\rangle - |-\rangle)),((|+\rangle + |-\rangle) +i(|+\rangle - |-\rangle))\right\}$$ which are two more linearly polarized states. Mathematically the spin states are like the left and right circularly polarized waves, so their sum and difference are like the $$x$$ and $$y$$ polarized waves but one of them shifted by a phase, and the $$45$$ degrees tilted ones really are literal sums and differences of the $$x$$ and $$y$$ (in phase) waves. Bounty Ended with 100 reputation awarded by Phonon occurred Jan 5 '15 at 12:03 2 Complete notation overhaul using MathJax. edit approved Jan 4 '15 at 12:55 Phonon 3,78022 gold badges1212 silver badges3434 bronze badges So I'll talk about polarization of classical electromagnetic waves just because you've already seen it. Imagine a wave travelling in the z$$z$$ direction with the electric field always pointing in the same direction, say plus or minus x$$\pm x$$. This is called a linearly polarized wave. Same if the wave travelledtraveled in the z$$z$$ direction and the electric field was in the plus or minus y direction. If those two waves were in phase and had the same magnitude, then their superposition would be a wave that oscillates at the same frequency/wavelength as the previous waves, and is still linearly polarized but this time not in the x$$x$$ or y$$y$$ direction but instead in the direction 45$$45$$ degrees (halfway) between them. Basically if the electric field always points in plus of minus the same direction, then that's linear polarization, and it could in theory be in any direction by adjusting the relative magnitude of an x$$x$$ polarized one and a y$$y$$ polarized one (that are in phase with each other). OK, that's it for classical waves. You could discuss how photons make up classical waves, but that's not really what the question is about. The question is about spin for photons. And spin states for the photon come in two kinds, and the names for the positive spin (spin hbar)$$|+\hbar\rangle$$ and the negative spin (spin -hbar)$$|-\hbar\rangle$$ are plus $$|+\rangle$$ and minus $$|-\rangle$$ and you can treat them just like the circularly polarized states. Now we're going to steal some math and some terminology. Think of multiplying by i$$i$$ as changing the phase of the wave by a quarter period, then we built up a circular polarization by X+iY$$X+iY$$ and the other circular polarization by X+iiiY=X-iY$$X+iiiY=X-iY$$ so given two circular polarization you see that we can add them to get a linearly polarized state |spin+>+|spin->$$|+\rangle + |-\rangle$$ gives one of the linearly polarized states and -i(|spin+>-i|spin->)$$-i(|+\rangle - i|-\rangle)$$ gives a linearly polarized state orthogonal to the other one. We can borrow all the math and terminology from the classical waves, and associate the spin states of the photon with the right and left circularly polarized waves. We are stealing the math and stealing the terminology, but the fact is that we have two vectors |+>$$|+\rangle$$ and |->$$|-\rangle$$ and they span a (complex) two space of possibilities and the basis {(|+> + |->), -i(|+> - |->) }$$\left\{(|+\rangle + |-\rangle), -i(|+\rangle - |-\rangle) \right\}$$ would work just as well. We could also use {((|+> + |->) - i(|+> - |->)),((|+> + |->) +i(|+> - |->))}$$\left\{((|+\rangle + |-\rangle) - i(|+\rangle - |-\rangle)),((|+\rangle + |-\rangle) +i(|+\rangle - |-\rangle))\right\}$$ which are two more linearly polarized states. Mathematically the spin states are like the left and right circularly polarized waves, so their sum and difference are like the x and y polarized waves but one of them shifted by a phase, and the 45$$45$$ degree tilted ones really are literal sums and differences of the x$$x$$ and y$$y$$ (in phase) waves. So { |+> , |->}$$\{ |+\rangle , |-\rangle \}$$ is one basis,{(|+> + |->), -i(|+> - |->) }$$\left\{(|+\rangle + |-\rangle), -i(|+\rangle - |-\rangle) \right\}$$ is another basis and{((|+> + |->) + i(|+> - |->)),((|+> + |->) -i(|+> - |->))}$$\left\{((|+\rangle + |-\rangle) - i(|+\rangle - |-\rangle)),((|+\rangle + |-\rangle) +i(|+\rangle - |-\rangle))\right\}$$ is a third basis. So I'll talk about polarization of classical electromagnetic waves just because you've already seen it. Imagine a wave travelling in the z direction with the electric field always pointing in the same direction, say plus or minus x. This is called a linearly polarized wave. Same if the wave travelled in the z direction and the electric field was in the plus or minus y direction. If those two waves were in phase and had the same magnitude, then their superposition would be a wave that oscillates at the same frequency/wavelength as the previous waves, and is still linearly polarized but this time not in the x or y direction but instead in the direction 45 degrees (halfway) between them. Basically if the electric field always points in plus of minus the same direction, then that's linear polarization, and it could in theory be in any direction by adjusting the relative magnitude of an x polarized one and a y polarized one (that are in phase with each other). OK, that's it for classical waves. You could discuss how photons make up classical waves, but that's not really what the question is about. The question is about spin for photons. And spin states for the photon come in two kinds, and the names for the positive spin (spin hbar) and the negative spin (spin -hbar) are plus and minus and you can treat them just like the circularly polarized states. Now we're going to steal some math and some terminology. Think of multiplying by i as changing the phase of the wave by a quarter period, then we built up a circular polarization by X+iY and the other circular polarization by X+iiiY=X-iY so given two circular polarization you see that we can add them to get a linearly polarized state |spin+>+|spin-> gives one of the linearly polarized states and -i(|spin+>-i|spin->) gives a linearly polarized state orthogonal to the other one. We can borrow all the math and terminology from the classical waves, and associate the spin states of the photon with the right and left circularly polarized waves. We are stealing the math and stealing the terminology, but the fact is that we have two vectors |+> and |-> and they span a (complex) two space of possibilities and the basis {(|+> + |->), -i(|+> - |->) } would work just as well. We could also use {((|+> + |->) - i(|+> - |->)),((|+> + |->) +i(|+> - |->))} which are two more linearly polarized states. Mathematically the spin states are like the left and right circularly polarized waves, so their sum and difference are like the x and y polarized waves but one of them shifted by a phase, and the 45 degree tilted ones really are literal sums and differences of the x and y (in phase) waves. So { |+> , |->} is one basis,{(|+> + |->), -i(|+> - |->) } is another basis and{((|+> + |->) + i(|+> - |->)),((|+> + |->) -i(|+> - |->))} is a third basis. So I'll talk about polarization of classical electromagnetic waves just because you've already seen it. Imagine a wave travelling in the $$z$$ direction with the electric field always pointing in the same direction, say $$\pm x$$. This is called a linearly polarized wave. Same if the wave traveled in the $$z$$ direction and the electric field was in the plus or minus y direction. If those two waves were in phase and had the same magnitude, then their superposition would be a wave that oscillates at the same frequency/wavelength as the previous waves, and is still linearly polarized but this time not in the $$x$$ or $$y$$ direction but instead in the direction $$45$$ degrees (halfway) between them. Basically if the electric field always points in plus of minus the same direction, then that's linear polarization, and it could in theory be in any direction by adjusting the relative magnitude of an $$x$$ polarized one and a $$y$$ polarized one (that are in phase with each other). OK, that's it for classical waves. You could discuss how photons make up classical waves, but that's not really what the question is about. The question is about spin for photons. And spin states for the photon come in two kinds, and the names for the positive spin $$|+\hbar\rangle$$ and the negative spin $$|-\hbar\rangle$$ are plus $$|+\rangle$$ and minus $$|-\rangle$$ and you can treat them just like the circularly polarized states. Now we're going to steal some math and some terminology. Think of multiplying by $$i$$ as changing the phase of the wave by a quarter period, then we built up a circular polarization by $$X+iY$$ and the other circular polarization by $$X+iiiY=X-iY$$ so given two circular polarization you see that we can add them to get a linearly polarized state $$|+\rangle + |-\rangle$$ gives one of the linearly polarized states and $$-i(|+\rangle - i|-\rangle)$$ gives a linearly polarized state orthogonal to the other one. We can borrow all the math and terminology from the classical waves, and associate the spin states of the photon with the right and left circularly polarized waves. We are stealing the math and stealing the terminology, but the fact is that we have two vectors $$|+\rangle$$ and $$|-\rangle$$ and they span a (complex) two space of possibilities and the basis $$\left\{(|+\rangle + |-\rangle), -i(|+\rangle - |-\rangle) \right\}$$ would work just as well. We could also use $$\left\{((|+\rangle + |-\rangle) - i(|+\rangle - |-\rangle)),((|+\rangle + |-\rangle) +i(|+\rangle - |-\rangle))\right\}$$ which are two more linearly polarized states. Mathematically the spin states are like the left and right circularly polarized waves, so their sum and difference are like the x and y polarized waves but one of them shifted by a phase, and the $$45$$ degree tilted ones really are literal sums and differences of the $$x$$ and $$y$$ (in phase) waves. So $$\{ |+\rangle , |-\rangle \}$$ is one basis,$$\left\{(|+\rangle + |-\rangle), -i(|+\rangle - |-\rangle) \right\}$$ is another basis and$$\left\{((|+\rangle + |-\rangle) - i(|+\rangle - |-\rangle)),((|+\rangle + |-\rangle) +i(|+\rangle - |-\rangle))\right\}$$ is a third basis. 1 answered Jan 4 '15 at 8:29 Timaeus 22.1k11 gold badge1616 silver badges5454 bronze badges The short answer is that the spin states of a photon come in two kinds, based on helicity, how the circular polarization tracks with the direction of the photons momentum. You can think of them as circularly polarized in the sense that we can define the relative relationship between the different polarizations the same way we do for classical electromagnetic waves (even though a single photon is not a classical electromagnetic wave), but we'll use the same math and the same terminology. So I'll talk about polarization of classical electromagnetic waves just because you've already seen it. Imagine a wave travelling in the z direction with the electric field always pointing in the same direction, say plus or minus x. This is called a linearly polarized wave. Same if the wave travelled in the z direction and the electric field was in the plus or minus y direction. If those two waves were in phase and had the same magnitude, then their superposition would be a wave that oscillates at the same frequency/wavelength as the previous waves, and is still linearly polarized but this time not in the x or y direction but instead in the direction 45 degrees (halfway) between them. Basically if the electric field always points in plus of minus the same direction, then that's linear polarization, and it could in theory be in any direction by adjusting the relative magnitude of an x polarized one and a y polarized one (that are in phase with each other). OK, what if they aren't in phase, what if they they are a quarter of a period out of phase, then when the x direction is big the y direction is zero, so it points entirely in the x direction, then later it is entirely in the y direction, and so its direction moves in a circle (if the magnitudes of the out of phase fields in the x and y direction are the same magnitude the head does move in a circle, otherwise the head moves in an ellipse). If instead you put them three quarters of a a period out of phase, they will go in a circle in the opposite direction. The waves where the head of the electric field move in a circle are called circularly polarized waves. OK, that's it for classical waves. You could discuss how photons make up classical waves, but that's not really what the question is about. The question is about spin for photons. And spin states for the photon come in two kinds, and the names for the positive spin (spin hbar) and the negative spin (spin -hbar) are plus and minus and you can treat them just like the circularly polarized states. Now we're going to steal some math and some terminology. Think of multiplying by i as changing the phase of the wave by a quarter period, then we built up a circular polarization by X+iY and the other circular polarization by X+iiiY=X-iY so given two circular polarization you see that we can add them to get a linearly polarized state |spin+>+|spin-> gives one of the linearly polarized states and -i(|spin+>-i|spin->) gives a linearly polarized state orthogonal to the other one. We can borrow all the math and terminology from the classical waves, and associate the spin states of the photon with the right and left circularly polarized waves. We are stealing the math and stealing the terminology, but the fact is that we have two vectors |+> and |-> and they span a (complex) two space of possibilities and the basis {(|+> + |->), -i(|+> - |->) } would work just as well. We could also use {((|+> + |->) - i(|+> - |->)),((|+> + |->) +i(|+> - |->))} which are two more linearly polarized states. Mathematically the spin states are like the left and right circularly polarized waves, so their sum and difference are like the x and y polarized waves but one of them shifted by a phase, and the 45 degree tilted ones really are literal sums and differences of the x and y (in phase) waves. So { |+> , |->} is one basis, {(|+> + |->), -i(|+> - |->) } is another basis and {((|+> + |->) + i(|+> - |->)),((|+> + |->) -i(|+> - |->))} is a third basis. Each basis can the property that it is equal parts any one from the other two basis sets. And that's what the key distribution is based on. Just having multiple basis for a two dimensional set of states. All I've done above is write everything in terms of the spin states. Mathematically any basis is fine, and all three of these are equally nice in that within a basis the two are orthogonal to each other, and if you pick one from one basis it has equal sized dot products with either of the ones from the other sets. Worrying about how these relate to classical waves is a distraction since it is the borrowing of the math and the terminology that is going on.


text:
# Discrete Math problem graphs

• Nov 6th 2006, 08:36 PM
clockingly
Discrete Math problem graphs
Okay, I'm having trouble figuring out this problem:

3. If d1, d2,...dV are the degrees of G, a graph on V vertices, what are the degrees of G complement?

I'm not exactly sure how to find an equation for that so I would greatly appreciate any help.
• Nov 6th 2006, 08:42 PM
ThePerfectHacker
Quote:

Originally Posted by clockingly
Okay, I'm having trouble figuring out this problem:

3. If d1, d2,...dV are the degrees of G, a graph on V vertices, what are the degrees of G complement?

I'm not exactly sure how to find an equation for that so I would greatly appreciate any help.

A definition a a complete graph $\mathcal{K}_V$ is $|V|$ vertices such that an edge exists between any two.
If $d_j$ is a degree at vertex $v_j$
Then the complement will have the degree of $|V|-d_j$
Thus,
V-d1,V-d2,...,V-dv

format: plain


text: 3. When the stopcock is opened, the gas spontaneously expands to fill both flasks. Second Law of Thermodynamics and entropy. The Standard Enthalpy and Entropy of Various Substances Substance CaC2(.) 1 General Chemistry II Jasperse Entropy, Spontaneity, and Free Energy. Define entropy in your own words, and list the variables or conditions that you must consider when comparing the entropy of two substances, or when trying to determine the relative change in entropy. Our tutors have indicated that to solve this problem you will need to apply the Entropy concept. The entropy change Suniverse must be > 0 for a spontaneous process, which means that (Ssystem + Ssurroundings) must be > 0 for a spontaneous process. Liquids and aqueous ions have more entropy because they move about more freely, and gases have an even larger amount of entropy. A.P. C 6 H 6 (l) + O 2 (g) → 6 CO 2 (g) + 3 H 2 O (l) ΔS > 0; entropy increases because there are more moles on the product side of the equation (9 vs 2) Uploaded by. Chapter 13 Entropy Practice Worksheet Entropy is the degree of randomness in a substance. Worksheets . For a reaction entropy change is given by 2. Some of the worksheets for this concept are Work entropy s dispersed energy levels increasing, , Entropy work, Ch302 work 13c answer keyproblems related to lecture, Spring 2005 entropy tutorial, , Unit 15 reaction energy reaction kinetics. 16 - Spontaneity, Entropy, and Free Energy MULTIPLE CHOICE. You will be tested on your comprehension of what entropy is and how it applies to real world situations. At constant P, I.e, entropy change of fusion: Predict whether entropy is increasing (ΔS > 0) or decreasing (ΔS < 0)? ... should consider in detail the meaning of entropy and the physi cal factors that determine the value of the entropy of vaporization. Solids are very ordered and have low entropy. - Facts & Calculation, What is Potential Energy? PV diagrams - … •Every energy transfer or transformation increases the entropy of the universe. Ludwig Boltzmann showed that the entropy of a system is related the number of microstates (W) available to the system. - Definition, Advantages & Disadvantages, What is Entropy? The change in entropy of the system when the temperature is raised at constant volume is given by equation DS = C vln(T f/T i) 3. Ch 17 Entropy Practice Test. The Standard Enthalpy and Entropy of Various Substances Substance CaC2(.) Choose an answer and hit 'next'. AP Review Answers: Review for Final ... Organic and Nuclear Practice Test Key. WORKSHEET: Entropy Changes ENTROPY - SOLUTIONS Entropy is a thermodynamic function that measures randomness or disorder. Practice: Thermodynamics questions. flashcard set, {{courseNav.course.topics.length}} chapters | Liquids and aqueous ions have more entropy because they move about more freely, and gases have an even larger amount of entropy… Qualitatively, entropy is simply a measure how much the energy of atoms and molecules become more spread out in a process and can be defined in terms of statistical probabilities of a system or in terms of the other thermodynamic quantities. Calculate the molar entropy of CO 2 gas at 350.K and 1.174 atm. 2 | Enthalpy Worksheet Use the following heat of formation table in questions 2 — 6. Visit Us. - Definition & Example, What is Heat Energy? It does not say anything about how fast a reaction will occur. {{courseNav.course.mDynamicIntFields.lessonCount}} lessons Ludwig Boltzmann showed that the entropy of a system is related the number of microstates (W) available to the system. Marginal entropy, joint entropy, conditional entropy, and the Chain Rule for entropy. 20-11 Figure 20.2 Spontaneous expansion of a gas. Zeroth law of thermodynamics. CO-2(0) N20(g) 03(g) AH; (kJ/mol) -126 -987 143 so (J/K • 310 201 214 131 189 193 211 2. Choose the one alternative that best completes the statement or answers the question. It is normally expressed in terms of change in entropy. BOTH ΔHf and ΔGf = 0 for elements in their standard state and both bear units of kJ/molrxn. ... Types of Chemical Reaction Worksheet Practice Answers. 4432 Bayview Road. This worksheet defines entropy, discusses the sign convention and its meaning for entropy, the 2nd and 3rd laws of thermodynamics, 12 practice problems assigning dS for reactions, and 6 practice problems assigning which has the lower entropy. That is, when one 1. Calculate the change in enthalpy of this system. Calculate the entropy change for a process in which 3.00 mol of liquid water at 0°C is combined with 1.00 mol of water at 100.0°C in a perfectly insulated container. HONORS CHEMISTRY 2 . So, calculating the standard molar free energy of formation is simply the same song, 3rd verse. Is the system becoming more random or less random? Worksheet — Entropy Entropy, S, is a measure of how dispersed or spread out a system's energy is among the available energy levels. Organic and Nuclear Worksheet Keys . A system at 776.5 K undergoes a change in enthalpy of –5.711 kJ/mol. Chapter 20: Entropy and the Second Law of Thermodynamics The Conservation of Energy law allows energy to flow bi-directionally between its various forms. (b) Bromine boils at 59.3 °C. Calculate the energy required to change the temperature of 1.000 mol of CO 2 from 298.0K to 350.0K, both at constant volume and at constant pressure. Explain. About This Quiz & Worksheet. Solutions to Thermodynamics Worksheet; Introductory TD stoichiometry problems; Latent heat of Vapourization; Calorimetry Worksheet q = mcΔt Class work When the stopcock is opened, the gas spontaneously expands to fill both flasks. Entropy Practice Worksheet **KEY** CONCLUDE whether entropy is INCREASING or DECREASING in the following examples & give a reason for your answer. - Definition, Law & Formula, {{courseNav.course.mDynamicIntFields.lessonCount}}, Positron: Definition, Symbol & Production, Thermal Equilibrium: Definition, Formula & Example, Thermal Expansion: Definition, Equation & Examples, What Is a Renewable Energy Source? POSITIVE –although it is 9 molecules going to 8, there is a liquid 3. First law of thermodynamics problem solving. Next lesson. This quiz and worksheet combo will help you evaluate your understanding of entropy and the number of ways a system can be arranged. As temperature is increased more of the excited state energy levels will The number of energy levels and how many are populated depend on a number of factors. As temperature is increased more of the excited state energy levels will Give the mathematical statement of the second law. Mutual information between ensembles of random variables. Spontaneity refers to a process that happens without any outside intervention. This comprehensive worksheet has chemistry apprentices perform calculations for several thermodynamic reactions. 2) The specific heat is the energy needed to raise the temperature of one gram of a nettexts. A.P. Entropy and Free Energy . Entropy Decreases; liquid water is less random than gaseous water (steam) 2) Solid CO 2 sublimes ! 2. Why entropy is a fundamental measure of information content. Earn Transferable Credit & Get your Degree, A crowd of people on the streets of New York, Create your account to access this entire worksheet, A Premium account gives you access to all lesson, practice exams, quizzes & worksheets. Abd_Sof_9491. Entropy is sometimes referred to as the “arrow of time”. | 1 S = 219.6 J/K CO-2(0) N20(g) 03(g) AH; (kJ/mol) -126 -987 143 so (J/K • 310 201 214 131 189 193 211 2. 1)The thermodynamic quantity that expresses the degree of disorder in a system is _____. Find out: a) The change in the Gibbs free energy at 25 °C. Worksheet — Entropy Entropy, S, is a measure of how dispersed or spread out a system's energy is among the available energy levels. 2. Solids are very ordered and have low entropy. Entropy Worksheet 1. | {{course.flashcardSetCount}} Explain. Chapter 13 Entropy Practice Worksheet Entropy is the degree of randomness in a substance. The laws of thermodynamics. The laws of thermodynamics. standard entropy standard enthalpy standard free energy number of moles standard reduction potential current (amperes) charge (coulombs) t q m c T S H G n E I q t D D D D ime (seconds) Faraday s constant , 96,485 coulombs per mole of electrons 1joule 1volt 1coulomb F. THERMODYNAMICS . Practice: Energy and thermodynamics. S = 219.6 J/K Gibb's Free Energy $$G$$ Spontaneity Equilibrium view all. Thermodynamics article. The molar entropy of CO 2 gas at 298.0 K and 1.000 atm is 213.64 J/mol K . Or if you need more Entropy practice, you can also practice Entropy practice problems. The number of energy levels and how many are populated depend on a number of factors. Specific heat and latent heat of fusion and vaporization. Choose the one alternative that best completes the statement or answers the question. Chemistry Practice Test: Ch. Uploaded by. S = k lnW . This is the currently selected item. This quiz/worksheet combo will help you test your understanding of calculations involving entropy. gibbs free energy enthalpy and entropy worksheet answers, You already know how to calculate enthalpy and entropy, just substitute free energy values using tables of standard values! All Chemistry Practice Problems Entropy Practice Problems Q. The entropy of a system at 337.1 K increases by 221.7 J/mol•K. The free energy value is found to be –717.5 kJ/mol. ΔS > 0; entropy increases because there are more moles on the product side of the equation (9 vs 2). 1. 1)The thermodynamic quantity that expresses the degree of disorder in a system is _____. 1. Entropy Worksheet 1. Extra Practice Problems General Types/Groups of problems: Evaluating Relative Molar Entropy for Chemicals Calculatingp1 ΔGfor Reactions (Math) p5 Evaluating ΔS for Reactions (non-math) p2 ΔG, ΔH, ΔS, Equilibrium, and Temperature p6 Calculating ΔS for Reactions (Math) p2 Answers p7 Does entropy increase or decrease with increase in temperature? Worksheets on thermodynamics practice problems. State the second law in words. First law of thermodynamics. Entropy Decreases; liquid water is less random than gaseous water (steam) 2) Solid CO 2 sublimes ! Source Coding Theorem; Preﬁx, Variable-, & Fixed-Length Codes. 1999 AP Exam Key . standard entropy standard enthalpy standard free energy number of moles standard reduction potential current (amperes) charge (coulombs) t q m c T S H G n E I q t D D D D ime (seconds) Faraday s constant , 96,485 coulombs per mole of electrons 1joule 1volt 1coulomb F. THERMODYNAMICS . Chemistry Practice Test: Ch. All other trademarks and copyrights are the property of their respective owners. This quiz and worksheet will test you on the following objectives: In this quiz and worksheet students practice the following skills: To learn more about entropy, read through the related lesson called What Is Entropy? Give the mathematical statement of the second law. A)entropy B)internal energy C)heat flow D)enthalpy E)bond energy Calculate the change in enthalpy of this system. Entropy, like enthalpy, is a state function and is therefore independent of the path taken between the final and initial states. State the second law in words. The entropy of a system at 337.1 K increases by 221.7 J/mol•K. Calculate the molar entropy of CO 2 gas at 350.0K and 1.000 atm. •Entropy is a measure of disorder or randomness. Once you find your worksheet, click on pop-out icon or print icon to worksheet to print or download. Practice Test Questions 6 . Chapter 20: Entropy and the Second Law of Thermodynamics The Conservation of Energy law allows energy to flow bi-directionally between its various forms. Biological and Biomedical Ch 17 Entropy, Free Energy ACS Answers . This worksheet is also available in the following formats: ΔS < 0; entropy decreases because liquid water is less random than gaseous water (steam), ΔS > 0; entropy increases because gaseous CO. ΔS > 0; entropy increases because two moles of a gas are more random than 1 mole of a gas. Enrolling in a course lets you earn progress by passing quizzes and exams. The standard enthalpy of this reaction is 42.05 kJ and its standard entropy is –18.28 J/K. Using data from the heat of formation table above, calculate the enthalpy of reaction Worksheets and Assignments on thermodynamics. Is entropy increasing or decreasing? Displaying top 8 worksheets found for - Entropy. All rights reserved. The entropy of the universe tends to increase OR DS univ > 0 or = 0 2. Free energy. Displaying top 8 worksheets found for - Enthalpy And Entrophy. This is the currently selected item. In order for a reaction to be spontaneous, Gibb's free energy must have a negative value. Enthalpy Worksheet With Answers 1 Enthalpy Worksheet Enthalpy Worksheet 890.4 kJ/mol of heat. Services, What Is Geothermal Energy? 1) Steam condenses to water ! Next lesson. Entropy is a state function that is often erroneously referred to as the 'state of disorder' of a system. As a member, you'll also get unlimited access to over 83,000 lessons in math, The symbol for change in entropy is ∆S. Predict the sign of the entropy change, ΔS°, for each of the reaction displayed.i) 2NO2 (g) → N2 … The molar entropy of CO 2 gas at 298.0 K and 1.000 atm is 213.64 J/mol K . 's' : ''}}. (Note, in this type of problem you'll either be given the molar entropy values of the reactants and products or you'll need to look them up in a table.) Why heat increases entropy. Predict the sign of the entropy change, ΔS°, for each of the reaction displayed.i) 2NO2 (g) → N2 … More is the disorder in system, higher is its entropy. Qualitatively, entropy is simply a measure how much the energy of atoms and molecules become more spread out in a process and can be defined in terms of statistical probabilities of a system or in terms of the other thermodynamic quantities. This quiz is based on the concepts in AP Chemistry Enduring understanding 5.E: "Chemical or physical processes are driven by a decrease in enthalpy or an increase in entropy, or both." Entropy and Gibbs Free energy (slides 10 -12) worksheet Wed 5/20 1/2 day Practice Problems finish problems Thurs 5/21 Lab - predicting precipiates from Gibb's Free Energy finish lab Fri 5/22 quest - H, S, G . Thermodynamics Worksheet Fill the blanks in the following sentences with the correct thermodynamics term: 1) The thing we measure when we want to determine the average kinetic energy of random motion in the particles of a substance is temperature. Using a Table of Thermochemical Data, calculate ΔS for the following reaction. Some of the worksheets for this concept are Work entropy s dispersed energy levels increasing, , Entropy work, Ch302 work 13c answer keyproblems related to lecture, Spring 2005 entropy tutorial, , Unit 15 reaction energy reaction kinetics. 3. • The net change in entropy of the system, ∆S , equals the sum of the entropy created during the spontaneous process and the change in energy associated with the heat flow. 5. Calculate the entropy change for a process in which 3.00 mol of liquid water at 0°C is combined with 1.00 mol of water at 100.0°C in a perfectly insulated container. Define entropy in your own words, and list the variables or conditions that you must consider when comparing the entropy of two substances, or when trying to determine the relative change in entropy. Second Law of Thermodynamics and entropy. Entropy is sometimes referred to as the “arrow of time”. benzokinon. First and Second Laws Combine •The quantity of the energy of the universe is constant •But the quality of that energy is not Marginal entropy, joint entropy, conditional entropy, and the Chain Rule for entropy. is Gibb's free energy, is enthalpy, and is entropy. More is the disorder in system, higher is its entropy. q V = 1505.4 J . Look up S° values for all reaction participants, taking care to look in the correct column of the table of thermochemical data. In this practice worksheet students will explore entropy concepts, predict if a process increases or decreases in entropy and finally use Gibb's Free Energy (with enthalpy and entropy) to determine if a process is spontaneous. Entropies Deﬁned, and Why They Are Measures of Information. Free energy. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. AP 2002 Part 1 Key . Entropy. 3. 2) The specific heat is the energy needed to raise the temperature of one gram of a 1) Steam condenses to water ! 20-11 Figure 20.2 Spontaneous expansion of a gas. Introduction Second Law Entropy $$S$$ Microstates and Boltzmann Entropy Change Entropy and Temperature Change Entropy and Phase Change Entropy Change of Surroundings Entropy of Reactions $$\Delta S_{\rm rxn}$$ Examples view all. q P = 1938.0 J Calculate the molar entropy of CO 2 gas at 350.0K and 1.000 atm. Solids are very ordered and have low entropy. Free Energy . Physical and Chemical Changes. Entropies Deﬁned, and Why They Are Measures of Information. Worksheet – Entropy Entropy, S, is a measure of how dispersed or spread out a system’s energy is among the available energy levels. The second law of thermodynamics states that in any spontaneous process the entropy of Calculate the enthalpy of vaporization of bromine at its boiling point. Calculate the molar entropy of CO 2 gas at 350.0K and 1.000 atm. Entropy Practice Worksheet **KEY** CONCLUDE whether entropy is INCREASING or DECREASING in the following examples & give a reason for your answer. This comprehensive worksheet has chemistry apprentices perform calculations for several thermodynamic reactions. The free energy value is found to be –717.5 kJ/mol. They will work with moles, limiting reactants, free energy, entropy, and enthalpy. Using data from the heat of formation table above, calculate the enthalpy of reaction - Definition & Examples, Drawing Conclusions from a Scientific Investigation, Intro to Biodiversity, Adaptation & Classification, Human Body Systems: Functions & Processes, Foundations of Chemical Compounds & Bonds, Foundations of Chemical Reactions, Acids, and Bases, Measurement & the Metric System Fundamentals, Planning a Scientific Investigation Or Experiment, Using Data for Investigation & Experimentation, Scientific Data: Organization, Analysis & Drawing Conclusions, Working Scholars® Bringing Tuition-Free College to the Community, Recognize reversible and irreversible processes, Understand how entropy applies to real world situations, Use the definition of multiplicity to compare items, Relate the laws of thermodynamics to entropy, Relate the second law of thermodynamics to entropy, Recall the definitions of reversible and irreversible processes. HONORS CHEMISTRY 2 . Does entropy increase or decrease with increase in temperature? It is normally expressed in terms of change in entropy. ΔS > 0; entropy increases as particles move about more when heated. The symbol for change in entropy is ∆S. 16 - Spontaneity, Entropy, and Free Energy MULTIPLE CHOICE. 4. Some of the worksheets for this concept are Entropy and gibbs energy wkst, Work entropy s dispersed energy levels increasing, , S for chemicals non math, Name date gibbs energy practice problems, Problems of thermochemistry gibbs energy and spontaneity, Ap chemistry entropy and energy. A)entropy B)internal energy C)heat flow D)enthalpy E)bond energy 1. Explanation: . Entropy is the degree of randomness in a substance. All Chemistry Practice Problems Entropy Practice Problems Q. Source Coding Theorem; Preﬁx, Variable-, & Fixed-Length Codes. 2 | Enthalpy Worksheet Use the following heat of formation table in questions 2 — 6. This worksheet defines entropy, discusses the sign convention and its meaning for entropy, the 2nd and 3rd laws of thermodynamics, 12 practice problems assigning dS for reactions, and 6 practice problems assigning which has the lower entropy. If the free energy value is 6.771 kJ/mol, what is the change in entropy? Why entropy is a fundamental measure of information content. For example in a pendulum, energy continually goes to/from kinetic energy and potential energy. One of the fundamental patterns of the universe is defined by entropy. ENTROPY and GIBBS FREE ENERGY . S = k lnW . © copyright 2003-2020 Study.com. Entropy and the Second Law of Thermodynamics • The second law of thermodynamics states that the total entropy of the universe always increases for a spontaneous process. Calculate the energy required to change the temperature of 1.000 mol of CO 2 from 298.0K to 350.0K, both at constant volume and at constant pressure. ΔS > 0; entropy increases because two moles of a gas are more random than 1 mole of a gas. Unit 4 Worksheets. This quiz and worksheet combo will help you evaluate your understanding of entropy and the number of ways a system can be arranged. Sciences, Culinary Arts and Personal A system at 776.5 K undergoes a change in enthalpy of –5.711 kJ/mol. - Definition, Law & Formula. For example in a pendulum, energy continually goes to/from kinetic energy and potential energy. The Third Law of Thermodynamics: Absolute entropy. AP exam keys: AP Part 2 Key 2014 . In this chemical reaction worksheet, learners define equilibrium, determine what affects the progress of a reaction, and compare and contrast entropy and enthalpy. kheil013. Practice: Energy and thermodynamics. Mutual information between ensembles of random variables. 437 lessons Based on entropy changes only, would you predict the reaction to be spontaneous or not. At constant P, I.e, entropy change of fusion: Liquids and aqueous ions have more entropy because they move about more freely, and gases have an even larger amount of entropy… Key Words: entropy, standard entropy, disorder, thermoch Some of the worksheets for this concept are Work entropy s dispersed energy levels increasing, , Chemistry 12 tutorial 2, S for chemicals non math, Ap chemistry entropy and energy, Entropy enthalpy changes a lab investigation, Thermodynamics practice problems key, Chapter 6 entropy and the laws of thermodynamics. This lesson covers the following objectives: 43 chapters | About This Quiz & Worksheet. 4HCN (l) + 5O 2(g) 2H 2 O (g) + 4CO 2(g) + 2N 2(g) •1) Determine, just by looking at this equation whether ∆S is positive or negative. The Trouton’s rule states that The entropies of vaporization of all liquids, except those with H-bonds, are approximately the same. flashcard set{{course.flashcardSetCoun > 1 ? b) The change in the Gibbs free energy at 169 °C assuming that the enthalpy and the entropy do not change at this temperature. Ch 18 Handout Keys . Entropy is a state function that is often erroneously referred to as the 'state of disorder' of a system. q V = 1505.4 J . The Third Law of Thermodynamics: Absolute entropy. The Gibb's free energy equation is used to determine the spontaneity of a reaction and is written as follows: . Entropy and Gibbs Free energy (slides 10 -12) worksheet Wed 5/20 1/2 day Practice Problems finish problems Thurs 5/21 Lab - predicting precipiates from Gibb's Free Energy finish lab Fri 5/22 quest - H, S, G . 4. water is heated from 25°C to 50°C : ΔS > 0; entropy increases as particles move about more when heated. You will receive your score and answers at the end. Uploaded by. Entropy: Entropy (S) is a state function and is a measure of degree of disorder of randomness of system. Entropy And Giibbs - Displaying top 8 worksheets found for this concept.. The combustion of methane, CH4, releases mole of methane is burned, 890.4 ICJ are given off to the surroundings This means that the products have 890.4 k.] less energy stored in the bonds than the reactants. The symbol for change in entropy is ∆S. Why heat increases entropy. English, science, history, and more. Download complete and submit the Worksheets and Assignments on the due dates. Thermodynamics Worksheet Fill the blanks in the following sentences with the correct thermodynamics term: 1) The thing we measure when we want to determine the average kinetic energy of random motion in the particles of a substance is temperature. Entropy, like enthalpy, is a state function and is therefore independent of the path taken between the final and initial states. 2002 Exam Part 2 Key . Key Words: entropy, standard entropy, disorder, thermoch For a reaction entropy change is given by 2.    This worksheet is also available in the following formats:  Word | RTF | PDF. This is the currently selected item. Multiply these values by any coefficients in the balanced equation, then simplify each side of the equation: A negative sign for ΔS° tells us that entropy is. The number of energy levels and how many are populated depend on a number of factors. Calculate the molar entropy of CO 2 gas at 350.K and 1.174 atm. If the free energy value is 6.771 kJ/mol, what is the change in entropy? q P = 1938.0 J Calculate the molar entropy of CO 2 gas at 350.0K and 1.000 atm. Answers to Practice Questions: Entropy 1. Entropy: Entropy (S) is a state function and is a measure of degree of disorder of randomness of system. You can view video lessons to learn Entropy. They will work with moles, limiting reactants, free energy, entropy, and enthalpy. Defined by entropy of –5.711 kJ/mol the stopcock is opened, the gas spontaneously expands to fill both flasks are... Disorder in system, higher is its entropy it does not say anything about how a! State energy levels and how many are populated depend on a number microstates... Spontaneity refers to a process that happens without any outside intervention flow bi-directionally between its Various forms entropy. And vaporization is normally expressed in terms of change in entropy Practice Worksheet entropy is a state that! Gas are more random or less random entropy increase or decrease with increase temperature! Formation table in Questions 2 — 6, get Practice tests, quizzes, and free energy value found... Energy \ ( G\ ) Spontaneity Equilibrium view all of factors to print or download of formation table in 2! Substance CaC2 (. = 0 for elements in their standard state and both bear units of.. Entropies of vaporization of all liquids, except those with H-bonds, are approximately the.! Energy continually goes to/from kinetic energy and potential energy in entropy s ) is a state that. At 350.K and 1.174 atm more moles on the product side of the state! 2 sublimes of Thermodynamics the Conservation of energy law allows energy to flow bi-directionally between Various! Also available in the Gibbs free energy, entropy change is given by 2 bear of. Particles move about more when heated exam keys: ap Part 2 Key 2014 source Theorem! Calculate ΔS for the following reaction becoming more random or less random to! Should consider in detail the meaning of entropy and Giibbs - displaying top 8 worksheets found for enthalpy... Worksheet is also available in the following reaction understanding of calculations involving entropy your! Marginal entropy, conditional entropy, conditional entropy, like enthalpy, is a fundamental of! Δhf and ΔGf = 0 for elements in their standard state and both bear of! Entropy: entropy this Worksheet is also available in the correct column of the entropy of Various Substances substance (! ; Introductory TD stoichiometry problems ; Latent heat of Vapourization ; Calorimetry Worksheet =... The worksheets and Assignments on the due dates & Worksheet often erroneously referred to as the 'state of disorder of... View all the excited state energy levels and how many are populated depend a... Both flasks patterns of the path taken between the final and initial states, quizzes and... Between the final and initial states Definition & example, What is heat energy process happens... Vs 2 ) Solid CO 2 gas at 350.0K and 1.000 atm is 213.64 J/mol K care to look the... Once you find your Worksheet, click on pop-out icon or print icon to Worksheet to print or.... Involving entropy top 8 worksheets found for - enthalpy and Entrophy you succeed and -. Reaction entropy practice worksheet be spontaneous or not Spontaneity, entropy, Spontaneity, change. Ap exam keys: ap Part 2 Key 2014 you succeed and aqueous have. Δhf and ΔGf = 0 for elements in their standard state and both bear units kJ/molrxn! Degree of randomness in a substance example, What is heat energy and Biomedical,... Energy and potential energy & Worksheet and entropy of a system entropy and the number energy... Care to look in the correct column of the fundamental patterns of the universe is defined by.... Test your understanding of entropy and Giibbs - displaying top 8 worksheets found for - enthalpy and Entrophy entropy... Are populated depend on a number of microstates ( W ) available to the system of Information and energy., joint entropy, like enthalpy, is a fundamental measure of Information.. To be –717.5 kJ/mol does not say anything about how fast a reaction to be spontaneous, 's... Enthalpy and Entrophy Spontaneity Equilibrium view all chapter 20: entropy ( s ) a! Various forms and potential energy is –18.28 J/K work with moles, limiting reactants, free energy value entropy practice worksheet to! Is its entropy ' of a gas are more random or less random ) the thermodynamic quantity expresses... Involving entropy predict the reaction to be –717.5 kJ/mol of vaporization of bromine at boiling. Worksheet enthalpy Worksheet enthalpy Worksheet enthalpy Worksheet enthalpy Worksheet with answers 1 enthalpy Worksheet with answers 1 enthalpy Worksheet Worksheet. For the following formats: Word | RTF | PDF passing quizzes and exams reaction will occur pendulum... 1 General Chemistry II Jasperse entropy, conditional entropy, like enthalpy, and the number of (. Your understanding of entropy and the physi cal factors that determine the value of the excited state levels... And potential energy a pendulum, energy continually goes to/from kinetic energy and potential energy is the.! Factors that determine the Spontaneity of a system is related the number of energy allows... ( W ) available to the system they will work with moles, limiting reactants, free value... Therefore independent of the path taken between the final and initial states worksheets and on... In entropy or download Chemistry apprentices perform calculations for several thermodynamic reactions heat fusion. One alternative that best completes the statement or answers the question aqueous ions have entropy. 776.5 K undergoes a change in entropy icon to Worksheet to print or download Worksheet Chemistry. Are approximately the same independent of the fundamental patterns of the universe defined! And vaporization with increase in temperature have more entropy Practice Worksheet entropy is and how many are populated depend a! System at 776.5 K undergoes a change in enthalpy of –5.711 kJ/mol by passing quizzes and exams a state that... Will A.P entropy increase or decrease with increase in temperature Practice entropy Practice Test with moles, reactants. All other trademarks and copyrights are the property of their respective owners receive! Personalized coaching to help you succeed other trademarks and copyrights are the property of respective... Is 42.05 kJ and its standard entropy is increasing ( ΔS > 0 ; entropy increases as particles about... Disadvantages, What is the degree of randomness of system ) the change in the correct column of fundamental! This quiz/worksheet combo will help you evaluate your understanding of entropy the due dates entropy s! Is often erroneously referred to as the 'state of disorder of randomness of system bear units kJ/molrxn... Have a negative value the entropies of vaporization to look in the following reaction 0 ) decreasing. Is simply the same song, 3rd verse of –5.711 kJ/mol entropy: entropy and physi. Your Worksheet, click on pop-out icon or print icon to Worksheet to print or download Class entropy. Work with moles, limiting reactants, free energy, is a state function and is written as follows.. Of vaporization of all liquids, except those with H-bonds, are approximately the same predict whether entropy and. To real world situations increasing ( ΔS > 0 ; entropy increases because are. Table in Questions 2 — 6 to look in the following reaction exam:. Conservation of energy law allows energy to flow bi-directionally between its Various.... And free energy must have a negative value Thermodynamics the Conservation of energy law allows to... Δs for the following formats: Word | RTF | PDF to print or download once you find your,! Factors that determine the Spontaneity of a gas are more random than gaseous (. Quizzes, and Why they are Measures of Information content Gibbs free,... And Biomedical Sciences, Culinary Arts and Personal Services, What is energy. Are more moles on the due dates law of Thermodynamics the Conservation of levels... 213.64 J/mol K less random boiling point, 3rd verse a state and... About how fast a reaction entropy change is given by 2 17 entropy Practice, can... Final... Organic and Nuclear Practice Test ( s ) is a state function and therefore. The one alternative that best completes the statement or answers the question is by! Is heat energy 25 °C data, calculate ΔS for the following reaction that determine the of! Must have a negative value allows energy to flow bi-directionally between its Various forms side of the entropy of 2! Personalized coaching to help you succeed passing quizzes and exams 350.K and 1.174 atm with,! Entropy, conditional entropy, and enthalpy Arts and Personal Services, What heat... K and 1.000 atm is 213.64 J/mol K at 776.5 K undergoes a change in following! Real world situations on pop-out entropy practice worksheet or print icon to Worksheet to print or.... Why entropy is the change in entropy choose the one alternative that best the... Be arranged final and initial states a fundamental measure of degree of disorder of randomness in a lets! The question entropies of vaporization Why they are Measures of Information physi cal factors determine... Culinary Arts and Personal Services, What is heat energy all Chemistry Practice problems entropy Practice problems Practice! The value of the universe Worksheet has Chemistry apprentices perform calculations for several thermodynamic reactions & Disadvantages, What entropy! Order for a reaction entropy change is given by 2, What is potential energy 8 worksheets found this... Conservation of energy levels and how it applies to real world situations or not on pop-out icon print. Move about more freely, and the number of microstates ( W ) available to system... 350.K and 1.174 atm work entropy is heat energy kJ/mol, What is heat energy the property of respective. Coaching to help you Test your understanding of entropy and the Chain Rule for entropy Various Substances substance (! Than gaseous water ( steam ) 2 ) Solid CO 2 gas at 350.0K 1.000... Will be tested on your comprehension of What entropy is the change in entropy Worksheet 890.4 kJ/mol of....


text:
# Practice Question on Computing the Fourier Transform of a Discrete-time Signal

Compute the Fourier transform of the signal

$x[n] = \cos \left( \frac{\pi}{6}n \right).\$

You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!

$\cos \left( \frac{\pi}{6}n \right)=\frac{1}{2}e^{j\frac{\pi}{6}n}+\frac{1}{2}e^{-j\frac{\pi}{6}n}$

$\mathcal X (\omega)=\sum_{m=-\infty}^\infty 2\pi \delta (\omega-k\omega_0+2\pi m)$

$\mathcal X (\omega)=\sum_{m=-\infty}^\infty 2\pi \delta (\omega-\frac{\pi}{6}+2\pi m)+\sum_{m=-\infty}^\infty 2\pi \delta (\omega+\frac{\pi}{6}+2\pi m)$

--Cmcmican 19:51, 28 February 2011 (UTC)


text:
explicit finite difference scheme for nonlinear diffusion

I've been trying to find an explicit method for the 1D heat-diffusion equation with variable thermal conductivity coefficient $\kappa$:

$$\frac{\partial T}{\partial t}=\frac{\partial}{\partial x} (\kappa \frac{\partial T}{\partial x})$$ However, it seems to me like there is no simple approach as in the case where $\kappa$=const. The explicit methods I found were quite complicated and to be honest, I haven't fully understood them while implicit methods look very similar to the linear ($\kappa$=const.) case.

Why is it so hard to use an explicit method in the nonlinear case? What is the problem?

regards.

• You said that $\kappa$ is variable. But variable as a function of what ? Function of $x$? of $t$? of both ? – JJacquelin Oct 2 '16 at 5:46
• As a function of x – OD IUM Oct 2 '16 at 11:26

When $k=k(x)$ is not constant, the PDE is still linear, just no longer constant coefficient. There are generally 2 approaches to explicit finite difference schemes in this case.

1) Expand the right hand side and write the PDE as

$\frac{\partial T}{\partial t} = k\frac{\partial^2 T}{\partial x^2} + k'\frac{\partial T}{\partial x}.$

Use the standard discretization for the first term, and use centered differences for $\frac{\partial T}{\partial x}$. It sounds like you already know how to solve the constant coefficient heat equation, so you should be good from here. If you need more details I can expand on this.

2) Discretize the equation in divergence form. I'll give more details here. Choose a time step $\Delta t$ and a spatial step $\Delta x$, and write $T^n_j = T(j\Delta x,n\Delta t)$ and $k_j = k(j\Delta x)$. Basically we will use a centered difference for the outer $\frac{\partial }{\partial x}$ using the mid-grid points $j+1/2$ and $j-1/2$. The scheme is

$T^{n+1}_j = T^n_j + \frac{\Delta t}{\Delta x^2} \left( k_{j+1/2}(T^n_{j+1} - T^n_j) - k_{j-1/2}(T^n_j - T^n_{j-1})\right).$

Since $k$ is a given function, there is no difficulty in evaluating $k$ off the grid.

EDIT: Let me add a few words about stability. Normally one does a Von Neumann stability analysis. For parabolic equations, there is a short cut based on monotonicity. For example, for method (2) if we write $s=\Delta t/\Delta x^2$ then

$T^{n+1}_j = (1-s(k_{j+1/2} + k_{j-1/2}))T^n_j + s(k_{j+1/2}T^n_{j+1} + k_{j-1/2}T^n_{j-1}).$

The scheme is monotone if all the coefficients on the right hand side (the numbers in front of $T^n_j$, $T^n_{j-1}$ and $T^n_{j+1}$) are all positive. Then the scheme satisfies a maximum principle and you can prove stability. Since $k$ is the thermal conductivity, $k>0$, so the only term in question is the first one. Hence the scheme is stable when

$1-s(k_{j+1/2} + k_{j-1/2}) \geq 0.$

Rearranging, we require

$\frac{\Delta t}{\Delta x^2} = s \leq \frac{1}{k_{j+1/2} + k_{j-1/2}}.$

If the PDE is uniformly elliptic, then there exists $\theta>0$ so that $k(x)\geq \theta$ for all $x$. Hence, the PDE is stable when $s \leq 1/(2\theta)$, or

$\Delta t \leq \frac{1}{2\theta} \Delta x^2.$

You can perform the same analysis for scheme (1), and you will get the same time step constraint as above, with the additional constraint that

$\Delta x \leq \min_x \frac{k(x)}{|k'(x)|}.$

If we let $M = \max_x |k'(x)|$, then we can also phrase this stability constraint as

$$\Delta x \leq \frac{\theta}{M}.$$

If $k$ is constant, we have $M=0$ and there is no constraint on $\Delta x$. In this respect, perhaps scheme (2) is slightly better, though the constraint on $\Delta x$ above is not burdensome, unless the PDE is degenerate $\theta =0$ or singular $k'=\infty$.

• Thx for your reply. Is there a name for the 1st approach so that I can Google for it? Never seen that before. And which of these methods is more stable? – OD IUM Oct 4 '16 at 8:14
• I don't think there is a specific name for the first scheme. Any book on numerical schemes for PDE will describe both. I've added a short summary of stability for each scheme. There is not much difference in terms of stability, except that (2) requires an additional constraint on $\Delta x$, but this is not a very strict constraint. – Jeff Oct 4 '16 at 16:21


text:
# Cannot find reference assemblies for .NET 3.5 or lower using core msbuild #1333

Open
opened this Issue Nov 10, 2016 · 34 comments

Projects
None yet
Member

### nguerrera commented Nov 10, 2016

dotnet/sdk supports targeting .NET 3.5 just fine if you use desktop msbuild, but it fails to find the reference assemblies using core msbuild. Both desktop and core work fine for NET4.0+.

## Steps to reproduce

Create a new project with dotnet new
Change the TargetFramework to net35 or lower
Add <RuntimeIdentifier>win10-x64</RuntimeIdentifier> to the .csproj
dotnet restore
dotnet build

I've been seeing this issue since the move to msbuild. Targeting .NET 3.5 and lower works in the Preview 2 release SDK. All the relevant targeting packs are installed.

## Expected behavior

A successful build

## Actual behavior

Fails with the error:
... error MSB3644: The reference assemblies for framework ".NETFramework,Version=v3.5" were not found. To resolve this, install the SDK or Targeting Pack for this framework version or retarget your application to a version of the framework for which you have the SDK or Targeting Pack installed. ...

## Environment data

dotnet --info output:

.NET Command Line Tools (1.0.0-preview3-004056)

Product Information:
Version: 1.0.0-preview3-004056
Commit SHA-1 hash: ccc4968bc3

Runtime Environment:
OS Name: Windows
OS Version: 10.0.14393
OS Platform: Windows
RID: win10-x64

Moved from dotnet/cli#4626 for @Thealexbarney.

Moved from dotnet/sdk#369 for @piotrpMSFT

@rainersigwald

Closed

Merged

Member

### dsplaisted commented Nov 11, 2016 • edited

 I think the problem is that there aren't any reference assemblies for .NET Framework 3.5. The only thing in my .NETFramework\3.5 reference assemblies folder is Profile\Client. When I target .NET 3.5 from a Class Library, the references are coming from C:\Windows\Microsoft.NET\Framework\v2.0.50727\.
Contributor

### rainersigwald commented Nov 16, 2016

This comes down to code in FrameworkLocationHelper.GetPathToDotNetFramework that is disabled on .NET Core:

Lines 1308 to 1310 in 63cf735

 #if !FEATURE_INSTALLED_MSBUILD return null; #else

There are some (probably avoidable) registry calls there, but the biggest obstacle seems to be that finding the v3.5 (or whatever) runtime depends on locating the current runtime (Path.GetDirectoryName(typeof(object).Module.FullyQualifiedName)) and looking around it to find the desired version of the framework. That won't work on Core where object comes from somewhere else entirely.

There may be another way to locate a framework (looks like HKEY_LOCAL_MACHINE\SOFTWARE\Microsoft\NET Framework Setup\NDP\v3.5 has an InstallPath value that looks promising) but we'd have to figure out the right mechanism (and test the heck out of it).

Closed

### KevinH-MS commented Feb 15, 2017

 What's the work-around for people who need to build such projects on VS 2017? I just ran into this today, and the only thing I could figure out was to compile on a separate machine with only VS 2015 installed.
Contributor

### rainersigwald commented Feb 15, 2017

 @KevinH-MS It sounds like you're seeing something different from this issue--as originally reported, projects build fine using MSBuild.exe. Can you describe your problem in more detail?

### KevinH-MS commented Feb 15, 2017

 I see. I think that's the answer to my question... 😄 (don't use dotnet build -f net35, use msbuild /p:TargetFramework=net35)

### ChrisMcKee commented Mar 1, 2017 • edited

 Issues a pain when you're trying to generate a multi-target nuget package; prior to the great cheese moving of 2016 I had no issues building a package that targeted 2/3.5/4/4.5/etc Is this likely to be sorted in the next few days when VS17s released? Seems ambitious... Admittedly I'm presuming this is msbuild related as it came with RC4 / VS2017 changes and the issue in the error being thrown is the same as the op (albeit for every version specified).  ... net20;net35;net40;net45;net451;net452;net462;netstandard1.3;netstandard1.6;netstandard1.6.1;netcoreapp1.1 ...  Hell, it might be that after all these changes and dotnet versions I need to set fire to my PC and start afresh to get things to build 🔥

Closed

### Thealexbarney commented Mar 1, 2017

 Right now the only options I see are: Stay on the old project.json system. (Can't use VS2017 as an IDE) Migrate to msbuild and be dependent on the full msbuild. (Can't build with dotnet SDK) Drop <= net35 support. I guess legacy support is a lower priority than other things for now. Ideally (IMO) there would be targeting packs on NuGet for each framework version, making it easier to target .NET Framework from the dotnet SDK.

### ghost commented Mar 2, 2017

 Our use case at work: targeting net20 is important for us as we have to make some of our libraries work on old Windows Server 2003 systems. We can't upgrade them right now. And we don't want to install anything (say, newer .NET framework versions) to avoid risks of breaking the systems. We love dotnet tooling, but while this issue is not fixed, we'll have to stay on the full msbuild for this use case.

Closed

### ChrisMcKee commented Mar 15, 2017 • edited

 @schotime I've been using RTW since it came out last week and this seems to be resolved.

### Thealexbarney commented Mar 15, 2017

 @ChrisMcKee In core MSBuild or full MSBuild?

### schotime commented Mar 15, 2017

 It works in vs for me, but not calling dotnet build manually.

### ChrisMcKee commented Mar 15, 2017 • edited

 @Thealexbarney Windows, so full. I'm not sure you could target .net 2 on non-win platforms as it would lack the required build targets. 2/3.5/4/4.5/4.6 predating it all. You're back in mono-land there.
Member

### drewnoakes commented Mar 16, 2017

 Also seeing this. Can build in VS2017, but dotnet build gives: MSB3644: The reference assemblies for framework ".NETFramework,Version=v3.5" were not found.

### ChrisMcKee commented Mar 16, 2017

 @drewnoakes use msbuild

### stevenvolckaert added a commit to stevenvolckaert/enterprise-library that referenced this issue Mar 18, 2017

 Temporarily dropping support for .NETFramework 3.5. The current versi…
…on of the dotnet CLI tooling contains a bug causing the build of this target to fail, breaking CI builds. See Microsoft/msbuild#1333 for more information.
 c050d5b

Merged

Open

### JoshClose commented Mar 24, 2017

 I'm getting the error with .NET 2.0. I'm trying to do dotnet pack. Is there a workaround for packaging NuGet packages that will work?
Contributor

### dasMulli commented Mar 24, 2017 • edited

 @JoshClose you can directly call msbuild /t:Pack /p:Configuration=Release  For version suffixes, add /p:VersionSuffix=beta-123 (however, note NuGet/Home#4337)

### JoshClose commented Mar 24, 2017

 How can I change the package output location? Sorry, I'm no too familiar with msbuild.
Contributor

### dasMulli commented Mar 24, 2017

 /p:PackageOutputPath=path\to\dir\

Closed

### skarllot commented Sep 22, 2018

 @JohHBrock have you tried to copy files and directories from 'C:\Program Files (x86)\Reference Assemblies\Microsoft\Framework\v3.5*' to 'C:\Program Files (x86)\Reference Assemblies\Microsoft\Framework.NETFramework\v3.5'?

Closed

Open

Merged


text:
## Sunday, February 20, 2011

### Generalized Distributive Law

With regular distributive law you can do things like

$$\sum_{x_1,x_2,x_3} \exp(x_1 + x_2 + x_3)=\sum_{x_1} \exp x_1 \sum_{x_2} \exp x_2 \sum_{x_3} \exp x_3$$

This breaks the original large sum into 3 small sums which can be computed independently.

A more realistic scenario requires factorization into overlapping parts. For instance take the following

$$\sum_{x1,x2,x3,x_4,x_5} \exp(x_1 x_2 + x_2 x_3 + x_3 x_4+x_4 x_5)$$

You can't use directly apply distributive law here. Naive evaluation of the sum is exponential in the number of variables, however this problem can be solved in time linear in the number of variables.

Consider the following factorization
$$\sum_{x1,x2} \exp x_1 x_2 \left( \sum_{x_3} \exp x_2 x_3 \left( \sum_{x_4} \exp x_3 x_4 \left( \sum_{x_5} \exp x_4 x_5 \right) \right) \right)$$

Now note that when computing inner sums, naive approach results in same sum being recomputed many times. Save those shared pieces and you get an algorithm that's quadratic in number of variable values and linear in number of variables

Viterbi and Forward-Backward are specific names of this kind of dynamic programming for chain structured sums as above.

To consider tree-structured sum, take a problem of counting independent sets in a tree-structured graph. In this case we can use the following factorization

This is the first step of the recursion and you can repeat it until your pieces have tractable size.

Now for a general case of non-tree-structured graph, consider the sum below

$$\sum_{x1,x2,x3,x_4,x_5} \exp( x_1 x_2 + x_2 x_3 + x_3 x_4+x_4 x_5 + x_5 x_1 )$$

Here, variables form a loop, so regular tree recursion like Viterbi does not apply. However, this sum can be found efficiently, linear in the number of variables, and cubically in the number of variable values. I have more in depth explanation of decomposing problems like above in an answer on cstheory site but the basic idea is similar to the approach with independent sets -- fix some variables and recurse on subparts.

This kind of general factorization has been re-invented many times and is known as Generalized Distributive Law, tree-decomposition, junction tree, clique tree, partial k-tree, and probably a few other names.

Notebook


text:
# Isaac rubs his back on non-existent doors

Haven't blogged in a while. Life's fine, project are a-plenty, but I just wanted to make a more lasting post about one particular issue that struck me as funny when programming Paper Isaac.

### Bugs, bugs, bugs

What's infuriating when letting others play an early prototype is that you hear constantly the same things. Some bugs are non-trivial to fix, some you're just not motivated to fix now... sometimes you just have your head elsewhere, gotta focus, or are elbow-deep in some other piece of code and the damn walls can wait.

But one particular issue became painful in testing recently against some tough enemies. When sliding along walls, Isaac got stuck on non-existent doors.

What?

### Everything is better with a schematic

Here, Isaac (in cyan) tries to escape from some danger on the right by carefully walking along the left wall. Unfortunately he got stuck around the middle of the room. The only way to make him move again would be to stop pressure against the wall, take a small sidestep, and keep going straight down.

As you can see, the walls are actually made of three parts, even when there's no door. There are two line segments (red), and a rectangle (orange). The rectangle corresponds to a door. If there's no door at this place in the room, or it's closed, or locked, or in anyway non-walkable, the shape is solid, and Isaac can't walk through it.

However, if it's open, Isaac can walk through it, the collision between types HERO and DOOR will be detected, we'll get the userData of the shape of type DOOR, figure out which direction we're trying to go, and then finally tell the game "Hey buddy, we're changing rooms, so get busy loading the state of the next room".

### Solutions, solutions, solutions

Like on many occasions in programming, you have two opportunities: do it right, or make it work. I chose to make it work. The "right" way, I think, would have been to have only one line segment when there's no door at all, and then our usual two segments and one rectangle when there's a door.

But wait! There are secret rooms and super-secret rooms, that you can open by bombing the wall where there's supposed to be a passage. And then we have to change the geometry of the wall, destroying the unique segment, recreating the two segments + rectangle, etc., etc. - we can get it right for example with a state machine and carefully controlled state changes, but is it worth the headache? Nah.

Instead, I did something much simpler... that I beat myself for not doing earlier. Notice how Isaac uses a square shape for collision detection up there? Well, why not use a circle instead? And indeed, using a circle does the job. Sure, Isaac bumps a little when he runs into the edges of doors (even nonexistent ones) - but it's not disruptive to the gameplay, and I'm pretty sure the geometry is always right.

### But wait, there's more!

While we have successfully fixed our above problem, our doors still kinda suck. Right now, even touching ever-so-slightly the door takes us to the next room. Sometimes it's really not what you want to do. When you end a combat and just happen to be next to a door, it doesn't mean you want to go through there.

What we need is to be much smarter at detecting when the player wants to change room. I haven't implemented that yet, but I have an idea... every frame, we can test if a player is in contact with a door, and actively moving in that direction (for example, pressing the S key while in contact with the bottom door). If that happens for more than, say, 15 consecutive frames (which would be 250ms), then we're pretty confident the player wants to change room.

The same logic can be applied to unlocking doors, but we can use a higher threshold here (e.g. 600ms) because the player does not want to waste a key - passing along a door doesn't mean you want to spend a key opening it.

### Conclusion

Anyway I'm going back to hacking more enemies in there, doing more drawing, and perhaps more music before the end of the month, that was fun, see you at the finish line.