The work done in heating one mole of an ideal gas as constant pressure from15∘C to 25∘C is
T1=15+273=288KT2=25+273=298K
We know that :PV1=nRT1 and PV2=nRT2
Hence V1=nRT1P and V2=nRT2P
∴ V2−V1=nRT2P−nRT1P i.e., V2−V1=nRPT2−T1
Or ΔV=nRPT2−T1
But W=−PΔV=−P×nRnT2−T1
=−nRTn−T1
=−nRT2−I1 W=−1mol×1.987calK−mol−×(298−288)K=19.87cal
So, the correct answer is (d)