50+ NEET MCQ Questions Thermodynamics with Solutions

Here we will provide you the 50+ MCQ Questions of Thermodynamics for NEET-UG. Thermodynamics is the chapter 1 in Class XI or Class 11 Physics NCERT Unit Thermodynamics NEET (conducted by NTA) is based on the NCERT book.

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These 50+ MCQ questions are selected by the experts of studyrate.in and these are more difficult questions, which will help you to better understand Thermodynamics NEET MCQ Questions with Answers.

Thermodynamics NEET MCQ


Two identical containers A and B of equal volume are filled with air at temperatures T1 and T2 (T1>T2) respectively. The gases are allowed to flow through a narrow tube until they reach a third container C of the same volume but evacuated previously. The temperature of the gases in C will be:
a) (T1 + T2) / 2
b) less than T2
c) greater than T1
d) less than (T1 + T2) / 2

Answer: d) less than (T1 + T2) / 2

Explanation: When the air flows from container A and B into container C, it will undergo adiabatic expansion, which means that no heat exchange takes place between the air and the surroundings. As a result, the temperature of the air will decrease due to the decrease in internal energy. Since the initial temperature of the air in container A is higher than that in container B, the final temperature of the air in container C will be less than the average temperature of the two initial containers, which is (T1 + T2) / 2

Answer: d) less than (T1 + T2) / 2 Explanation: When the air flows from container A and B into container C, it will undergo adiabatic expansion, which means that no heat exchange takes place between the air and the surroundings. As a result, the temperature of the air will decrease due to the decrease in internal energy. Since the initial temperature of the air in container A is higher than that in container B, the final temperature of the air in container C will be less than the average temperature of the two initial containers, which is (T1 + T2) / 2

A heat engine absorbs 100 J of heat from a high-temperature reservoir at 500 K and releases 60 J of heat to a low-temperature reservoir at 300 K. What is the efficiency of the heat engine?
a) 40%
b) 60%
c) 67%
d) 83%

Answer: c) 67% Explanation: The efficiency of a heat engine is given by (1 – Qc / Qh), where Qh is the heat absorbed from the high-temperature reservoir and Qc is the heat released to the low-temperature reservoir. Substituting the given values, we get: Efficiency = (1 – 60 J / 100 J) x 100% Efficiency = 40% Efficiency = 100% – 40% Efficiency = 60% Therefore, the efficiency of the heat engine is 60%.

An ideal gas undergoes a reversible isothermal expansion. The work done by the gas is W and the heat absorbed by the gas is Q. Which of the following expressions is correct?
a) W > Q
b) W < Q
c) W = Q
d) It depends on the initial and final states of the gas.

Answer: c) W = Q Explanation: In a reversible isothermal expansion of an ideal gas, the temperature remains constant, and the internal energy of the gas remains constant. Therefore, the heat absorbed by the gas is equal to the work done by the gas, i.e., Q = W.

A cylinder of height 20 cm contains an ideal gas at 27°C. The cylinder is fitted with a piston of mass 0.2 kg and area 20 cm2. The atmospheric pressure is 1.01 x 105 Pa. The minimum mass of the additional load that should be placed on the piston so that the gas is compressed adiabatically to half its initial volume is:
a) 0.5 kg
b) 1 kg
c) 2 kg
d) 4 kg

Answer: b) 1 kg Explanation: The pressure of an ideal gas undergoing an adiabatic compression is given by: PfVf^γ = PiVi^γ, where Pf is the final pressure, Vf is the final volume, Pi is the initial pressure, Vi is the initial volume, and γ is the ratio of the specific heats of the gas. Since the initial volume is halved, the final volume is Vi/2. We can rearrange the above equation to get: Pf/P


In a cyclic process, the heat added to the system is 800 J and the work done by the system is 500 J. What is the change in the internal energy of the system?
A) 300 J
B) -300 J
C) 1300 J
D) -1300 J

Answer: A

A gas is compressed isothermally to half its volume. If the initial pressure of the gas was P and the final pressure is 2P, what is the work done on the gas?
A) P/2V
B) PV/2
C) PV
D) 2PV

Answer: B

A heat engine operates between two reservoirs at temperatures T1 and T2, where T2 is greater than T1. What is the maximum efficiency of the engine?
A) (T2 – T1)/T2
B) (T2 – T1)/T1
C) T1/(T2 – T1)
D) T2/(T2 – T1)

Answer: B

The efficiency of a Carnot engine operating between two temperatures is 40%. What is the efficiency of another Carnot engine operating between the same temperatures but with a different working substance?
A) less than 40%
B) greater than 40%
C) equal to 40%
D) cannot be determined without additional information

Answer: C

A cylinder with a movable piston contains an ideal gas. The gas is expanded isothermally by increasing the temperature of a heat reservoir. What is the work done by the gas during the expansion?
A) zero
B) positive
C) negative
D) cannot be determined without additional information

Answer: B

A refrigerator operates between two temperatures, T1 and T2, where T2 is less than T1. What is the coefficient of performance of the refrigerator?
A) T1/(T2 – T1)
B) (T2 – T1)/T1
C) T2/(T1 – T2)
D) (T1 – T2)/T2

Answer: D

A gas is compressed adiabatically. What is the relationship between the pressure and volume of the gas?
A) PV = constant
B) P^2V = constant
C) PV^2 = constant
D) P^(1/3)V^(4/3) = constant

Answer: D

An ideal gas undergoes a reversible isothermal expansion. What is the change in entropy of the gas?
A) positive
B) negative
C) zero
D) cannot be determined without additional information

Answer: C

An ideal gas is compressed at constant temperature. What is the change in entropy of the gas?
A) positive
B) negative
C) zero
D) cannot be determined without additional information

Answer: B

A Carnot engine operates between two reservoirs at temperatures T1 and T2, where T2 is greater than T1. What is the maximum work that can be extracted from the engine?
A) (T2 – T1)/T2
B) (T2 – T1)/T1
C) T1/(T2 – T1)
D) T2/(T2 – T1)

Answer: D


Two reversible engines are working between the same temperature limits. The efficiency of one engine is 25% and the efficiency of the other engine is 50%. Which engine consumes more heat?
a) Both consume the same amount of heat
b) The engine with 25% efficiency consumes more heat
c) The engine with 50% efficiency consumes more heat
d) Cannot be determined with the given information

Answer: b) The engine with 25% efficiency consumes more heat Explanation: Efficiency of an engine is given by: Efficiency = (Work output / Heat input) For a reversible engine, the efficiency can also be given by: Efficiency = (T1 – T2) / T1 where T1 and T2 are the temperatures of the hot and cold reservoirs, respectively. Let the heat input for the engine with 25% efficiency be Q1 and the heat input for the engine with 50% efficiency be Q2. Efficiency of engine 1 = 25% = 0.25 = (Q1 – W) / Q1, where W is the work output. Solving for Q1, we get Q1 = (4/3)W Efficiency of engine 2 = 50% = 0.5 = (Q2 – W) / Q2, where W is the work output. Solving for Q2, we get Q2 = 2W We see that Q1 = (4/3)W is greater than Q2 = 2W. Hence, the engine with 25% efficiency consumes more heat. Answer: b) The engine with 25% efficiency consumes more heat

A Carnot engine operates between temperatures 500 K and 300 K. The work output of the engine is 100 J. What is the heat absorbed by the engine from the high-temperature reservoir?
a) 200 J
b) 300 J
c) 400 J
d) 500 J

Answer: c) 400 J Explanation: The efficiency of a Carnot engine is given by: Efficiency = (T1 – T2) / T1 where T1 and T2 are the temperatures of the hot and cold reservoirs, respectively. Let the heat absorbed by the engine be Q1 and the heat rejected be Q2. Efficiency of the Carnot engine = Work output / Heat input = (Q1 – Q2) / Q1 = 1 – (Q2 / Q1) Also, we have: Work output = Q1 – Q2 Given: Work output = 100 J, T1 = 500 K, and T2 = 300 K So, we can find the efficiency: 0.2 = 1 – (Q2 / Q1) Q1 – Q2 = 100 J Q1 / Q2 = 5/3 Solving these equations simultaneously, we get: Q1 = 400 J and Q2 = 100 J Therefore, the heat absorbed by the engine from the high-temperature reservoir is 400 J. Answer: c) 400 J

An ideal gas is taken through a process in which its pressure is increased from 1 atm to 2 atm and its volume is decreased from 5 L to 2 L. What is the change in internal energy of the gas?
a) 0 J
b) +3.9 J
c) -3.9 J
d) -15.6 J

Answer: a) 0 J Explanation: The change in internal energy of an ideal gas is given by: ΔU = Q – W


A Carnot engine is operating between two reservoirs having temperatures 900 K and 300 K. The efficiency of the engine will be:
a) 50%
b) 33.33%
c) 66.67%
d) 25%

Answer: b) 33.33%

An ideal gas is taken through the process ABC as shown in the figure. The amount of heat absorbed by the gas in the process AB is 50 J, and the work done by the gas in the process BC is 30 J. Find the heat absorbed in the process CA.
(Temperature-Volume graph with lines AB, BC and CA, with AB being an isobaric process, BC being an adiabatic process and CA being an isochoric process)

a) 20 J
b) 30 J
c) 10 J
d) 40 J

Answer: c) 10 J

A vessel of volume V contains a mixture of 1 mole of ideal gas A and 3 moles of ideal gas B at a temperature T. The molar specific heats of A and B are C_{vA} and C_{vB} respectively. The mixture is subjected to an isothermal process where 2 moles of gas B are transferred to another vessel of the same volume kept at the same temperature. The molar specific heat of the new mixture is:
a) 5/4 (C_{vA} + C_{vB})
b) (3C_{vA} + C_{vB})/4
c) C_{vA} + 3C_{vB}
d) (C_{vA} + 3C_{vB})/4

Answer: b) (3C_{vA} + C_{vB})/4

A gas mixture contains 3 moles of oxygen and 7 moles of nitrogen. The specific heats of oxygen and nitrogen are R/2 and (5/2)R respectively, where R is the gas constant. The mixture is subjected to a process where its pressure is increased from 1 atm to 5 atm. The amount of heat supplied to the gas during the process is Q. The value of Q/(nRT) for the process is:
a) 10/7
b) 5/14
c) 1/2
d) 2/7

Answer: a) 10/7

An ideal gas is taken through a cyclic process ABCDA as shown in the figure. The amount of heat absorbed by the gas in the process AB is Q1, and the amount of heat absorbed in the process BC is Q2. The amount of work done by the gas in the process CD is W, and the amount of work done in the process DA is W/2. If the net heat supplied to the gas during the cycle is zero, then the value of Q2/Q1 is:
(Temperature-Volume graph with lines AB, BC, CD, and DA forming a square-shaped cycle.)
a) 1/4
b) 1/2
c) 1
d) 2

Answer: a) 1/4


Which of the following statements is false regarding Carnot cycle?
a. All the processes are reversible.
b. Efficiency is the highest for any heat engine operating between two given temperatures.
c. It consists of two isothermal and two adiabatic processes.
d. The heat source and the heat sink are at different temperatures.

Answer: d. The heat source and the heat sink are at different temperatures.

Which of the following processes is not adiabatic?
a. Isothermal process
b. Isobaric process
c. Isochoric process
d. Polytropic process

Answer: a. Isothermal process

A heat engine takes in heat at 400 K and rejects heat at 300 K. What is the maximum efficiency of this engine?
a. 25%
b. 33.3%
c. 50%
d. 75%

Answer: b. 33.3%

Which of the following statements is true regarding the Second Law of Thermodynamics?
a. It states that heat always flows from a hotter body to a colder body.
b. It states that the entropy of a system always increases over time.
c. It states that the work done on a system is equal to the change in its internal energy.
d. It states that the efficiency of a heat engine cannot exceed 100%.

Answer: b. It states that the entropy of a system always increases over time.

An ideal gas undergoes a process in which its internal energy increases by 50 J and 100 J of heat is supplied to it. What is the work done by the gas during this process?
a. 50 J
b. 100 J
c. 150 J
d. It cannot be determined without additional information.

Answer: d. It cannot be determined without additional information.


Two identical containers A and B each contain an ideal gas at temperature T. The container A is placed on a hot plate while the container B is placed on a cold plate. The temperatures of the containers are measured as T1 and T2 respectively. Which of the following is true?
A) T1 > T2
B) T1 < T2

C) T1 = T2

D) Insufficient information

.

Answer: A) T1 > T2 Explanation: The container A is on a hot plate, so its temperature will increase, while the container B is on a cold plate, so its temperature will decrease. Since the two containers have the same initial temperature and are identical, the temperature of container A will be higher than that of container B

An ideal gas undergoes a process in which its volume increases by a factor of 2 and its pressure decreases by a factor of 4. The change in internal energy of the gas is:
A) Zero
B) Positive
C) Negative
D) Insufficient information

Answer: A) Zero Explanation: The change in internal energy of an ideal gas only depends on its temperature, which is not given in this problem. However, since the gas undergoes an isentropic process (PV^γ = constant, where γ is the ratio of specific heats), the change in internal energy is zero.

A heat engine operates between two reservoirs at temperatures 500 K and 300 K. The efficiency of the engine is:
A) 40%
B) 50%
C) 60%
D) Insufficient information
.

Answer: A) 40% Explanation: The efficiency of a heat engine is given by (1 – T2/T1), where T1 and T2 are the temperatures of the hot and cold reservoirs, respectively. Plugging in the values given in the problem, we get: (1 – 300/500) = 0.4 or 40%

An ideal gas undergoes an isothermal process. Which of the following statements is true?
A) The change in internal energy is zero.
B) The change in entropy is zero.
C) The heat absorbed by the gas is equal to the work done by the gas.
D) All of the above.

Answer: C) The heat absorbed by the gas is equal to the work done by the gas. Explanation: In an isothermal process, the temperature remains constant, so the change in internal energy is zero. However, the change in entropy is not necessarily zero. The heat absorbed by the gas is equal to the work done by the gas, as per the first law of thermodynamics.

An ideal gas is compressed adiabatically. Which of the following statements is true?
A) The temperature of the gas increases.
B) The temperature of the gas decreases.
C) The temperature of the gas remains constant.
D) Insufficient information.

Answer: B) The temperature of the gas decreases. Explanation: In an adiabatic process, no heat is exchanged between the gas and its surroundings. Therefore, the first law of thermodynamics reduces to: dU = -W, where dU is the change in internal energy of the gas and W is the work done on the gas. Since the gas is being compressed, W is negative, so dU is negative as well. Since the internal energy of an ideal gas is proportional to its temperature, the temperature of the gas must decrease.

We hope there NEET MCQ of Class 11 Thermodynamics will help you to score an excellent rank in NEET-UG. If you have any queries feel free to write in the comments section. We at Study Rate are always ready to serve our students.

Sneha

Master's in Biology, Skilled in vocational training. Strong Analytical and creative knowledge.

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