Q. 500 J of heat energy is removed from 4 moles of a monoatomic ideal gas at constant volume. The temperature drops by
A) $40℃$
B) $30℃$
C) $10℃$
D) $0℃$
Explanation
(c) For monoatomic gas,
$C_{v} = \frac{3}{2}R = \frac{3}{2} \times 8.31\ Jmol^{- 1{^\circ}}C^{- 1}$
$Q = 500\ J,\ n = 4\theta = ?$
$\theta = \frac{Q}{nC_{v}} = \frac{500}{4 \times \frac{3}{2} \times 8.31} = 10℃$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.