Skip to main content
Back to THERMODYNAMICS

JEE PHYSICS MCQ: THERMODYNAMICS

1 Medium
1 of 1
Application +4 | -1
Work done by a system under isothermal change from a volume V1V_1 to V2V_2 for a gas which obeys Vander Waal's equation

(V−βn)(P+αn2V)=nRT(V-\beta n)\left(P+\frac{\alpha n^2}{V}\right)=n R T
View answer and explanation
Correct option: A
nRTlog⁡e(V2−nβV1−nβ)+αn2(V1−V2V1V2)n R T \log _e\left(\frac{V_2-n \beta}{V_1-n \beta}\right)+\alpha n^2\left(\frac{V_1-V_2}{V_1 V_2}\right)
Option A (nRTlog⁡e(V2−nβV1−nβ)+αn2(V1−V2V1V2)n R T \log _e\left(\frac{V_2-n \beta}{V_1-n \beta}\right)+\alpha n^2\left(\frac{V_1-V_2}{V_1 V_2}\right)) is correct because it is the work done calculated by integrating the pressure from the van der Waals equation over the volume change.

More Questions from THERMODYNAMICS

Practice All Chapter MCQs →
Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.