Problem Set 2. Chapter 1 Numerical:

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Chapter 1 Nuerical: roble Set 16. The atoic radius of xenon is 18 p. Is that consistent with its b paraeter of 5.15 1 - L/ol?

8 1-3 3. The 3 conversion fro 3 to L is 1d/ 1d/ 1d/=1 L/ 3, which akes the volue of a single xenon 5.8 1-7 L. The volue of a ole of xenon is thus 5.8 1-7 L 6. 1 3 = 3.18 1-3 L/ol.

1 Chapter 1 Nuerical: roble Set 16. The atoic radius of xenon is 18 p. Is that consistent with its b paraeter of L/ol? Hint: what is the volue of a ole of xenon atos and how does that copare to the b paraeter? Double hint: a picoeter is Answer: First, using V = π r, the volue of a single xenon ato is The 3 conversion fro 3 to L is 1d/ 1d/ 1d/=1 L/ 3, which akes the volue of a single xenon L. The volue of a ole of xenon is thus L = L/ol. So they are about a factor of ~1 different. 17. It ay not be fair to copare the volue of an ato to the b paraeter as there ust be soe in-between space when packing a ole of atos as close as possible. This ay ake the volue of the b paraeter appear a bit over ~1 greater than the volue of the ato. For instance, in the hexagonal close pack structure, the volue taken up by a sphere of radius r is: Vhcp=8 π r 3. Can you now show that the volue of hexagonally close packed xenon atos is only about ~4% less than the b paraeter (eaning that they are actually pretty siilar)? Answer: First, using Vhcp=8 p r 3 with r=18 p gives a volue of or L. Multiply by Avogadro s nuber and you see that the volue per ol is L/ol. 18. a. The van der Waals equation of state predicts the residual volue of a real gas is: li ( V ) = b a. If this gas was actually a perfect gas, can you tell e what is the value of a that akes this true? Hint: it should be a function of b, R, and T, and you do know what the residual volue of a perfect gas is, don t you? (4 pts) b. Take the a and b constants of nitrogen, and calculate how a copares to b at 16 C. Now do the sae for heliu. Explain why this explains why nitrogen is very perfect gas-like and heliu is not. Be very careful about units! exaple: Joule Liter/ol is the sae as ka Liter /ol! (4 pts)

2 Answer: a. A perfect gas ust have no residual volue, so given: b a =, then it ust be true that a = b. b. The a constant of N is 1.37 bar L /ol and b is.387 L/ol. Now I have to be careful and realize that the real teperature in Kelvin is is ( ) K = K and that a needs to be in ka to use J/K/ol as R. This akes a = 137. ka L /ol. Now copared to b =.387 L/ol J/K/ol K = J L/ol, which is very close to a and thus nitrogen is like a perfect gas. The sae for heliu is a = 3.46 ka L /ol and b = 85.7 J L/ol, which eans that heliu has a large residual volue and is not very perfect gaslike. Chapter 1 Theoretical: 19. I cae up with y own equation of state. It works like: = b, where "b" is an epirical constant. Can you calculate the for of the copression factor Z =, where is the perfect gas volue per ol? It should be a function of "b" and. Answer: First divide by and recognize that = : Z = = b = 1 b = 1 b.. It turns out van der Waals was very careful when designing the van der Waals equation: = a ( b) V. If you re not sure, let s see why he divided the "a" ter by and not. Specifically, can you calculate the residual volue of an alternative equation of state = a, and show that the residual volue is - L/ol (which doesn t ake sense)? ( b) Answer: Note the definition of residual volue: li. Thus, if we start with the above equation and ultiply by (V b) and divide be : b = a ( b). Now a bit ore algebra yields: = b a ( b) = b a + ab. Now when we take the liit of low pressure, we can use we li = 1 and show that the residual volue is li b a + ab.

3 Thus, the iddle ter will fly off to - L/ol. 4. Now let s see if we can derive an alternative expression for the copression factor for the van der Waals equation of state: = Z = = 1 a ( b) + b? a ( b) V. Can you show that: Hint: start by ultiplying the left and right-hand sides by (V b ). Answer: First ultiply everything by b as so: ( b) = a ( b) V. Now do soe rearranging: b = a ( b) V and now: = a ( b) V + b. Next, divide by : ZZ = = 1 a ( b) + b. 3. a. If I was using Dieterici equation of state: = e a ( b) always positive, I would be able to deterine that the copression factor is: Z = = e a + b. Can you show that this is indeed correct?, where the constants a and b are b. If a gas that follows the Dieterici equation of state had a= and b=, it would be a perfect gas. Is that case what is the copression factor Z equal to? Answer: a. First ultiply by ( b) and divide by : b = e a Z = as so: and bring the b over: = e a + b. Next, ultiply by to get b. If a= then e a = V = e a + b = e a V + b = 1 and the b ter goes away, eaning that Z=1.

4 Chapter Nuerical: 1. The piston in a car engine has to push out the exhaust gas by copressing it. To do so, it oves its 1 c area by 1. c against an external pressure of 1 bar (that s the tailpipe pressure). How uch work does this take? Answer. First, calculate the change in volue: 1 c 1 c (1d/1c) 3 =.1L. Technically, as the gas is being pushed, the change in volue ust be negative (its being copressed out of the piston). Next, recognize that this is a constant pressure proble, aking the work -ext V = ka (-.1 L) =.11 J. In the version I initially assigned I realize I didn t word the proble well enough to tell that this is a copression, so we will accept a negative work if you thought we were describing an expansion out of the piston (but reeber that s not really the case). Chapter Theoretical: 1. Shown here is a V graph of a gas expanding (negative work), I can see that the area under the curve, the work energy, is greatest for the reversible process. This is consistent with the idea that a reversible process provides the ost negative work. a. However, there sees to be a proble when I copress a gas as shown here, where the irreversible process has the greatest area under the curve. Can you tell e how these graphs are still consistent with the fact that a reversible process provides the ost negative work? b. Using the graph of expansion above, explain why expanding against no pressure (i.e. a vacuu) eans no work is done. ( pts) c. How do I calculate the work energy created by a piston if I a pressing as hard as possible against it, but its jaed and unovable? ( pts) Answer. a. There are two ways to answer this question. One is to note that in the copression, the direction that V changes is reversed, which essentially adds another negative sigh to the integral equation for work: V f V i ext V. This akes the total work positive for copression, and the

5 irreversible is the ost positive. This is the sae thing as saying that the reversible is the ost negative. Here is another way to answer it: since V is negative, the work is autoatically positive (all copressions are). Thus, the criterion that a reversible process provides the ost negative work is the sae thing as saying that reversible process provides the least positive work. Consequently, a reversible copression provides the least positive work is fully consistent with what we know about reversible processes. b. If ex =, the graph under the curve has no area, and thus no energy. c. In this case, V isn t changing, so there is no work done. The answer is always the sae, J.. a. Assue isotheral conditions: if the reversible work is the ost negative work out of a syste, reversible heat is the ost positive heat. Why? ( pts) b. Assue isotheral conditions: if I reversibly do work on a syste by expanding it then the heat q is +. Does the cylinder feel hot or cold when you touch it and why? ( pts) Answer. a. The hint is at the beginning. Since this is an isotheral condition, then U = J = q + w. As a result, q = - w and basically heat will do the exact opposite of whatever work does. b. Expansion would naturally ake a gas cold, so to war it to keep teperature constant heat has to flow into the cylinder. If you touch it, then the heat is coing out of your hand which would feel cold. 19, reston T. Snee This work is licensed under a Creative Coons Attribution-NonCoercial 4. International License.

PHY 171. Lecture 14. (February 16, 2012)

PHY 171. Lecture 14. (February 16, 2012) PHY 171 Lecture 14 (February 16, 212) In the last lecture, we looked at a quantitative connection between acroscopic and icroscopic quantities by deriving an expression for pressure based on the assuptions