a. 4.2x10-4 m 3 b. 5.5x10-4 m 3 c. 1.2x10-4 m 3 d. 1.4x10-5 m 3 e. 8.8x10-5 m 3

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The following two problems refer to this situation: #1 A cylindrical chamber containing an ideal diatomic gas is sealed by a movable piston with cross-sectional area A = 0.0015 m 2. The volume of the chamber is initially V=10-3 m 3 ; the pressure outside the chamber constant at 10 5 Pa. Initially the gas pressure inside is 10 5 Pa and its temperature is 27 C. 1. A 20-kg mass placed on the piston causes the gas to undergo adiabatic compression. What is the final volume after the mass comes to rest? a. 4.2x10-4 m 3 b. 5.5x10-4 m 3 c. 1.2x10-4 m 3 d. 1.4x10-5 m 3 e. 8.8x10-5 m 3

The following two problems refer to this situation: #2 A cylindrical chamber containing an ideal diatomic gas is sealed by a movable piston with cross-sectional area A = 0.0015 m 2. The volume of the chamber is initially V=10-3 m 3 ; the pressure outside the chamber constant at 10 5 Pa. Initially the gas pressure inside is 10 5 Pa and its temperature is 27 C. 2. If instead of putting the weight on the piston, we had cooled the gas in the container from 27 C by T = 1 C, which of these formulas correctly describes how far,, the piston falls? a. A D = V (DT / 27) b. 0 c. A D = V (DT / 300)

3. A rubber ball with specific heat c 150J kg 1 K 1 is dropped from a height of 5 m. After bouncing from a hard floor, it reaches a maximum height of 4.0 m. Assuming that all the energy lost in the collision goes into heating the ball, what is the change in temperature of the ball? a. DT = 0.60 K b. DT = 0.020 K c. DT = 0.42 K d. DT = 0.065 K e. Not enough information is given

The next two problems pertain to the following situation: #1 A movable partition separates two sides of a container in equilibrium. The right compartment contains 3.0 moles of hydrogen gas, while the left hand side contains 3.0 moles of helium gas. Both sides are at the same temperature. (Volumes in figure are not necessarily drawn correctly.) 4. What can you say about the volume of the two compartments? a. V b. V c. V right left right V left V V right left The number of molecules on both sides are the same, and the translational K.E. per molecule is the same; so the pressure on both sides is the same, and therefore the volumes are the same.

The next two problems pertain to the following situation: #2 A movable partition separates two sides of a container in equilibrium. The right compartment contains 3.0 moles of hydrogen gas, while the left hand side contains 3.0 moles of helium gas. Both sides are at the same temperature. (Volumes in figure are not necessarily drawn correctly.) 5. What can you say about the internal energy of the gas in the two compartments? a. U left U right b. c. U left U right U left U right H 2 has translational K.E. and rotational K.E.; He is a noble gas and has only translational energy. Both sides have the same number of moles. The K.E. per molecule of H 2 is (5/2)kT, and for He is (3/2) kt. So He has less internal energy.

6. A mass M = 1.0 kg is connected to a rigid wall with a spring with spring constant k = 10 N/m. If everything is at T=27 C, what is the root-mean-square displacement of the mass with respect to its equilibrium position? Assume the mass can only move in the x-direction. a. b. c. d. e. x 2 6.1 10 12 m x 2 2.8 10 11 m x 2 4.2 10 12 m x 2 1.1 10 10 m x 2 2.0 10 11 m x

The next two problems pertain to the following situation: #1 A sample of monatomic gas expands from its initial pressure, volume, and temperature to a final pressure, volume, and temperature along the dashed path shown in the figure. Each solid curve represents an isotherm. 7. What is the change in the internal energy of the gas as it goes along the dashed path? a. b. c. d. e. DU 3 2 Nk(T f T i )ln( V f V i ) DU 3 2 Nk(P f P i )(V f V i ) DU 5 2 Nk(P fv f P i V i ) DU 0 DU 3 2 Nk(T f T i ) Internal energy U is a state function, and only depends on the state of a system, and not on the history. U = αnkt = αpv. So, the change in the internal energy is just: 3 3 NkTi DU NkT 2 2 3 DU Nk Tf Ti 2 f

The next two problems pertain to the following situation: #2 A sample of monatomic gas expands from its initial pressure, volume, and temperature to a final pressure, volume, and temperature along the dashed path shown in the figure. Each solid curve represents an isotherm. 8. If the gas were instead allowed to expand adiabatically from to, find the work done by the gas. a. b. c. d. e. W 3 2 (P iv i P f V f ) W (P i V i P f V f ) W Nk(T i T f )( V f V i ) W Nk(T i T f ) ln( V f V i ) The work done cannot be found not enough information is given We had in lecture that in the adiabatic case (Q=0), the amount of work is equal to W = ΔU i f i i f f Work DU Nk T T PV P V The last "=" because it is an ideal gas; and This can also be derived from And calculating the work V PV constant 2 PdV V 1 And remembering that 1 3 2

9. A well insulated uniform rod of cross section 1 cm 2 and length 3d = 30 cm has temperature probes attached at positions d = 10 cm and 2d = 20 cm as illustrated. A 16-Watt heating element is attached at the left end. The rod and the heat sink at the right end are initially at a temperature of zero C. After the heating element is switched on (and held on), the temperatures at the two positions increase and approach, after a long time, values of T 1 = 32 C and T 2 = 16 C. What is the thermal conductivity of the material of the rod? a. 235 Watt m -1 K -1 b. 401 Watt m -1 K -1 c. 1000 Watt m -1 K -1

10. Compare the heat capacities of 1 kg of copper, 1 kg of the gas Argon, and 1 kg of the gas H 2, all at room temperature and 1 atm pressure. a. C Cu > C Ar > C H2 b. C H2 = C Ar = C Cu c. C Cu > C Ar = C H2 d. C H2 > C Cu > C Ar e. C Ar = C H2 > C Cu Here we have calculated C V. Actually the problem says that the pressure is constant (I did not notice this). If we calculate C p (for gases, C p /C V = γ; for solids C p =C V ). For the right calculation, see the next page.

For the last problem (10 ) using Cp (which is correct because the pressure is constant) instead of Cv. Remember that Cp 1 C V

11. Five oscillators have a total energy U = 5. Which of these energy distributions is more probable? (you are not to consider any other distributions.) Distribution b is shown in the figure. 5 4 3 a. They are equally probable. b. (1,1,1,1,1) is more probable. c. (5,0,0,0,0) is more probable. 2 1 0 All microstates are equally probable

The following two problems refer to this situation: #1 A 10-cm cubical box is filled with gas. Suppose a particular atom starts at the center as shown. It is observed that when the temperature is 300 K, the average time, <t>, before the atom strikes a wall is 50 sec. atom 12. If the temperature is increased to 330 K, what is the new value of <t>? a. <t> = 45.5 sec b. <t> = 52.5 sec c. <t> = 47.7 sec d. <t> = 55.0 sec e. <t> = 60.5 sec

The following two problems refer to this situation: #2 A 10-cm cubical box is filled with gas. Suppose a particular atom starts at the center as shown. It is observed that when the temperature is 300 K, the average time, <t>, before the atom strikes a wall is 50 sec. atom 13. Suppose that we repeat the measurement in a box of 20 cm on a side, with the gas pressure and temperature not changed (i.e., T remains at 300 K). What is the value of <t> in this case? a. <t> = 35 sec b. <t> = 100 sec c. <t> = 200 sec d. <t> = 71 sec e. <t> = 400 sec

The following two problems refer to this situation: #1 We have three objects. Two are indistinguishable (the B s), and one is distinguishable (the A ). We arrange these three objects in three multiple-occupancy bins. One microstate is shown. A B B 14. What is the dimensionless entropy of this system? a. s tot = 2.30 b. s tot = 2.89 c. s tot = 1.79 d. s tot = 2.20 e. s tot = 3.30

The following two problems refer to this situation: #2 We have three objects. Two are indistinguishable (the B s), and one is distinguishable (the A ). We arrange these three objects in three multiple-occupancy bins. One microstate is shown. A B B 15. Suppose, instead, that all three objects are indistinguishable. (Replace the A with a B.) What happens to the total entropy? a. s tot is smaller. b. s tot is larger. c. s tot is the same.

16. A box contains five distinguishable particles in equilibrium. They are free to move between the left and right halves. Assuming that every microstate is equally likely, calculate the ratio, P A /P B, of the probabilities that five distinguishable particles will be in these two macrostates: A with three particles on the left and two on the right, and B with one on the left a. P A /P B = 24 b. P A /P B = 1 c. P A /P B = 5 d. P A /P B = 2 e. P A /P B = 10 Macrostate A: N L = 3, N R = 2. Macrostate B: N L = 1, N R = 4. A B

The following two problems are related #1 10 grams of Hydrogen (an -ideal diatomic gas) are initially confined to the left half of a 30-liter insulated container. The initial temperature is 280 K. The partition separating left and right is then quickly removed, and the gas then expands to fill the container. 17. How large is the change, DS, of the entropy of the gas? a. DS = 0 b. DS = +8.75 J/K c. DS = +83.0 J/K d. DS = +59.6 J/K e. DS = +28.8 J/K

The following two problems are related #2 10 grams of Hydrogen (an -ideal diatomic gas) are initially confined to the left half of a 30-liter insulated container. The initial temperature is 280 K. The partition separating left and right is then quickly removed, and the gas then expands to fill the container. 18. What is the final temperature, T f, of the gas? a. T f = 280 K b. T f = 198 K c. T f = 140 K There is no work done, and no Q; so there is no temperature change.

19. Which of the choices below is correct? A system evolving toward thermal equilibrium a. is evolving toward the most likely macrostate. b. is evolving toward the most likely microstate. c. must be sharing energy or volume with other systems. Systems always move toward the most probable macro state.

20. Consider an isolated system that has dimensionless entropy, s. If we put another identical system nearby (still isolated), the total entropy, s tot, of the two systems is: a. s tot = s 2 b. s tot = 2 s c. s tot = 2s

The next three problems are related. #1 In a recent experiment, physicists at UCSB were able to cool a macroscopic mechanical oscillator so that it was essentially in its quantum mechanical ground state with E = 0. 21. Assuming the frequency of the system is 9 GHz (corresponding to energy levels separated by a fixed amount ε = 6 x 10-24 Joules, as illustrated here), how low must the temperature have been if the probability to be out of the ground state was less than 1%? a. 0.06 K b. 0.12 K c. 0.29 K d. 0.09 K e. Cannot be determined from the information given. See the next page for the approximation

The probability at any temperature of being in any level of energy is kt P 1 e e n n kt The probability of being in the lowest excited state is according to the problem: kt kt P1 1 e e.01 So P1 kt e The same approximation can be used in the next two problems.

The next three problems are related. #2 In a recent experiment, physicists at UCSB were able to cool a macroscopic mechanical oscillator so that it was essentially in its quantum mechanical ground state with E = 0. 22. If the temperature is such that the probability of having exactly one quantum of energy (i.e., E = ) is 1%, what is the probability of being in the state with E = 2e? a. 2% b. 0.5% c. 0.01%

The next three problems are related. #3 In a recent experiment, physicists at UCSB were able to cool a macroscopic mechanical oscillator so that it was essentially in its quantum mechanical ground state with E = 0. 23. Suppose that there were two states with E = ε and one with E = 0 (i.e., the first excited level is degenerate). How would the probability P(E=0) to be in the ground state change (compared to the case of a non-degenerate first-excited level)? a. P(E=0) would be smaller than if the first excited level had only one state. b. P(E=0) would be the same as if the first excited level had only one state. c. P(E=0) would be greater than if the first excited level had only one state.

The next two problems are related: #1 On a particular night in Champaign, a low-lying mist sits within ~5m of the ground (i.e., at 5 m, the mist density (number of mist droplets per m 3 ) is 10 times lower than it is at ground level). 24. Assuming T = 20 C, and that the vertical mist density distribution is adequately modeled using a Boltzmann distribution, what is the approximate average number of H 2 O molecules per droplet? a. 6300 b. 3200 c. 1830 d. 440 e. 50

The next two problems are related: #2 On a particular night in Champaign, a low-lying mist sits within ~5m of the ground (i.e., at 5 m, the mist density (number of mist droplets per m 3 ) is 10 times lower than it is at ground level). 25. Compare the density of mist at a height of 10 m to the density at 5 meters. Hint: Sketch the density as a function of altitude. a. The density at 10 m is smaller than it is at 5 m, by a factor of about 10. b. The density at 10 m is smaller than it is at 5 m, by a factor of about 2. c. The density at 10 m is larger than it is at 5 m, by a factor of about 2. This is the probability of being at a height of 10 meters, which is 2x5 meters. We could also have just calculated P 10 /P 5 = 0.1 directly.