First Law of Thermodynamics — MH-CET Physics MCQs with Solutions
Free MH-CET Physics First Law of Thermodynamics MCQs with step-by-step solutions (34 questions). Part of Thermodynamics. Practise online on Prepizo — no login needed.
▶ Practise First Law of Thermodynamics online (free)
Questions with solutions
Q1 — 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.
Q2 — 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.
Q3 — 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 $Q = W$.
Q4 — 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: $\Delta U = Q - W = 100 - 40 = 60\text{ J}$.
Q5 — 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: $W = Q - \Delta U = 500 - 300 = 200\text{ J}$.
Q6 — First Law of Thermodynamics · medium · theory
A gas is compressed adiabatically. Its internal energy:
A. Decreases, since work is done on the gas
B. Increases, since work is done on the gas ✓ Correct
C. Remains unchanged
D. First increases then decreases
Solution: With $Q = 0$, the first law gives $\Delta U = -W$. In compression $W$ (work by the gas) is negative, so $\Delta U$ is positive and the gas heats up.
Q7 — 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.
Q8 — First Law of Thermodynamics · hard · numerical
A gas is compressed such that $200\text{ J}$ of work is done on it while it releases $150\text{ J}$ of heat. The change in its internal energy is:
A. $+50\text{ J}$ ✓ Correct
B. $+350\text{ J}$
C. $-350\text{ J}$
D. $-50\text{ J}$
Solution: Here $Q = -150\text{ J}$ (released) and $W = -200\text{ J}$ (done on the gas). So $\Delta U = Q - W = -150 - (-200) = +50\text{ J}$.
Q9 — 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.
Q10 — First Law of Thermodynamics · hard · theory
In the free expansion of an ideal gas into an evacuated insulated chamber:
A. The internal energy increases
B. $Q = 0$ but the gas cools appreciably
C. $Q = 0$, $W = 0$ and hence the temperature remains unchanged ✓ Correct
D. Work is done by the gas against the vacuum
Solution: There is nothing to push against, so $W = 0$; the chamber is insulated, so $Q = 0$. Hence $\Delta U = 0$ and, for an ideal gas, the temperature is unchanged.
Q11 — 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.
Q12 — First Law of Thermodynamics · medium · theory
A limitation of the first law of thermodynamics is that it:
A. Does not indicate the direction in which a process will occur ✓ Correct
B. Does not conserve energy
C. Cannot be applied to cyclic processes
D. Applies only to ideal gases
Solution: The first law would be equally satisfied if heat flowed spontaneously from a cold body to a hot one. It is the second law that forbids this and fixes the direction.
Q13 — First Law of Thermodynamics · medium · numerical
A system undergoes a process in which it absorbs $300\text{ J}$ of heat and its internal energy decreases by $100\text{ J}$. The work done by the system is:
A. $100\text{ J}$
B. $200\text{ J}$
C. $400\text{ J}$ ✓ Correct
D. $300\text{ J}$
Solution: $W = Q - \Delta U = 300 - (-100) = 400\text{ J}$.
Q14 — 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.
Q15 — First Law of Thermodynamics · hard · theory
The internal energy of a real gas, unlike that of an ideal gas, also depends on volume because:
A. Its molecules have no kinetic energy
B. It obeys Boyle's law exactly
C. Its temperature is always constant
D. Intermolecular forces contribute potential energy ✓ Correct
Solution: Changing the volume alters the average separation of molecules and hence the intermolecular potential energy, which an ideal gas is assumed not to possess.
Q16 — 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 $Q = \Delta U = n C_v \Delta T$.
Q17 — First Law of Thermodynamics · medium · numerical
In a thermodynamic process, $150\text{ J}$ of work is done on a gas and its internal energy rises by $150\text{ J}$. The heat exchanged is:
A. Zero, so the process is adiabatic ✓ Correct
B. $150\text{ J}$ absorbed
C. $300\text{ J}$ absorbed
D. $300\text{ J}$ released
Solution: With $W = -150\text{ J}$ and $\Delta U = +150\text{ J}$, $Q = \Delta U + W = 150 - 150 = 0$. No heat is exchanged, so the process is adiabatic.
Q18 — First Law of Thermodynamics · easy · numerical
A system absorbs $250\text{ J}$ of heat and does $100\text{ J}$ of work. The increase in its internal energy is:
A. $250\text{ J}$
B. $350\text{ J}$
C. $150\text{ J}$ ✓ Correct
D. $100\text{ J}$
Solution: $\Delta U = Q - W = 250 - 100 = 150\text{ J}$.
Q19 — First Law of Thermodynamics · medium · numerical
The internal energy of a gas decreases by $80\text{ J}$ while it does $120\text{ J}$ of work. The heat absorbed is:
A. $200\text{ J}$
B. $-40\text{ J}$
C. $40\text{ J}$ ✓ Correct
D. $-200\text{ J}$
Solution: $Q = \Delta U + W = -80 + 120 = 40\text{ J}$.
Q20 — First Law of Thermodynamics · hard · numerical
A gas is compressed with $300\text{ J}$ of work done on it while it rejects $200\text{ J}$ of heat. The change in its internal energy is:
A. $+500\text{ J}$
B. $+100\text{ J}$ ✓ Correct
C. $-100\text{ J}$
D. $-500\text{ J}$
Solution: Here $Q = -200\text{ J}$ and $W = -300\text{ J}$, so $\Delta U = Q - W = -200 + 300 = +100\text{ J}$.
Q21 — First Law of Thermodynamics · hard · numerical
Heat of $600\text{ J}$ is supplied at constant volume to $2\text{ moles}$ of a monatomic ideal gas. The rise in its temperature is ($R = 8.314$):
A. $24\text{ K}$ ✓ Correct
B. $12\text{ K}$
C. $36\text{ K}$
D. $48\text{ K}$
Solution: $Q = nC_v\Delta T = 2 \times 12.47 \times \Delta T = 24.94\Delta T$, so $\Delta T = \dfrac{600}{24.94} \approx 24\text{ K}$.
Q22 — First Law of Thermodynamics · easy · numerical
A system absorbs $400\text{ J}$ of heat and does $250\text{ J}$ of work. The increase in its internal energy is:
A. $250\text{ J}$
B. $150\text{ J}$ ✓ Correct
C. $400\text{ J}$
D. $650\text{ J}$
Solution: $\Delta U = Q - W = 400 - 250 = 150\text{ J}$.
Q23 — First Law of Thermodynamics · easy · numerical
A gas absorbs $1000\text{ J}$ of heat while its internal energy rises by $600\text{ J}$. The work done by the gas is:
A. $400\text{ J}$ ✓ Correct
B. $600\text{ J}$
C. $1000\text{ J}$
D. $1600\text{ J}$
Solution: $W = Q - \Delta U = 1000 - 600 = 400\text{ J}$.
Q24 — First Law of Thermodynamics · easy · numerical
The internal energy of a gas rises by $200\text{ J}$ while it does $300\text{ J}$ of work. The heat absorbed is:
A. $-100\text{ J}$
B. $600\text{ J}$
C. $100\text{ J}$
D. $500\text{ J}$ ✓ Correct
Solution: $Q = \Delta U + W = 200 + 300 = 500\text{ J}$.
Q25 — First Law of Thermodynamics · hard · numerical
A gas is compressed with $400\text{ J}$ of work done on it while it releases $250\text{ J}$ of heat. The change in its internal energy is:
A. $+150\text{ J}$ ✓ Correct
B. $-650\text{ J}$
C. $+650\text{ J}$
D. $-150\text{ J}$
Solution: Here $Q = -250\text{ J}$ and $W = -400\text{ J}$, so $\Delta U = Q - W = -250 + 400 = +150\text{ J}$.
Q26 — First Law of Thermodynamics · hard · numerical
Heat of $300\text{ J}$ is given at constant volume to $3\text{ moles}$ of a monatomic ideal gas. The rise in temperature is ($R = 8.314$):
A. $12.0\text{ K}$
B. $24.0\text{ K}$
C. $4.8\text{ K}$
D. $8.0\text{ K}$ ✓ Correct
Solution: $Q = nC_v\Delta T = 3 \times 12.47 \times \Delta T = 37.41\Delta T$, so $\Delta T = \dfrac{300}{37.41} \approx 8.0\text{ K}$.
Q27 — First Law of Thermodynamics · medium · numerical
An ideal gas absorbs $800\text{ J}$ of heat during an isothermal expansion. The work done by the gas is:
A. Zero
B. $1600\text{ J}$
C. $400\text{ J}$
D. $800\text{ J}$ ✓ Correct
Solution: In an isothermal process $\Delta U = 0$, so the first law gives $W = Q = 800\text{ J}$.
Q28 — First Law of Thermodynamics · medium · numerical
A gas does $500\text{ J}$ of work during an adiabatic expansion. The change in its internal energy is:
A. $-1000\text{ J}$
B. Zero
C. $-500\text{ J}$ ✓ Correct
D. $+500\text{ J}$
Solution: With $Q = 0$, $\Delta U = -W = -500\text{ J}$: the work is done at the cost of the internal energy, so the gas cools.
Q29 — First Law of Thermodynamics · medium · numerical
A gas absorbs a net $250\text{ J}$ of heat over one complete cycle. The net work it does is:
A. $125\text{ J}$
B. $250\text{ J}$ ✓ Correct
C. $500\text{ J}$
D. Zero
Solution: Over a cycle $\Delta U = 0$, so $W = Q = 250\text{ J}$.
Q30 — First Law of Thermodynamics · hard · numerical
A gas releases $600\text{ J}$ of heat while $400\text{ J}$ of work is done on it. The change in its internal energy is:
A. $-200\text{ J}$ ✓ Correct
B. $-1000\text{ J}$
C. $+200\text{ J}$
D. $+1000\text{ J}$
Solution: Here $Q = -600\text{ J}$ and $W = -400\text{ J}$, so $\Delta U = -600 + 400 = -200\text{ J}$.