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Thermodynamics Part 5

Section: Physics

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Question : 1 of 86

Marks: +1, -0

One mole of diatomic gas having rotational modes only is kept is a cylinder with a piston system. The cross-section area of the cylinder is

4 \text{cm}^2

. The gas is heated slowly to raise the temperature by 1.2 °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)

[JEE Main 8 apr 2026 shift 2]

24.8
25
15.04
29.98

Solution:

1st Method

Q = \Delta U + W

Q = \frac{f}{2} nR \Delta T + nR \Delta T = \left(\frac{f}{2} + 1\right) nR \Delta T

If

f = 2 \Rightarrow Q = 2 nR \Delta T \Rightarrow 2 \times 1 \times 8.3 \times 1.2 = 19.92

(No option is matching)

If

f = 5 \Rightarrow Q = 3.5 nR \Delta T

\Rightarrow 3.5 \times 1 \times 8.3 \times 1.2 = 34.86

(No option is matching)

2nd Method

Q = \Delta U + W

Q = \frac{f}{2} nR \Delta T + P \Delta V

If

f = 2

\Rightarrow Q = nR \Delta T + P \Delta V

= 1 \times 8.3 \times 1.2 + 10^{5} \times 25 \times 10^{-3} \times 4 \times 10^{-4}

= 10.96

(No option is matching)

If f = 5

\Rightarrow Q = 2.5 nR \Delta T + P \Delta V

2.5 \times 1 \times 8.3 \times 1.2 + 10^{5} \times 25 \times 10^{-3} \times 4 \times 10^{-4}

= 25.9

(closest is 25)

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