JEE Main 2026 — Thermodynamics Question with Solution

From the given $P-V$ diagram, the process from $P_1$ to $P_2$ is an isochoric process because the volume remains constant at $V=1{\text{ m}}^3$.
For an isochoric process, the work done $W=0$.
According to the first law of thermodynamics, the heat involved is $Q=\Delta U+W=\Delta U$.
The change in internal energy is given by $\Delta U=n{C}_v\Delta T$.
Using the ideal gas equation $PV=nRT$, we have $n\Delta T=\frac{V\Delta P}{R}$.
Substituting this into the expression for $Q$:
$Q=C_v\left(\frac{V(P_2-P_1)}{R}\right)$
Given values: $n=10$ moles, $V=1{\text{ m}}^3$, $P_1=21.7\text{ Pa}$, $P_2=30\text{ Pa}$, $C_v=21\text{ J/mol}\cdot \text{K}$, and $R=8.3\text{ J/mol}\cdot \text{K}$.
$Q=21\times \frac{1\times (30-21.7)}{8.3}$
$Q=21\times \frac{8.3}{8.3}=21\text{ J}$.
Thus, the value of $\alpha$ is 21.