Answer to Question #109068 in Molecular Physics | Thermodynamics for RAJESH

Question

For two states defined by dV = 0 and dT  0, prove that the thermodynamic variables are connected through relation
(dv/dp)T(dp/dT)V(dT/dv)P= -1

Answer 1

We know for the simple compressible substance,

$du=C_vdT+(\dfrac{du}{dv})_T dv$

$ds=\dfrac{C_v}{T}dT+\dfrac{1}{T}[(\dfrac{du}{dv})_T +P]dv$

Now, taking entropy as a function of temperature and volume,

$\dfrac{\partial s}{\partial T}_v=\dfrac{C_v}{T}$

Now, using maxwell relation,

$(\dfrac{\partial s}{\partial V})_T=(\dfrac{\partial P}{\partial T})_s=\dfrac{1}{T}[(\dfrac{\partial u}{\partial v})_T+P]$

From the above, we will get,

$(\dfrac{\partial u}{\partial v})_T=T(\dfrac{\partial P}{\partial T})_v-P$

Now, we know that PV=nRT

$(\dfrac{\partial P}{\partial T})_v=\dfrac{R}{V}$

$(\dfrac{\partial u}{\partial v})_T=T(\dfrac{R}{V})_v-P=P-P=0$

So,

$ds=\dfrac{C_p}{T}dT+[(\dfrac{dV}{dT})_P]dP$

From the above,

$(\dfrac{dP}{dT})_v=\dfrac{C_p-C_v}{T(\partial V/\partial T)_P}$

now, $\beta=\dfrac{1}{V}(\dfrac{\partial V}{\partial T})_P$

$k=-\dfrac{1}{V}(\dfrac{\partial V}{\partial T})_T$

Hence, from the above, we can write it as

$(\dfrac{\partial P}{\partial T})_v(\dfrac{\partial T}{\partial V})_P(\dfrac{\partial V}{\partial P})=-1$