JEE Main 2026 — Physics Thermodynamics

2026mcqmedium

One mole of diatomic gas having rotational modes only is kept in a cylinder with a piston system. The cross-section area of the cylinder is $4$ cm $^2$ . The gas is heated slowly to raise the temperature by $1.2\,^\circ$ C during which the piston moves by $25$ mm. The amount of heat supplied to the gas is ________ J. (Atmospheric pressure $=100$ kPa, $R=8.3$ J/mol·K) (Neglect mass of the piston)

Official previous-year question

Held on 8 Apr 2026 · Verified 6 Jul 2026.

Options

A
$24.8$
B
$25$
C
$15.04$
D
$29.98$

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Solution

Using the first law of thermodynamics:

$Q = \Delta U + W$

Since the gas is diatomic with rotational modes only, the degrees of freedom are $f = 5$ , so:

$C_V = \dfrac{5}{2}R$

For $n = 1\text{ mole}$ and $\Delta T = 1.2\text{ K}$ :

$\Delta U = nC_V\Delta T = 1 \times \dfrac{5}{2} \times 8.3 \times 1.2 = 24.9\text{ J}$

Work done by the gas:

$W = P\,\Delta V$

Cross-sectional area of the piston:

$A = 4\text{ cm}^2 = 4 \times 10^{-4}\text{ m}^2$

Piston displacement:

$x = 25\text{ mm} = 2.5 \times 10^{-2}\text{ m}$

Change in volume:

$\Delta V = A \cdot x = (4 \times 10^{-4})(2.5 \times 10^{-2}) = 1 \times 10^{-5}\text{ m}^3$

With $P = 100\text{ kPa} = 10^5\text{ Pa}$ :

$W = (10^5)(1 \times 10^{-5}) = 1\text{ J}$

Therefore, the heat supplied is:

$Q = \Delta U + W = 24.9 + 1 = 25.9\text{ J}$

Hence, the heat supplied to the gas is $25.9\text{ J}$ .

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Solution

Using the first law of thermodynamics:

$Q = \Delta U + W$

Since the gas is diatomic with rotational modes only, the degrees of freedom are $f = 5$ , so:

$C_V = \dfrac{5}{2}R$

For $n = 1\text{ mole}$ and $\Delta T = 1.2\text{ K}$ :

$\Delta U = nC_V\Delta T = 1 \times \dfrac{5}{2} \times 8.3 \times 1.2 = 24.9\text{ J}$

Work done by the gas:

$W = P\,\Delta V$

Cross-sectional area of the piston:

$A = 4\text{ cm}^2 = 4 \times 10^{-4}\text{ m}^2$

Piston displacement:

$x = 25\text{ mm} = 2.5 \times 10^{-2}\text{ m}$

Change in volume:

$\Delta V = A \cdot x = (4 \times 10^{-4})(2.5 \times 10^{-2}) = 1 \times 10^{-5}\text{ m}^3$

With $P = 100\text{ kPa} = 10^5\text{ Pa}$ :

$W = (10^5)(1 \times 10^{-5}) = 1\text{ J}$

Therefore, the heat supplied is:

$Q = \Delta U + W = 24.9 + 1 = 25.9\text{ J}$

Hence, the heat supplied to the gas is $25.9\text{ J}$ .

JEE Main 2026 — Physics Thermodynamics

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Solution

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JEE Main 2026 — 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