Physics · Thermodynamics

JEE Main 2026 — 4 April, Morning Shift — Question 10

An ideal gas undergoes a process P = P₀[1 + (V₀/V)²]⁻¹. Two samples A and B (2 moles each) with initial volumes V₀ and 3V₀ undergo the process and attain same pressure. The difference T_B - T_A is

Option A:
$\frac{9P_0V_0}{8R}$
Option B:
$\frac{11P_0V_0}{10R}$
Correct
Option C:
$\frac{7P_0V_0}{6R}$
Option D:
$\frac{13P_0V_0}{11R}$

Answer: B

Step-by-step solution

At V=V₀: P=P₀/2 ⇒ T_A = (P₀/2)V₀/(2R)=P₀V₀/(4R). At V=3V₀: P=9P₀/10 ⇒ T_B = (9P₀/10)×3V₀/(2R)=27P₀V₀/(20R). Difference = (27/20 - 1/4)P₀V₀/R = (27/20-5/20)=22/20=11/10 ⇒ (11P₀V₀)/(10R)

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 4 April, Morning Shift
Subject: Physics
Chapter: Thermodynamics
Topic: First Law of Thermodynamics