Respuesta :
Answer:
Joule-Thomson coefficient for an ideal gas:
[tex]\mu_{J.T} = 0[/tex]
Explanation:
Joule-Thomson coefficient can be defined as change of temperature with respect to pressure at constant enthalpy.
Thus,
[tex]\mu_{J.T} = \left [\frac{\partial T}{\partial P} \right ]_H[/tex]
Also,
[tex]H= H (T,P)[/tex]
[tex]Differentiating\ it,[/tex]
[tex]dH= \left [\frac{\partial H}{\partial T}\right ]_P dT + \left [\frac{\partial H}{\partial P}\right ]_T dT[/tex]
Also, [tex] C_p[/tex] is defined as:
[tex]C_p = \left [\frac{\partial H}{\partial T}\right ]_P[/tex]
[tex]So,[/tex]
[tex]dH= C_p dT + \left [\frac{\partial H}{\partial P}\right ]_T dT[/tex]
Acoording to defination, the ethalpy is constant which means [tex]dH = 0[/tex]
[tex]So,[/tex]
[tex]\left [\frac{\partial H}{\partial P}\right ]_T = -C_p\times \left [\frac{\partial T}{\partial P}\right ]_H[/tex]
[tex]Also,[/tex]
[tex]\mu_{J.T} = \left [\frac{\partial T}{\partial P}\right ]_H[/tex]
[tex]So,[/tex]
[tex]\left [\frac{\partial H}{\partial P}\right ]_T =-\mu_{J.T}\times C_p[/tex]
For an ideal gas,
[tex]\left [\frac{\partial H}{\partial P}\right ]_T = 0[/tex]
So,
[tex]0 =-\mu_{J.T}\times C_p[/tex]
Thus, [tex]C_p[/tex] ≠0. So,
[tex] \mu_{J.T} = 0[/tex]
