Match List I with List II : List I List II A Isothermal Process I Work done by the gas decreases internal energy B Adiabatic Process II No change in internal energy C Isochoric Process III The heat absorbed goes partly to increase internal energy and partly to do work D Isobaric Process IV No work is done on or by the gas Choose the correct answer from the options given below :

(A) Change in internal energy is expressed as$\Delta U=n{C}_{v}\Delta T$, here, $\Delta T$ is change in temperature. Since, in an isothermal process temperature remains constant, thus, $\Delta U=0$ $A\rightarrow II$ (B) In an adiabatic process, heat transfer, $Q=0$. So, from first law of thermodynamics, $Q=\Delta U+W$ $\Delta U=-W$ Since, work done by gas is positive, thus, $\Delta U$ is negative $B\rightarrow I$ (C) In an isochoric process, volume remains constant, so work done by or on the gas $W=P\Delta V=0$ $C\rightarrow IV$ (D) In an isobaric process, pressure remains constant, so work done$W=P\Delta V\neq 0$ and change in internal energy $\Delta U=n{C}_{v}\Delta T\neq 0$ Thus, heat absorbed goes partly to increase internal energy and partly do work. $D\rightarrow III$

JEE Main 2023 — Physics Thermodynamics

2023mcqeasy

Match List I with List II :

	List I		List II
A	Isothermal Process	I	Work done by the gas decreases internal energy
B	Adiabatic Process	II	No change in internal energy
C	Isochoric Process	III	The heat absorbed goes partly to increase internal energy and partly to do work
D	Isobaric Process	IV	No work is done on or by the gas

Choose the correct answer from the options given below :

Official previous-year question

Held on 25 Jan 2023 · Verified 6 Jul 2026.

Options

A
A-II, B-I, C-III, D-IV
B
A-II, B-I, C-IV, D-III
C
A-I, B-II, C-IV, D-III
D
A-I, B-II, C-III, D-IV

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Solution

(A) Change in internal energy is expressed as $\Delta U=n{C}_{v}\Delta T$ , here, $\Delta T$ is change in temperature.

Since, in an isothermal process temperature remains constant, thus, $\Delta U=0$

$A\rightarrow II$

(B) In an adiabatic process, heat transfer, $Q=0$ .

So, from first law of thermodynamics,

$Q=\Delta U+W$

$\Delta U=-W$

Since, work done by gas is positive, thus, $\Delta U$ is negative

$B\rightarrow I$

$C\rightarrow IV$

(D) In an isobaric process, pressure remains constant, so work done $W=P\Delta V\neq 0$ and change in internal energy $\Delta U=n{C}_{v}\Delta T\neq 0$

Thus, heat absorbed goes partly to increase internal energy and partly do work.

$D\rightarrow III$

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Solution

(A) Change in internal energy is expressed as $\Delta U=n{C}_{v}\Delta T$ , here, $\Delta T$ is change in temperature.

Since, in an isothermal process temperature remains constant, thus, $\Delta U=0$

$A\rightarrow II$

(B) In an adiabatic process, heat transfer, $Q=0$ .

So, from first law of thermodynamics,

$Q=\Delta U+W$

$\Delta U=-W$

Since, work done by gas is positive, thus, $\Delta U$ is negative

$B\rightarrow I$

$C\rightarrow IV$

(D) In an isobaric process, pressure remains constant, so work done $W=P\Delta V\neq 0$ and change in internal energy $\Delta U=n{C}_{v}\Delta T\neq 0$

Thus, heat absorbed goes partly to increase internal energy and partly do work.

$D\rightarrow III$

JEE Main 2023 — Physics Thermodynamics

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JEE Main 2023 — Physics Thermodynamics

Options

Did you get this right?

Solution

Your note

JEE Main Thermodynamics in other years

More JEE Main Thermodynamics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Optics

Waves & Oscillations

Keep practicing Thermodynamics