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Thermodynamics — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Thermodynamics MCQs with step-by-step solutions covering First Law of Thermodynamics, Thermodynamic Systems & Zeroth Law, Isothermal & Adiabatic Processes, Thermodynamic Efficiency & Carnot Cycle, Heat Engines & Refrigeration, Entropy & Second Law. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — Thermodynamic Systems & Zeroth Law · easy · theory
The zeroth law of thermodynamics states that if two systems are each in thermal equilibrium with a third system, then:
A. They are in thermal equilibrium with each other ✓ Correct
B. They must have equal volumes
C. Heat flows spontaneously between them
D. They must have equal internal energies
Solution: This law establishes temperature as a valid physical property: bodies in mutual thermal equilibrium share a common temperature, which is what a thermometer measures.
Q2 — Thermodynamic Systems & Zeroth Law · easy · theory
The work done by a gas expanding at constant pressure $P$ from volume $V_1$ to $V_2$ is:
A. $P(V_2 - V_1)$ ✓ Correct
B. $\dfrac{1}{2}P(V_2 - V_1)$
C. $\dfrac{P}{V_2 - V_1}$
D. $P V_1 V_2$
Solution: For an isobaric process $W = \int P\,dV = P\int dV = P\,\Delta V = P(V_2 - V_1)$.
Q3 — Thermodynamic Systems & Zeroth Law · easy · theory
On a $P$-$V$ indicator diagram, the work done by a gas during a process is represented by:
A. The intercept on the pressure axis
B. The area under the curve ✓ Correct
C. The length of the curve
D. The slope of the curve
Solution: Since $W = \int P\,dV$, the work is the area between the process curve and the volume axis.
Q4 — Thermodynamic Systems & Zeroth Law · easy · theory
For a gas taken around a complete cyclic process, the change in internal energy is:
A. Equal to the heat absorbed
B. Always positive
C. Equal to the work done
D. Zero ✓ Correct
Solution: Internal energy is a state function, so returning to the initial state means $\Delta U = 0$, whatever path the cycle follows.
Q5 — Thermodynamic Systems & Zeroth Law · easy · theory
In a cyclic process, the net heat absorbed by the system is equal to:
A. Zero
B. The change in temperature
C. The net work done by the system ✓ Correct
D. The change in internal energy
Solution: The first law gives $Q = \Delta U + W$, and with $\Delta U = 0$ over a cycle, $Q = W$.
Q6 — Thermodynamic Systems & Zeroth Law · easy · theory
In an isochoric (constant volume) process, the work done by the gas is:
A. Equal to $nR\Delta T$
B. Equal to $P\Delta V$
C. Zero ✓ Correct
D. Equal to the heat supplied
Solution: With $\Delta V = 0$, $W = P\,\Delta V = 0$. All the heat supplied therefore goes into raising the internal energy.
Q7 — Thermodynamic Systems & Zeroth Law · easy · theory
Two bodies are said to be in thermal equilibrium when:
A. They have equal heat capacities
B. There is no net flow of heat between them ✓ Correct
C. They have equal masses
D. They have equal internal energies
Solution: Thermal equilibrium means both are at the same temperature, so heat flows equally in both directions and there is no net transfer.
Q8 — Thermodynamic Systems & Zeroth Law · easy · theory
A wall that permits no exchange of heat between a system and its surroundings is called:
A. A diathermic wall
B. An isothermal wall
C. A conducting wall
D. An adiabatic wall ✓ Correct
Solution: An adiabatic (perfectly insulating) wall blocks heat flow, whereas a diathermic wall conducts heat freely and allows the two sides to reach a common temperature.
Q9 — Thermodynamic Systems & Zeroth Law · easy · theory
By the usual sign convention in thermodynamics, work done by the system and heat absorbed by the system are taken respectively as:
A. Negative and positive
B. Positive and positive ✓ Correct
C. Positive and negative
D. Negative and negative
Solution: Heat added to the system and work done by the system are both counted positive, which is why the first law is written $Q = \Delta U + W$.
Q10 — First Law of Thermodynamics · easy · theory
The first law of thermodynamics is expressed as:
A. $\Delta Q = \Delta U - \Delta W$
B. $\Delta Q = \Delta U + \Delta W$ ✓ Correct
C. $\Delta U = \Delta Q + \Delta W$
D. $\Delta W = \Delta Q + \Delta U$
Solution: The heat supplied to a system is partly used to raise its internal energy and partly spent as work done by the system against the surroundings.
Q11 — First Law of Thermodynamics · easy · theory
The first law of thermodynamics is essentially a statement of the conservation of:
A. Mass
B. Momentum
C. Entropy
D. Energy ✓ Correct
Solution: It asserts that energy supplied as heat is fully accounted for by the change in internal energy plus the work done — none is created or destroyed.
Q12 — First Law of Thermodynamics · easy · theory
During an isothermal expansion of an ideal gas, the change in its internal energy is:
A. Positive
B. Zero ✓ Correct
C. Negative
D. Equal to the work done
Solution: The internal energy of an ideal gas depends only on temperature. Since $\Delta T = 0$ in an isothermal process, $\Delta U = 0$ and hence $Q = W$.
Q13 — First Law of Thermodynamics · easy · numerical
A gas absorbs $100\text{ J}$ of heat and does $40\text{ J}$ of work on its surroundings. The increase in its internal energy is:
A. $140\text{ J}$
B. $40\text{ J}$
C. $100\text{ J}$
D. $60\text{ J}$ ✓ Correct
Solution: $\Delta U = Q - W = 100 - 40 = 60\text{ J}$.
Q14 — First Law of Thermodynamics · easy · numerical
An ideal gas absorbs $500\text{ J}$ of heat while its internal energy increases by $300\text{ J}$. The work done by the gas is:
A. $200\text{ J}$ ✓ Correct
B. $300\text{ J}$
C. $500\text{ J}$
D. $800\text{ J}$
Solution: $W = Q - \Delta U = 500 - 300 = 200\text{ J}$.
Q15 — First Law of Thermodynamics · easy · theory
In an isochoric process, the heat supplied to the system is equal to:
A. The work done by the system
B. The increase in its internal energy ✓ Correct
C. The work done on the system
D. Zero
Solution: Since $W = 0$ at constant volume, $Q = \Delta U$ entirely.
Q16 — First Law of Thermodynamics · easy · theory
The internal energy of a given mass of an ideal gas depends only on its:
A. Pressure
B. Volume
C. Temperature ✓ Correct
D. Density
Solution: For an ideal gas there is no intermolecular potential energy, so $U$ is purely the molecular kinetic energy, which is a function of temperature alone.
Q17 — First Law of Thermodynamics · easy · theory
For an isobaric process, the first law may be written as:
A. $Q = P\Delta V$
B. $Q = \Delta U + P\Delta V$ ✓ Correct
C. $Q = \Delta U$
D. $Q = 0$
Solution: At constant pressure the work term is $P\,\Delta V$, so the heat supplied both raises $U$ and performs expansion work.
Q18 — First Law of Thermodynamics · easy · theory
For an ideal gas undergoing an isothermal process, the heat absorbed is:
A. Entirely converted into work done by the gas ✓ Correct
B. Zero
C. Entirely used to raise the internal energy
D. Equally shared between work and internal energy
Solution: With $\Delta U = 0$, the first law reduces to $Q = W$, so all the heat taken in appears as external work.
Q19 — First Law of Thermodynamics · easy · theory
Heat supplied to a gas at constant volume raises its temperature by $\Delta T$. The heat required is:
A. $n R \Delta T$
B. $n (C_p + C_v) \Delta T$
C. $n C_p \Delta T$
D. $n C_v \Delta T$ ✓ Correct
Solution: At constant volume the relevant molar heat capacity is $C_v$, so $Q = \Delta U = n C_v \Delta T$.
Q20 — Isothermal & Adiabatic Processes · easy · theory
For an adiabatic process involving an ideal gas, the pressure and volume are related by:
A. $\dfrac{P}{V^\gamma} = \text{constant}$
B. $TV^\gamma = \text{constant}$
C. $PV^\gamma = \text{constant}$ ✓ Correct
D. $PV = \text{constant}$
Solution: Combining the first law with $Q = 0$ and the ideal gas equation gives $PV^\gamma = $ constant, where $\gamma = \dfrac{C_p}{C_v}$.
Q21 — Isothermal & Adiabatic Processes · easy · theory
For an isothermal process involving a fixed mass of an ideal gas:
A. $\dfrac{P}{T} = \text{constant}$
B. $\dfrac{V}{T} = \text{constant}$
C. $PV = \text{constant}$ ✓ Correct
D. $PV^\gamma = \text{constant}$
Solution: At constant $T$, $PV = nRT$ makes the product $PV$ a constant — this is Boyle's law.
Q22 — Isothermal & Adiabatic Processes · easy · theory
When an ideal gas expands adiabatically, its temperature:
A. Rises, because no heat escapes
B. Remains constant
C. Falls, because the gas does work at the cost of its internal energy ✓ Correct
D. Rises and then falls
Solution: With $Q = 0$ the work of expansion can only be paid for out of internal energy, so $U$ and therefore $T$ decrease.
Q23 — Isothermal & Adiabatic Processes · easy · theory
The sudden bursting of a cycle tyre is an example of a process that is very nearly:
A. Isobaric
B. Isochoric
C. Isothermal
D. Adiabatic ✓ Correct
Solution: The expansion is far too rapid for any appreciable heat exchange, so $Q \approx 0$ and the escaping air cools noticeably.
Q24 — Thermodynamic Efficiency & Carnot Cycle · easy · theory
The efficiency of a Carnot engine working between source temperature $T_H$ and sink temperature $T_C$ is:
A. $1 - \dfrac{T_H}{T_C}$
B. $1 - \dfrac{T_C}{T_H}$ ✓ Correct
C. $\dfrac{T_H - T_C}{T_C}$
D. $\dfrac{T_C}{T_H}$
Solution: For a Carnot cycle $\dfrac{Q_C}{Q_H} = \dfrac{T_C}{T_H}$, so $\eta = 1 - \dfrac{Q_C}{Q_H} = 1 - \dfrac{T_C}{T_H}$, with temperatures in kelvin.
Q25 — Thermodynamic Efficiency & Carnot Cycle · easy · numerical
The efficiency of a Carnot engine operating between $600\text{ K}$ and $300\text{ K}$ is:
A. $25\%$
B. $50\%$ ✓ Correct
C. $33.3\%$
D. $75\%$
Solution: $\eta = 1 - \dfrac{300}{600} = 1 - 0.5 = 0.50 = 50\%$.
Q26 — Thermodynamic Efficiency & Carnot Cycle · easy · numerical
An ideal Carnot heat engine operates between $500\text{ K}$ and $300\text{ K}$. Its efficiency is:
A. $40\%$ ✓ Correct
B. $50\%$
C. $60\%$
D. $25\%$
Solution: $\eta = 1 - \dfrac{300}{500} = 1 - 0.6 = 0.40 = 40\%$.
Q27 — Thermodynamic Efficiency & Carnot Cycle · easy · theory
The Carnot cycle consists of:
A. Two isobaric and two isochoric processes
B. Two isothermal and two adiabatic processes ✓ Correct
C. Two isothermal and two isochoric processes
D. Four adiabatic processes
Solution: Heat is taken in during isothermal expansion at $T_H$ and rejected during isothermal compression at $T_C$, with two adiabatic steps connecting the two isotherms.
Q28 — Thermodynamic Efficiency & Carnot Cycle · easy · theory
The efficiency of a heat engine in terms of the heat absorbed $Q_H$ and the work done $W$ is:
A. $\dfrac{Q_H}{W}$
B. $\dfrac{Q_H - W}{Q_H}$
C. $\dfrac{W}{Q_C}$
D. $\dfrac{W}{Q_H}$ ✓ Correct
Solution: Efficiency is the useful output divided by the input: $\eta = \dfrac{W}{Q_H} = 1 - \dfrac{Q_C}{Q_H}$.
Q29 — Thermodynamic Efficiency & Carnot Cycle · easy · numerical
A Carnot engine works between a source at $400\text{ K}$ and a sink at $300\text{ K}$. Its efficiency is:
A. $25\%$ ✓ Correct
B. $50\%$
C. $75\%$
D. $33\%$
Solution: $\eta = 1 - \dfrac{300}{400} = 1 - 0.75 = 0.25 = 25\%$.
Q30 — Heat Engines & Refrigeration · easy · theory
A heat engine is a device that converts:
A. Mechanical work entirely into heat
B. Heat from a cold body into a hot body without work
C. Electrical energy into heat
D. Heat energy into mechanical work, operating in a cycle ✓ Correct
Solution: A heat engine takes in heat from a hot reservoir, converts part of it into work, and rejects the remainder to a cold reservoir, repeating the cycle indefinitely.