Answer to Question #141386 in Physics for Vijay

Question

Show that (Cp-Cv) d²T/dPdV + (dCp/dP)v (dT/dV)p - (dCv/dV)p (dT/dP)v =1 using maxwell's equation

Answer 1

I am assuming that gas is ideal and number of moles is 1.

then, by ideal gas equation, I can write it as $PV = RT$

Now,

$(\frac{\partial }{\partial V})_P [(\frac{\partial }{\partial P})_V(PV)] =(\frac{\partial }{\partial V})_P [(\frac{\partial }{\partial P})_V(TR)]$

Since, $C_P - C_V = R$

Then equation will be,

$(\frac{\partial }{\partial V})_P [(\frac{\partial }{\partial P})_V(PV)] =(\frac{\partial }{\partial V})_P [(\frac{\partial }{\partial P})_V[T(C_P-C_V) ]]$

Differentiating both sides with respect to P keeping V as constant,

$(\frac{\partial }{\partial V})_P [(\frac{\partial P}{\partial P})_VV] =(\frac{\partial }{\partial V})_P [(\frac{\partial T }{\partial P})_V(C_P-C_V)+T(\frac{\partial C_P }{\partial P})_V]$

$(\frac{\partial }{\partial V})_P [V] =(\frac{\partial }{\partial V})_P [(\frac{\partial T }{\partial P})_V(C_P-C_V)+T(\frac{\partial C_P }{\partial P})_V]$

Again differentiating with respect to V keeping P as constant,

$(\frac{\partial V}{\partial V})_P =(\frac{\partial^2 T}{\partial V \partial P})(C_P-C_V)+(\frac{\partial C_P }{\partial P})_V (\frac{\partial T}{\partial V})_P -(\frac{\partial C_V }{\partial V})_P(\frac{\partial T}{\partial P})_V$

It can be written as,

$1 =(\frac{\partial^2 T}{\partial V \partial P})(C_P-C_V)+(\frac{\partial C_P }{\partial P})_V (\frac{\partial T}{\partial V})_P -(\frac{\partial C_V }{\partial V})_P(\frac{\partial T}{\partial P})_V$

which is the required result.