10 mole of an ideal gas is undergoing the process shown in the figure. The heat involved in the process from $\mathrm{P}_{1}$ to $\mathrm{P}_{2}$ is $\alpha$ Joule $(\mathrm{P}_{1}=21.7 \mathrm{~Pa}$ and $\mathrm{P}_{2}=30 \mathrm{~Pa}, \mathrm{C}_{\mathrm{v}}=21 \mathrm{~J} / \mathrm{K} . \mathrm{mol}, R=8.3 \mathrm{~J} / \mathrm{mol} . \mathrm{K})$. The value of $\alpha$ is ____.

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Heat & Thermodynamics

10 mole of an ideal gas is undergoing the process shown in the figure. The heat involved in the process from

P_{1}

P_{2}

α

Joule

(P_{1} = 21.7 Pa

and

P_{2} = 30 Pa, C_{v} = 21 J / K . mol, R = 8.3 J / mol . K)

. The value of

α

is ____.

(A)24
(B)15
(C)21✓ Correct
(D)28

Answer: (C)

Solution

The step from

P_{1}

P_{2}

runs vertically at the fixed volume

V = 1 m^{3}

, so it is an isochoric process and the heat supplied is

Δ Q = n C_{v} Δ T

. At constant volume the gas law gives

n R Δ T = V Δ P

, so

n Δ T = \frac{V ( P _{2} - P _{1} )}{R} = \frac{1 \times ( 30 - 21.7 )}{8.3} = 1

mol

\cdot

K. Hence

Δ Q = C_{v} (n Δ T) = 21 \times 1 = 21

J. The value of

α

21

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Single correct option

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