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=298KWe know that  :PV1=nRT1  and  PV2=nRT2Hence  V1=nRT1P  and V2=nRT2P           ∴ V2−V1=nRT2P−nRT1P i.e., V2−V1=nRPT2−T1Or        ΔV=nRPT2−T1But        W=−PΔV=−P×nRnT2−T1           =−nRTn−T1 =−nRT2−I1 W=−1mol×1.987calK−mol−×(298−288)K=19.87calSo, the correct answer is (d)