JEE Main 2026 — Apr 2, Shift 1 — Physics solution

JEE Main2026Apr 2, Shift 1Physics

A vessel contains $0.15$ m $^{3}$ of a gas at pressure $8$ bar and temperature $14 0^{\circ}$ C with $c_{p} = 3 R$ and $c_{v} = 2 R$ . It is expanded adiabatically till pressure falls to $1$ bar. The work done during this process is ____ kJ. ( $R$ is gas constant)

Your answer

Solution

The adiabatic index is $γ = C_{P} / C_{V} = 3 R /2 R = 1.5$ .

From $P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$ :

$V_{2} = V_{1} (\frac{P _{1}}{P _{2}})^{1/ γ} = 0.15 \times (\frac{8}{1})^{2/3} = 0.15 \times 4 = 0.6$ m $^{3}$

Work done in an adiabatic process:

$W = \frac{P _{1} V _{1} - P _{2} V _{2}}{γ - 1} = \frac{( 8 \times 0.15 - 1 \times 0.6 ) \times 1 0 ^{5}}{1.5 - 1} = \frac{0.6 \times 1 0 ^{5}}{0.5} = 1.2 \times 1 0^{5}$ J

$∴ W = 120$ kJ

← Back to full paper

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →