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JEE Main 2022 (28 Jul, Shift 2) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2022 (28 Jul, Shift 2)

Q164JEE Main 2022 · 28 JUL · SHIFT 2Medium

Thermodynamic Processes

At a certain temperature, the degrees of freedom per molecule for gas is 8. The gas performs 150 J of work when it expands under constant pressure. The amount of heat absorbed by the gas will be _________ J.

Numerical answer type. Work it out, then open the answer.

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Answer: 750

$f = 8$ : $C_v = \frac{f}{2}R = 4R$ , $C_p = 5R$ .
At constant pressure $\frac{Q}{W} = \frac{C_p}{R} = 5$
$Q = 5 \times 150 = 750\text{ J}$
So the answer is 750 J.

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