Thermodynamics Study Materials for Competitive Exams

Thermodynamics

First Law of Thermodynamics Heat & Temperature Heat Transfer

Q. During a phase change, the temperature of a substance:

A. Increases
B. Decreases
C. Remains constant
D. Varies unpredictably

Solution

During a phase change, the temperature of a substance remains constant while the substance absorbs or releases heat.

Correct Answer: C — Remains constant

Q. During an isochoric process, the volume of the system:

A. Increases
B. Decreases
C. Remains constant
D. Varies with temperature

Solution

In an isochoric process, the volume of the system remains constant.

Correct Answer: C — Remains constant

Q. If 100 g of water at 0°C is mixed with 100 g of water at 100°C, what will be the final temperature?

A. 50°C
B. 25°C
C. 75°C
D. 0°C

Solution

The final temperature will be 50°C due to equal masses and specific heat capacities.

Correct Answer: A — 50°C

Q. If 100 g of water at 0°C is mixed with 100 g of water at 100°C, what will be the final temperature of the mixture?

A. 50°C
B. 25°C
C. 75°C
D. 0°C

Solution

Using the principle of conservation of energy, the final temperature will be 50°C.

Correct Answer: A — 50°C

Q. If 500 J of heat is added to a system and 200 J of work is done by the system, what is the change in internal energy?

A. 300 J
B. 500 J
C. 700 J
D. 200 J

Solution

According to the first law of thermodynamics, ΔU = Q - W. Here, ΔU = 500 J - 200 J = 300 J.

Correct Answer: A — 300 J

Q. If a gas expands against a constant external pressure, the work done by the gas is given by:

A. W = P_ext * ΔV
B. W = ΔU + Q
C. W = Q - ΔU
D. W = P_ext / ΔV

Solution

The work done by the gas during expansion against a constant external pressure is given by W = P_ext * ΔV.

Correct Answer: A — W = P_ext * ΔV

Q. If a gas expands and does 150 J of work while absorbing 100 J of heat, what is the change in internal energy?

A. -50 J
B. 50 J
C. 250 J
D. 100 J

Solution

Using the first law of thermodynamics, ΔU = Q - W = 100 J - 150 J = -50 J.

Correct Answer: A — -50 J

Q. If a system absorbs 100 J of heat and does 40 J of work, what is the change in internal energy?

A. 60 J
B. 40 J
C. 100 J
D. 140 J

Solution

Using the first law of thermodynamics, ΔU = Q - W = 100 J - 40 J = 60 J.

Correct Answer: A — 60 J

Q. If the temperature of an ideal gas is doubled at constant volume, what happens to its pressure?

A. It halves
B. It remains the same
C. It doubles
D. It quadruples

Solution

According to Gay-Lussac's law, pressure is directly proportional to temperature at constant volume, so it doubles.

Correct Answer: C — It doubles

Q. If the temperature of an object increases, what happens to the rate of heat radiation from that object?

A. Decreases
B. Increases
C. Remains constant
D. Becomes zero

Solution

According to Stefan-Boltzmann law, the rate of heat radiation increases with the fourth power of the temperature, so as the temperature increases, the rate of radiation also increases.

Correct Answer: B — Increases

Q. If the temperature of an object increases, what happens to the rate of heat transfer by radiation?

A. Decreases
B. Increases
C. Remains constant
D. Becomes zero

Solution

The rate of heat transfer by radiation increases with the fourth power of the absolute temperature, according to Stefan-Boltzmann law.

Correct Answer: B — Increases

Q. If the temperature of an object is doubled, how does its thermal radiation change according to the Stefan-Boltzmann law?

A. It doubles
B. It quadruples
C. It remains the same
D. It increases eightfold

Solution

According to the Stefan-Boltzmann law, if the temperature is doubled, the thermal radiation increases by a factor of 2^4 = 16, not quadruples.

Correct Answer: B — It quadruples

Q. If two objects are in thermal contact, what will happen to their temperatures over time?

A. They will remain the same
B. They will equalize
C. One will increase, the other will decrease
D. They will both decrease

Solution

According to the second law of thermodynamics, heat will flow from the hotter object to the cooler one until they reach thermal equilibrium.

Correct Answer: B — They will equalize

Q. In a closed system, if 500 J of heat is added and 200 J of work is done by the system, what is the change in internal energy?

A. 300 J
B. 500 J
C. 700 J
D. 200 J

Solution

According to the first law of thermodynamics, ΔU = Q - W. Here, ΔU = 500 J - 200 J = 300 J.

Correct Answer: A — 300 J

Q. In a cyclic process, the change in internal energy is:

A. Positive
B. Negative
C. Zero
D. Depends on the path taken

Solution

In a cyclic process, the system returns to its initial state, so the change in internal energy is zero.

Correct Answer: C — Zero

Q. In a cyclic process, the change in internal energy of the system is:

A. Positive
B. Negative
C. Zero
D. Depends on the work done

Solution

In a cyclic process, the system returns to its initial state, so the change in internal energy is zero.

Correct Answer: C — Zero

Q. In a cyclic process, what is the net change in internal energy of the system?

A. Positive
B. Negative
C. Zero
D. Depends on the path taken

Solution

In a cyclic process, the system returns to its initial state, so the net change in internal energy is zero.

Correct Answer: C — Zero

Q. In a cyclic process, what is the net change in internal energy?

A. Positive
B. Negative
C. Zero
D. Depends on the process

Solution

In a cyclic process, the system returns to its initial state, so the net change in internal energy is zero.

Correct Answer: C — Zero

Q. In a heat engine, if the work done is 200 J and the heat absorbed is 500 J, what is the efficiency?

A. 40%
B. 50%
C. 60%
D. 80%

Solution

Efficiency = (Work done / Heat absorbed) * 100 = (200 J / 500 J) * 100 = 40%.

Correct Answer: B — 50%

Q. In a heat engine, the work done is equal to:

A. Heat absorbed from the hot reservoir
B. Heat rejected to the cold reservoir
C. Heat absorbed minus heat rejected
D. Heat absorbed plus heat rejected

Solution

The work done by a heat engine is equal to the heat absorbed from the hot reservoir minus the heat rejected to the cold reservoir.

Correct Answer: C — Heat absorbed minus heat rejected

Q. In a process where 100 J of heat is added to a system and the internal energy increases by 40 J, how much work is done by the system?

A. 60 J
B. 40 J
C. 100 J
D. 140 J

Solution

Using the first law of thermodynamics, ΔU = Q - W, we have 40 J = 100 J - W, thus W = 100 J - 40 J = 60 J.

Correct Answer: A — 60 J

Q. In a process where 300 J of heat is added to a system and the internal energy increases by 100 J, how much work is done by the system?

A. 200 J
B. 100 J
C. 300 J
D. 400 J

Solution

Using the first law of thermodynamics, ΔU = Q - W. Rearranging gives W = Q - ΔU. Here, W = 300 J - 100 J = 200 J.

Correct Answer: A — 200 J

Q. In an isochoric process, what happens to the internal energy of a gas when heat is added?

A. It decreases
B. It remains constant
C. It increases
D. It depends on the gas

Solution

In an isochoric process, the volume remains constant, and any heat added to the system increases the internal energy.

Correct Answer: C — It increases

Q. In an isochoric process, what happens to the internal energy of an ideal gas when heat is added?

A. It decreases.
B. It remains constant.
C. It increases.
D. It depends on the amount of heat added.

Solution

In an isochoric process, the volume remains constant, and any heat added increases the internal energy of the gas.

Correct Answer: C — It increases.

Q. In an isothermal process for an ideal gas, which of the following is true?

A. The internal energy remains constant.
B. The temperature increases.
C. The pressure decreases.
D. The volume remains constant.

Solution

In an isothermal process, the temperature remains constant, which implies that the internal energy of an ideal gas also remains constant.

Correct Answer: A — The internal energy remains constant.

Q. In an isothermal process, how does the internal energy of an ideal gas change?

A. Increases
B. Decreases
C. Remains constant
D. Depends on the amount of gas

Solution

In an isothermal process for an ideal gas, the temperature remains constant, and thus the internal energy also remains constant.

Correct Answer: C — Remains constant

Q. In an isothermal process, the change in internal energy of an ideal gas is:

A. Positive
B. Negative
C. Zero
D. Depends on the amount of gas

Solution

In an isothermal process for an ideal gas, the temperature remains constant, hence the change in internal energy is zero.

Correct Answer: C — Zero

Q. In an isothermal process, the temperature of the system:

A. Increases
B. Decreases
C. Remains constant
D. Varies linearly

Solution

In an isothermal process, the temperature of the system remains constant.

Correct Answer: C — Remains constant

Q. In which of the following processes does the internal energy of a system remain constant?

A. Isothermal process
B. Adiabatic process
C. Isobaric process
D. Isochoric process

Solution

In an isothermal process, the internal energy of an ideal gas remains constant.

Correct Answer: A — Isothermal process

Q. In which of the following processes does the temperature of a substance remain constant?

A. Heating
B. Cooling
C. Phase change
D. Compression

Solution

During a phase change, the temperature of a substance remains constant despite heat being added or removed.

Correct Answer: C — Phase change

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