Question

Show that the Joule-Thompson Coefficient is zero for ideal gas.

Answer 1

Answer:

Joule-Thomson coefficient for an ideal gas:

$\mu_{J.T} = 0$

Explanation:

Joule-Thomson coefficient can be defined as change of temperature with respect to pressure at constant enthalpy.

Thus,

$\mu_{J.T} = \left [\frac{\partial T}{\partial P} \right ]_H$

Also,

$H= H (T,P)$

$Differentiating\ it,$

$dH= \left [\frac{\partial H}{\partial T}\right ]_P dT + \left [\frac{\partial H}{\partial P}\right ]_T dT$

Also, $C_p$ is defined  as:

$C_p = \left [\frac{\partial H}{\partial T}\right ]_P$

$So,$

$dH= C_p dT + \left [\frac{\partial H}{\partial P}\right ]_T dT$

Acoording to defination, the ethalpy is constant which means $dH = 0$

$So,$

$\left [\frac{\partial H}{\partial P}\right ]_T = -C_p\times \left [\frac{\partial T}{\partial P}\right ]_H$

$Also,$

$\mu_{J.T} = \left [\frac{\partial T}{\partial P}\right ]_H$

$So,$

$\left [\frac{\partial H}{\partial P}\right ]_T =-\mu_{J.T}\times C_p$

For an ideal gas,

$\left [\frac{\partial H}{\partial P}\right ]_T = 0$

So,

$0 =-\mu_{J.T}\times C_p$

Thus, $C_p$ ≠0. So,

$\mu_{J.T} = 0$