Review the following concepts if necessary:

- Heat, Temperature, and Power for AP Physics B
- Thermal Expansion, Ideal Gas Law, and Kinetic Theory of Gases for AP Physics B
- First Law of Thermodynamics for AP Physics B
- PV Diagrams for AP Physics B
- Heat Engines and the Second Law of Thermodynamics for AP Physics B

### Problems

### Multiple Choice:

- To increase the diameter of an aluminum ring from 50.0 mm to 50.1 mm, the temperature of the ring must increase by 80°C. What temperature change would be necessary to increase the diameter of an aluminum ring from 100.0 mm to 100.1 mm?
- 20°C
- 40°C
- 80°C
- 110°C
- 160°C

- A gas is enclosed in a metal container with a moveable piston on top. Heat is added to the gas by placing a candle flame in contact with the container's bottom. Which of the following is true about the temperature of the gas?
- The temperature must go up if the piston remains stationary.
- The temperature must go up if the piston is pulled out dramatically.
- The temperature must go up no matter what happens to the piston.
- The temperature must go down no matter what happens to the piston.
- The temperature must go down if the piston is compressed dramatically.

- A small heat engine operates using a pan of 100°C boiling water as the high temperature reservoir and the atmosphere as a low temperature reservoir. Assuming ideal behavior, how much more efficient is the engine on a cold, 0°C day than on a warm, 20°C day?
- 1.2 times as efficient
- 2 times as efficient
- 20 times as efficient
- infinitely more efficient
- just as efficient

- A 1-m
^{3}container contains 10 moles of ideal gas at room temperature. At what fraction of atmospheric pressure is the gas inside the container?- 1/40 atm
- 1/20 atm
- 1/10 atm
- 1/4 atm
- 1/2 atm

### Free Response:

- A small container of gas undergoes a thermodynamic cycle. The gas begins at room temperature. First, the gas expands isobarically until its volume has doubled. Second, the gas expands adiabatically. Third, the gas is cooled isobarically; finally, the gas is compressed adiabatically until it returns to its original state.
- The initial state of the gas is indicated on the
*PV*diagram below. Sketch this process on the graph. - Is the temperature of the gas greater right before or right after the adiabatic expansion? Justify your answer.
- Is heat added to or removed from the gas in one cycle?
- Does this gas act as a refrigerator or a heat engine?

- The initial state of the gas is indicated on the

### Solutions

**B**—Solve the formula for length expansion for temperature change:**A**—Use the first law of thermodynamics, Δ*U = Q + W*. The candle adds heat to the gas, so*Q*is positive. Internal energy is directly related to temperature, so if Δ*U*is positive, then temperature goes up (and vice versa). Here we can be sure that Δ*U*is positive if the work done on the gas is either positive or zero. The only possible answer is A—for B, the work done on the gas is negative because the gas expands. (Note that just because we add heat to a gas does NOT mean that temperature automatically goes up!)**A**—The equation for efficiency of an ideal heat engine is**D**—Estimate the answer using the ideal gas law. Solving for pressure,*P = nRT/V*. Plug in: (10 moles)(8)(300)/(1) = 24,000 N/m^{2}. (Just round off the ideal gas constant to 8 J/mol·K and assume 300 K. Remember, you don't have a calculator, so make computation easy.) Well, atmospheric pressure is listed on the constant sheet and is 100,000 N/m^{2}. Thus, the pressure of the container is about 1/4 of atmospheric.-
- The temperature is greater right before the expansion. By definition, in an adiabatic process, no heat is added or removed. But because the gas expanded, work was done
*by*the gas, meaning the W term in the first law of thermodynamics is negative. So, by Δ*U = Q + W, Q = 0*and*W*is negative; so Δ*U*is negative as well. Internal energy is directly related to temperature. Therefore, because internal energy decreases, so does temperature. - In a full cycle, the gas begins and ends at the same state. So the total change in internal energy is
*zero*. Now consider the total work done on or by the gas. When the gas expands in the first and second process, there's more area under the graph than when the gas compresses in the second and third processes. So, the gas does more work expanding than compressing; the net work is thus done by the gas, and is*negative*. To get no change in internal energy, the Q term in the first law of thermodynamics must be positive; heat must be added to the gas. - This is a heat engine. Heat is added to the gas, and net work is done by the gas; that's what a heat engine does. (In a refrigerator, net work is done on the gas, and heat is removed from the gas.)

- The temperature is greater right before the expansion. By definition, in an adiabatic process, no heat is added or removed. But because the gas expanded, work was done

The only term on the right-hand side that is different from the original situation is L_{0}, which has doubled (from 50.0 mm to 100.0 mm). If you double the denominator of an equation, the entire equation is cut in half—thus the answer is 40°C.

The temperatures must be in kelvins. The denominator doesn't change from a hot day to a cold day, so look at the numerator only. Because a change of 1°C is equivalent to a change of 1 K, the numerator is 100 K on a cold day and 80 K on a warm day. So the engine is 100/80 ~ 1.2 times as efficient on a warm day as on a cold day.

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