An ideal gas has a specific heat at constant pressure CP = (5/2)R. The gas is kept in a closed vessel of volume 8.3 x 10-3 m3 at a temperature of 300 K and a pressure of 1.6 x 106 N/m2. An amount of 2.49 x 104 J of heat energy is supplied to the gas. Calculate the final pressure

difference of specific heats is CP-CV=R CV=CP−R CV=(5/2)R_−R CV=(3/2)R given pressure P=1.6 x   106 pascal; volume V=8.3 x   10-3m3;temperature T=300K;universal gas constant=8.3J/mol-Knumbe of moles =μ=PVRT=1.6 x   106× 8.3 x   10-38.3×300μ=163amount of heat given is,ΔU=(ΔQ)V=μCVdT∵Work done,∆W =0 for closed rigid container as volume is constant 2.49 x   104=163×32×8.3dT ⇒dT=T2-T1=375T2=T1+375⇒T2=300+375=675we know P α TP1P2=T1T21.6 x   106P2=300675⇒P2=3.6×106N/m2