JEE Main 2026 — Jan 24, Shift 2 — Physics solution

JEE Main2026Jan 24, Shift 2Physics

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 $α$ 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
D
28

Solution

From the graph the process $P_{1} \to P_{2}$ is isochoric at $V = 1 m^{3}$ .

At constant volume: $Δ Q = n C_{v} Δ T$ , and from $P V = n R T$ we have $Δ T = \frac{V Δ P}{n R}$ .

$Δ Q = n C_{v} \cdot \frac{V Δ P}{n R} = \frac{C _{v} V Δ P}{R} = \frac{21 \times 1 \times ( 30 - 21.7 )}{8.3} = \frac{21 \times 8.3}{8.3} = 21 J$ .

← Back to full paper

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →