Answers: 2
![\mu_{J.T} = \left [\frac{\partial T}{\partial P} \right ]_H](/tpl/images/0086/1612/b4ba4.png)
![dH= \left [\frac{\partial H}{\partial T}\right ]_P dT + \left [\frac{\partial H}{\partial P}\right ]_T dT](/tpl/images/0086/1612/3977b.png)
is defined as:![C_p = \left [\frac{\partial H}{\partial T}\right ]_P](/tpl/images/0086/1612/f03db.png)
![dH= C_p dT + \left [\frac{\partial H}{\partial P}\right ]_T dT](/tpl/images/0086/1612/06d3f.png)
![\left [\frac{\partial H}{\partial P}\right ]_T = -C_p\times \left [\frac{\partial T}{\partial P}\right ]_H](/tpl/images/0086/1612/f9876.png)
![\mu_{J.T} = \left [\frac{\partial T}{\partial P}\right ]_H](/tpl/images/0086/1612/32957.png)
![\left [\frac{\partial H}{\partial P}\right ]_T =-\mu_{J.T}\times C_p](/tpl/images/0086/1612/156ea.png)
![\left [\frac{\partial H}{\partial P}\right ]_T = 0](/tpl/images/0086/1612/baba1.png)
≠0. So,
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Show that the joule-thompson coefficient is zero for ideal gas....
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