Chapter 11 Super Review

1. A two mole sample of a gas has a temperature of 1000 K and a volume of 6 m 3. What is the pressure?

PV = nRT P x 6 = 2 x 8.31x 1000 P = 2770 Pa

2. If an ideal gas does work on its surroundings, how does the volume change?

If the gas does work ON the environment, the gas must be expanding.

3. What value MUST change in order for work to be done on or by the system?

For work to be done, the volume must change.

4. If the pressure on a gas is Pa and the volume is 4 m 3, how much work is done by the gas if it is allowed to expand at constant pressure to three times its initial volume?

The gas expanded to three times 4 m 3, which is 12 m 3. W = P  V W = x (12-4) W = J Since the gas is expanding, the gas did work on the environment, so we call this negative work: J.

5. During what type of process is there no change in the internal energy of the system?

If  U = 0, then there is no change in temperature, so it is an isothermic change.

6. During what type of process is the temperature held constant?

If the temperature is held constant it is an isothermic change.

7. During what type of process is there no change in volume?

If the volumeis held constant it is an isovolumetric change. (Note: No work is done in an isovolumetric change.)

8. During compression of a gas, 235 J of work is done on the gas. If the internal energy increases by 222 J, what is the total amount of energy transferred as heat?

8.  U = Q + W 222 = Q Q = -13 J

9. The internal energy of a gas is 72 J. If 83 J is added as heat while the system does 62 J of work. What is the system’s final internal energy?

9.  U = Q + W  U = 83 + (-62)  U = 21 J Since U was initially 72 J, the final U is = 93 J.

10. An engine absorbs 2300 J of heat from a hot reservoir and gives off 820 J of heat to a cold reservoir. How much work is done?

10. W = Q H - Q C W = 2300 – 820 W = 1480 J

11. According to the second law of thermodynamics, can all of the heat received by heat engine be converted to work?

11. No, there must always be some energy released to the environment as Q H.

12. A heat engine performs 2500 J of net work while adding 5500 J of heat to the cold- temperature reservoir. What is the efficiency of the engine?

12. e = W/Q H Q H = W + Q C Q H = Q H = 8000 J e = 2500/8000 e = 0.31 or 31%

13. An engine adds 4000 J of energy as heat and removes 1000 J of energy as heat. What is the engine’s efficiency?

13. e = (Q H - Q C )/Q H e = ( )/4000 e = 0.75 or 75%

14. When a system’s disorder is increased, (more, no, less) energy is available to do work.

14. Less, the greater the disorder the less energy available to do work.

15. A teacher pushes quickly on a plunger of a pump. No air escapes. Describe any transfer of energy as heat and any work done on or by the air in the system.

15. No heat is transferred because the compression happens too quickly. But, since the gas is compressed, work is done on the gas.

16. A book is on an inflated balloon. The sun shines on the balloon and the balloon expands. The book rises. Describe any movement of heat and any work done?

16. Heat from the environment is transferred to the gas in the balloon; the balloon does work on the book as the gas expands.

17. According to the first law of thermodynamics, How can the internal energy of a system be increased?

17.  U = Q + W Increase Q and/or W. In other words, add heat to the system or do work to the system.

18. Name a room that has lower entropy than this classroom.

18. Entropy is disorder, so the more orderly the room is, the lower the entropy. My classroom is so disorderly, that almost any room would be have less entropy. A good example is the library; the library is very orderly so it has lower entropy.

Problem 19: One mole of a gas is taken through the cycle abca shown above. State A has a volume of 2 m 3 and pressure 400 Pa. State C has a volume of 8 m 3 and pressure 100 Pa. Process CA lies along an isotherm.

A) What is the temperature at state A?

PV = nRT 400 x 2 = 1 x 8.31 x T T = 96 K

B) What is the temperature at state B?

P 1 V 1 /T 1 = P 2 V 2 /T 2 (400)(2)/96 = (400)(8)/T 2 T 2 = 384 J

C) What is the temperature at state C?

State C is on the isotherm with state A, so they must be the same temperature: 96 K.

D) How much work is done on the gas during process AB?

W = P  V W = (400)(8-2) W = 2400 J The gas is expanding, so this is negative work on the gas: J.

E) How much work is done on the gas during process BC?

None, the gas is remaining at a constant volume. The volume must change for work to be done.

F) If the internal energy U at A is 1200 joules and U at B is 4800 joules, what is the heat Q added to the gas during AB?

 U = Q + W ( ) = Q + (-2400) 3600 = Q + (-2400) Q = 6000 J

G) What is the heat Q added to the gas during BC?

The change in internal energy  U is the opposite of the change from A to B because the gas is returning to the same temperature, the work is zero because there is no change in volume.  U = Q + W = Q + 0 Q = J

H) If the work done on the gas in process CA is 1100 joules, what is the heat added to the gas in process CA?

This is an isotherm so  U is 0.  U = Q + W 0 = Q Q = J

I) What is the net work for the entire cycle?

A to B is J, B to C is 0 J, C to A is 1100 J. The total work is J.

J) What is the net heat transferred for the entire cycle?

In a cycle  U is 0.  U = Q + W 0 = Q + (-1300) Q = J

K) Is this a heat engine or a refrigeration system?

Since heat Q is added (+1300 J) and work W (-1300 J) is removed, it is a heat engine. You can also tell it is a heat engine because it is a clockwise cycle. Counterclockwise cycles are refrigeration systems.