10 mole of an ideal gas is undergoing the process shown in the figure. The heat involved in the process from P₁ to P₂ is α Joule ( P₁ = 21.7 Pa and P₂ = 30 Pa, Cv = 21 J/K.mol, R = 8.3 J/mol.K). The value of α is ______. (a) 21 (b) 28 (c) 24 (d) 15

Answer: (a). P₁ → P₂ is at constant V = 1 m³ (isochoric), so Q = nCvΔ T. Δ T = (VΔ P)/(nR) = (1 × (30 - 21.7))/(10 × 8.3) = 0.1 K Q = 10 × 21 × 0.1 = 21 J So the answer is (a) 21.

10 mole of an ideal gas is undergoing the process shown in the… | JEE Main 2026 (24 Jan, Shift 2)

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P-V Diagrams & Cyclic Processes

10 mole of an ideal gas is undergoing the process shown in the figure. The heat involved in the process from $P_1$ to $P_2$ is $\alpha$ Joule ( $P_1 = 21.7\text{ Pa}$ and

$P_2 = 30\text{ Pa}, C_v = 21\text{ J/K.mol}, R = 8.3\text{ J/mol.K})$ . The value of $\alpha$ is ______.

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Answer: (a)

$P_1 \to P_2$ is at constant $V = 1\ \text{m}^3$ (isochoric), so $Q = nC_v\Delta T$ .
$\Delta T = \frac{V\Delta P}{nR} = \frac{1 \times (30 - 21.7)}{10 \times 8.3} = 0.1\ \text{K}$
$Q = 10 \times 21 \times 0.1 = 21\ \text{J}$
So the answer is (a) 21.

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