If pressure of a gas contained in a closed vessel is increased by 0.4% when heated by 1oC, the initial temperature must be

P1=P, T1=T, P2=P+(0.4% of P)⇒ P2=P+0.4100P=P+P250 and T2=T+1From Gay-Lussac's lawP1P2=T1T2⇒PP+P250=TT+1 (as V= constant for closed vessel) By solving, we get T = 250 K