AP Physics 2: Algebra-Based Practice Quiz: The First Law of Thermodynamics

Correct Answer: B

The content explicitly states, 'The internal energy of a system is the sum of the kinetic energy of the objects that make up the system and the potential energy of the configuration of those objects.'

Correct Answer: C

The content states, 'The first law of thermodynamics is a restatement of conservation of energy that accounts for energy transferred into or out of a system by work, heating, or cooling.'

Correct Answer: C

The equation for the internal energy of an ideal monatomic gas is given as U = (3/2)nRT. Since n and R are constants for a given amount of gas, the internal energy U is directly proportional to the absolute temperature T.

Correct Answer: B

The work done on a system is defined as W = -PΔV. When a gas expands, its final volume is greater than its initial volume, so the change in volume (ΔV) is positive. Therefore, W = -P(positive value), which results in a negative value for W.

Correct Answer: B

The content provides the relevant equation for the first law of thermodynamics as ΔU = Q + W, where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done on the system.

Correct Answer: A

Using the first law of thermodynamics, ΔU = Q + W. Substituting the given values: ΔU = (+200 J) + (+150 J) = +350 J. The internal energy of the system increases by 350 J.

Correct Answer: C

The provided content explicitly states, 'Pressure-volume graphs (also known as PV diagrams) are representations used to represent thermodynamic processes.'

Correct Answer: D

The work done on the system is W = -PΔV. Since the container is rigid, the change in volume ΔV is zero. Therefore, the work done W is zero. According to the first law, ΔU = Q + W. Since W = 0, it simplifies to ΔU = Q.

Correct Answer: C

Using the formula W = -PΔV. The change in volume is ΔV = V_final - V_initial = 3.0 m³ - 5.0 m³ = -2.0 m³. The work done on the system is W = -(100 Pa)(-2.0 m³) = +200 J. The positive sign indicates work was done on the system.

Correct Answer: B

The change in internal energy is ΔU = -400 J. If the system does 150 J of work on its surroundings, the work done ON the system is W = -150 J. Using the first law, ΔU = Q + W, we have -400 J = Q + (-150 J). Solving for Q gives Q = -400 J + 150 J = -250 J. The negative sign indicates that 250 J of heat was removed from the system.

Correct Answer: C

The internal energy of an ideal monatomic gas is given by U = (3/2)nRT. Since U is directly proportional to the absolute temperature T, doubling T will also double the internal energy U.

Correct Answer: B

The provided text specifies that for an ideal monatomic gas, the potential energy between atoms is considered negligible. Therefore, 'The internal energy of an ideal monatomic gas is the sum of the kinetic energies of the constituent atoms in the gas.'

Correct Answer: D

The internal energy of an ideal monatomic gas depends only on its temperature (U = 3/2 nRT). In an isothermal process, the temperature (T) is constant. Therefore, the internal energy (U) does not change, and ΔU = 0.

Correct Answer: B

Heat absorbed by the system is Q = +50 J. The system does 50 J of work on its surroundings, which means the work done ON the system is W = -50 J. Using the first law, ΔU = Q + W = (+50 J) + (-50 J) = 0 J. This describes an isothermal process for an ideal gas.

Correct Answer: B

For an ideal monatomic gas, internal energy depends only on temperature. In an isothermal (constant temperature) process, the change in internal energy ΔU is zero. According to the first law of thermodynamics, ΔU = Q + W. If ΔU = 0, then 0 = Q + W, which rearranges to Q = -W. This means the heat added to the system is equal to the work done by the system on its surroundings.

Correct Answer: C

In an adiabatic process, Q = 0. The system is compressed, so its volume decreases (ΔV is negative). The work done on the system is W = -PΔV = -P(negative value), which is positive. According to the first law, ΔU = Q + W = 0 + W, so ΔU = W. Since W is positive, ΔU is positive, meaning the internal energy increases. For an ideal gas, internal energy is directly proportional to temperature (U = 3/2 nRT), so an increase in internal energy means an increase in temperature.