Calculate q (heat transfered) for an isothermal reversible expansion of 1mole of an ideal gas from an initial pressure of 1.0bar to a final pressure of 0.1bar at a constant temperature of 273K.
A
7kJ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
5.2kJ
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
4.3kJ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
6kJ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B5.2kJ As in isothermal process as temperature remains constant so △U is zero (as it is a function of temperature only) △U = 0
According to first law of thermodynamics, △U=W+q 0=W+q q=−W as W=−2.303nRT(logP1P2) q=−W=2.303nRT(logP1P2) Given : P1=1bar P2=0.1bar T=273K n=1mol