A gas is expanded from volume $V_{0} = to 2 V_{0}$ under three different processes. Process 1 is isobaric process, process 2 is isothermal and process 3 is adiabatic. Let $∆ U_{1}, ∆ U_{2} and ∆ U_{3}$ be ihe change in internal energy of the gas in these three processes. Then:

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$∆ U_{2} < ∆ U_{1} < ∆ U_{3}$

$∆ U_{1} > ∆ U_{2} > ∆ U_{2}$

$∆ U_{2} < ∆ U_{3} < ∆ U_{1}$

$∆ U_{1} < ∆ U_{2} < ∆ U_{2}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Process 2 is an isothermal process. Hence, $∆ U_{2} = 0$ .
Process I is an isobathic (P = constant) expansion.
Hence, temperature of the gas will increases;
or $∆ U_{1}$ = positive
Process 3 is an adiabatic expansion. Hence, temperature will decrease;
or $∆ U_{3} =$ negative
Therefore, $∆ U_{1} > ∆ U_{2} > ∆ U_{3}$ is the correct option.