JEE Main 2026 — Physics Thermodynamics

2026integermedium

A certain gas is isothermally compressed to $\left(\dfrac{1}{3}\right)^{rd}$ of its initial volume ( $V_o = 3$ litre) by applying required pressure. If the bulk modulus of the gas is $3 \times 10^5$ N/m $^2$ , the magnitude of work done on the gas is _______ J.

Official previous-year question

Held on 6 Apr 2026 · Verified 6 Jul 2026.

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Solution

For an isothermal process, the isothermal bulk modulus equals the gas pressure:

$B_T = -V\dfrac{dP}{dV}$

Since $PV = \text{constant}$ , differentiating gives $P\,dV + V\,dP = 0$ , so $\dfrac{dP}{dV} = -\dfrac{P}{V}$ , which yields $B_T = P$ .

Therefore, the initial pressure is:

$P_o = 3 \times 10^5\text{ N/m}^2$

Volumes:

$V_o = 3\text{ L} = 3 \times 10^{-3}\text{ m}^3$

$V_f = \dfrac{1}{3}V_o = 10^{-3}\text{ m}^3$

Work done on the gas during an isothermal compression:

$W_{\text{on}} = -W_{\text{by}} = -P_o V_o \ln\left(\dfrac{V_f}{V_o}\right) = P_o V_o \ln\left(\dfrac{V_o}{V_f}\right)$

$W_{\text{on}} = (3 \times 10^5)(3 \times 10^{-3})\ln 3$

$W_{\text{on}} = 900 \ln 3\text{ J}$

Using $\ln 3 \approx 1.0986$ :

$W_{\text{on}} \approx 900 \times 1.0986 \approx 988.74\text{ J}$

Rounded to the nearest integer:

$W_{\text{on}} \approx 989\text{ J}$

Hence, the work done on the gas is approximately $989\text{ J}$ .

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JEE Main 2026 — Physics Thermodynamics

2026integermedium

Official previous-year question

Held on 6 Apr 2026 · Verified 6 Jul 2026.

Did you get this right?

Solution

For an isothermal process, the isothermal bulk modulus equals the gas pressure:

$B_T = -V\dfrac{dP}{dV}$

Since $PV = \text{constant}$ , differentiating gives $P\,dV + V\,dP = 0$ , so $\dfrac{dP}{dV} = -\dfrac{P}{V}$ , which yields $B_T = P$ .

Therefore, the initial pressure is:

$P_o = 3 \times 10^5\text{ N/m}^2$

Volumes:

$V_o = 3\text{ L} = 3 \times 10^{-3}\text{ m}^3$

$V_f = \dfrac{1}{3}V_o = 10^{-3}\text{ m}^3$

Work done on the gas during an isothermal compression:

$W_{\text{on}} = -W_{\text{by}} = -P_o V_o \ln\left(\dfrac{V_f}{V_o}\right) = P_o V_o \ln\left(\dfrac{V_o}{V_f}\right)$

$W_{\text{on}} = (3 \times 10^5)(3 \times 10^{-3})\ln 3$

$W_{\text{on}} = 900 \ln 3\text{ J}$

Using $\ln 3 \approx 1.0986$ :

$W_{\text{on}} \approx 900 \times 1.0986 \approx 988.74\text{ J}$

Rounded to the nearest integer:

$W_{\text{on}} \approx 989\text{ J}$

Hence, the work done on the gas is approximately $989\text{ J}$ .

Your note

JEE Main Thermodynamics in other years

Keep practicing Thermodynamics

Work through every JEE Main Thermodynamics PYQ, year by year.

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JEE Main 2026 — Physics Thermodynamics

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Solution

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JEE Main Thermodynamics in other years

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JEE Main 2026 — Physics Thermodynamics

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