JEE Main 2022 — Physics Thermodynamics

2022mcqeasy

A sample of an ideal gas is taken through the cyclic process $ABCA$ as shown in figure. It absorbs, $40J$ of heat during the part $AB$ , no heat during $BC$ and rejects $60J$ of heat during $CA$ . A work of $50J$ is done on the gas during the part $BC$ . The internal energy of the gas at $A$ is $1560J$ . The work done by the gas during the part $CA$ is

JEE Main 2022 Physics, Thermodynamics — question figure

Official previous-year question

Held on 28 Jun 2022 · Verified 6 Jul 2026.

Options

A
$20J$
B
$30J$
C
$-30J$
D
$-60J$

Did you get this right?

Solution

In the part $A\rightarrow B$ , the volume remains constant. Thus, the work done by the gas is zero. It is given that the heat absorbed by the gas is $40J$ . From first law of thermodynamics, $Q=\Delta U+W$

$\Rightarrow 40=\Delta U+0\Rightarrow \Delta U=40J$

or $\Delta U={U}_{B}-{U}_{A}=40J$

The increase in internal energy from $A$ to $B$ is $40J$ . The internal energy is $1560J$ at $A$ , then

${U}_{B}-1560=40J\Rightarrow {U}_{B}=1600J$

In the part $B\rightarrow C$ , the work done by the gas is $W=50J$ and no heat is given to the system. Again using $Q=\Delta U+W$

$0=\Delta U-50$ or $\Delta U={U}_{C}-{U}_{B}=50J$

${U}_{C}-1600=50J\Rightarrow {U}_{C}=1650J$

The change in internal energy from $C\rightarrow A$ ,

$\Delta U=1560-1650=-90J$

The heat given to the system is $Q=-60J$

Using $Q=\Delta U+W$ , we get $W=Q-\Delta U=-60-(-90)=30J$

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Solution

$\Rightarrow 40=\Delta U+0\Rightarrow \Delta U=40J$

or $\Delta U={U}_{B}-{U}_{A}=40J$

The increase in internal energy from $A$ to $B$ is $40J$ . The internal energy is $1560J$ at $A$ , then

${U}_{B}-1560=40J\Rightarrow {U}_{B}=1600J$

In the part $B\rightarrow C$ , the work done by the gas is $W=50J$ and no heat is given to the system. Again using $Q=\Delta U+W$

$0=\Delta U-50$ or $\Delta U={U}_{C}-{U}_{B}=50J$

${U}_{C}-1600=50J\Rightarrow {U}_{C}=1650J$

The change in internal energy from $C\rightarrow A$ ,

$\Delta U=1560-1650=-90J$

The heat given to the system is $Q=-60J$

Using $Q=\Delta U+W$ , we get $W=Q-\Delta U=-60-(-90)=30J$

JEE Main 2022 — Physics Thermodynamics

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JEE Main 2022 — 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

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