Given, pressure (p)∝kV3 T1=100∘C,T2=300∘C ∆T=T2−T1=300−100=200∘C ⇒ By using ideal gas equation, pV=nRT kV3⋅V=nRT⇒kV4=nRT On differentiating both sides w.r.t temperature, we get 4kV3
dV
dT
=nR ⇒4kV3dV=nRdT⇒kV3dV=nRdT∕4 ⇒pdV=nRdT∕4 As, work done (W)=pdV=nRdT∕4 =