For a Van der Walls gas find :
(a) the equation of the adiabatic curve in the variables `T, V` ,
the difference of the molar heat capacities `C_p - C_v` as a function of `T` and `V`.

Explanation

(a) From the first law for an adiabatic
`dQ = dU + pd V = 0`
From the previous problem
`d U =((del U)/(del T))_V dT + ((del U)/(del V))_T dV = C_V dT + (a)/(V^2) dV`
So, `0 = C_V dT + (RT dV)/(V - b)`
This equation can be integrated if we assume that `C_V` and `b` are constanr then
`(R)/(C_V) (dV)/(V - b) + (dT)/(T) = 0` or, `1n T +(R)/(C_V) 1n (V - b) = constant`
or, `T(V - b)^(R//c_v) = constant`
(b) We use
`dU = C_V dT + (a)/(V^2) dV`
Now, `dQ = C_V dT + (RT)/(V - b)dV`
So along constant `p`,
`C_p = C_V + (RT)/(V - b) ((del V)/(del T))`
Thus `C_p - C_V = (RT)/(V - b)((del V)/(del T))_p`, But `p = (RT)/(V - b)- (a)/(V^2)`
On differentiating, `0 = (-(RT)/((V - b)^2) +(2 a)/(V^2))((del V)/(del T))_p + (R)/(V - b)`
or, `T((del V)/(del T))_p = (RT//V -b)/((RT)/((V - b)^2 )-(2 a)/(V^3)) = (V -b)/(1-(2a(V - b)^2)/(RT V^3))`
and `C_p - C_V = (R)/(1 -(2a(V - b)^2)/(RTV^3))`.