At any position of the piston, say at a distance x from the bottom, net force on it is zero PA=Fext+mg+kx3 ∴P=Fext+mg+kx3 ∴ External work done by the gas is ∆wgas=∫PdV=∫
(Fext+mg+kx3)
A
Adx =
L1
∫
L0
Fextdx+mg
L1
∫
L0
dx+k
L1
∫
L0
x3dx =∆Wext+mg(L1−L0)+
K
4
(L14−L04) =nRTln(
L1
L0
)+mg(L1−L0)+
K
4
(L14−L04) According to I law of thermodynamics for the gas ∆Q=∆U+∆Wgas =0+∆Wgas ⇒∆Q=∆Wgas ∆Q=nRTln(
