JEE Main 2026 Jan 24 Shift 2, Physics Q21: Thermodynamic Processes

JEE Main2026Jan 24, Shift 2Physics

When $300$ J of heat given to an ideal gas with $C_{p} = \frac{7}{2} R$ its temperature raises from $20^{\circ} C$ to $50^{\circ} C$ keeping its volume constant. The value of $\frac{m * 1000}{M}$ is: _____ g (nearest integer). ( $R = 8.314$ J/mol·K, M is molar mass of gas, m is the mass of the gas).

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Solution

At constant volume: $Δ Q = n C_{v} Δ T$ , with $C_{v} = C_{p} - R = \frac{5 R}{2}$ and $Δ T = 30 K$ .

$300 = n \cdot \frac{5}{2} \cdot 8.314 \cdot 30 \Rightarrow n \approx 0.48$ mol.

$\frac{m}{M} = 0.481$ .

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JEE Main 2026 Jan 24 Shift 2, Physics Q21: Thermodynamic Processes

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