The correct option is A −11.4 kJ
work done in an isothermal reversible process is given as:
w=−2.303nRT(logV2V1)
Here, n=2 mol
T=27+273=300 K
V1=50 L, V2=5 atm
R=8.314 JK−1mol−1
Using the formula we get,
w=−2.303×2×8.314×300×log550
w=11488 J
For isothermal process, △U=0 (as temperature is constant)
so, from first law
△U=q+w
q=−w=−11488 J
or
q=−11.4 kJ