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JEE Main 20268 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
8 April
Shift
Evening Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2026 · 8 April · Evening Shift · Physics · Thermodynamics

One mole of diatomic gas having rotational modes only is kept in a...

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 44 cm2^2. The gas is heated slowly to raise the temperature by 1.2\,^\circC during which the piston moves by 2525 mm. The amount of heat supplied to the gas is ________ J. (Atmospheric pressure =100=100 kPa, R=8.3R=8.3 J/mol·K) (Neglect mass of the piston)

Options

  1. A24.824.8
  2. B2525
  3. C15.0415.04
  4. D29.9829.98

Correct answer

B. 2525

Step-by-step solution

Using the first law of thermodynamics:
Q=ΔU+WQ = \Delta U + W

Since the gas is diatomic with rotational modes only, the degrees of freedom are f=5f = 5, so:
CV=52RC_V = \dfrac{5}{2}R

For n=1 molen = 1\text{ mole} and ΔT=1.2 K\Delta T = 1.2\text{ K}:
ΔU=nCVΔT=1×52×8.3×1.2=24.9 J\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 ΔVW = P\,\Delta V

Cross-sectional area of the piston:
A=4 cm2=4×10−4 m2A = 4\text{ cm}^2 = 4 \times 10^{-4}\text{ m}^2

Piston displacement:
x=25 mm=2.5×10−2 mx = 25\text{ mm} = 2.5 \times 10^{-2}\text{ m}

Change in volume:
ΔV=A⋅x=(4×10−4)(2.5×10−2)=1×10−5 m3\Delta V = A \cdot x = (4 \times 10^{-4})(2.5 \times 10^{-2}) = 1 \times 10^{-5}\text{ m}^3

With P=100 kPa=105 PaP = 100\text{ kPa} = 10^5\text{ Pa}:
W=(105)(1×10−5)=1 JW = (10^5)(1 \times 10^{-5}) = 1\text{ J}

Therefore, the heat supplied is:
Q=ΔU+W=24.9+1=25.9 JQ = \Delta U + W = 24.9 + 1 = 25.9\text{ J}

Hence, the heat supplied to the gas is 25.9 J25.9\text{ J}.
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