Physics: AP Physics 2: Thermodynamics and Heat Engines Worksheet

Question 1

A gas absorbs 500 J of heat from its surroundings and does 200 J of work on the surroundings. What is the change in the gas's internal energy?

Accepted Answer

Using the first law of thermodynamics, ΔU = Q - W. The gas absorbs Q = 500 J and does W = 200 J of work, so ΔU = 500 J - 200 J = 300 J. The internal energy of the gas increases by 300 J.

Question 2

An ideal gas is compressed in a cylinder. During the compression, 150 J of work is done on the gas, and the gas releases 60 J of heat to the surroundings. What is the change in internal energy of the gas?

Accepted Answer

Work done on the gas means the work done by the gas is W = -150 J. Heat released means Q = -60 J. Using ΔU = Q - W, ΔU = -60 J - (-150 J) = 90 J. The internal energy of the gas increases by 90 J.

Question 3

A gas expands at a constant pressure of 2.0 x 10^5 Pa from a volume of 0.010 m^3 to 0.030 m^3. How much work does the gas do?

Accepted Answer

For a constant-pressure process, W = PΔV. The change in volume is 0.030 m^3 - 0.010 m^3 = 0.020 m^3. Therefore, W = (2.0 x 10^5 Pa)(0.020 m^3) = 4.0 x 10^3 J. The gas does 4000 J of work.

Question 4

An ideal gas undergoes an isochoric process. Its pressure increases while its volume remains constant. Explain why the work done by the gas is zero.

Accepted Answer

The work done by a gas is W = PΔV for a constant-pressure process, and more generally it is the area under the curve on a PV diagram. In an isochoric process the volume does not change, so ΔV = 0. Therefore, the gas does no work.

Question 5

A sample of ideal gas is heated at constant volume. The gas absorbs 240 J of heat. How much work is done by the gas, and what is the change in internal energy?

Accepted Answer

At constant volume, the gas does no work because ΔV = 0. Therefore W = 0 J. Using ΔU = Q - W, ΔU = 240 J - 0 J = 240 J. The internal energy increases by 240 J.

Question 6

A heat engine absorbs 1200 J of energy from a hot reservoir and exhausts 750 J of energy to a cold reservoir during each cycle. How much work does the engine do per cycle, and what is its efficiency?

Accepted Answer

The work output is W = QH - QC = 1200 J - 750 J = 450 J. The efficiency is e = W/QH = 450 J/1200 J = 0.375. The engine does 450 J of work per cycle and has an efficiency of 37.5%.

Question 7

A Carnot engine operates between a hot reservoir at 600 K and a cold reservoir at 300 K. What is the maximum possible efficiency of the engine?

Accepted Answer

The Carnot efficiency is emax = 1 - TC/TH, using absolute temperatures in kelvins. Substituting the values gives emax = 1 - 300 K/600 K = 0.50. The maximum possible efficiency is 50%.

Question 8

A student claims that a heat engine operating between 500 K and 250 K can have an efficiency of 60%. Is this possible? Explain your answer.

Accepted Answer

This is not possible. The maximum efficiency is the Carnot efficiency, emax = 1 - TC/TH = 1 - 250 K/500 K = 0.50, or 50%. Since 60% is greater than the Carnot limit, the claimed engine violates the second law of thermodynamics.

Question 9

A cyclic process is shown on a PV diagram as a closed clockwise loop. Explain whether the net work done by the gas over one cycle is positive, negative, or zero.

Accepted Answer

The net work done by the gas over one complete cycle is the area enclosed by the loop on the PV diagram. A clockwise loop represents positive net work done by the gas. Therefore, the gas does positive net work over the cycle.

Question 10

A refrigerator removes 300 J of heat from its cold interior while 100 J of work is done on it. How much heat is expelled to the warmer room, and what is the refrigerator's coefficient of performance?

Accepted Answer

Energy conservation gives QH = QC + W = 300 J + 100 J = 400 J. The coefficient of performance for a refrigerator is COP = QC/W = 300 J/100 J = 3.0. The refrigerator expels 400 J to the room and has a COP of 3.0.

Question 11

During an irreversible process, 600 J of heat flows from a hot object at 400 K to a cold object at 300 K. Calculate the total entropy change of the two-object system.

Accepted Answer

The hot object loses heat, so its entropy change is ΔShot = -600 J/400 K = -1.5 J/K. The cold object gains heat, so its entropy change is ΔScold = 600 J/300 K = 2.0 J/K. The total entropy change is 0.5 J/K, so the entropy of the two-object system increases.

Question 12

A PV diagram shows an ideal gas moving from state A to state B along two different paths. Path 1 has a larger area under the curve than Path 2. Compare the work done by the gas along the two paths, and explain whether the change in internal energy depends on the path.

Accepted Answer

The work done by the gas is larger along Path 1 because the area under the curve on a PV diagram is larger for that path. The change in internal energy does not depend on the path because internal energy is a state function. If both paths begin at the same state A and end at the same state B, both paths have the same ΔU.

Physics: AP Physics 2: Thermodynamics and Heat Engines

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