Question

be calculated using The isothermal Joule-Thomson coefficient jer may Derive an expression for Joule-Thomson coefficient μ for a gas with this eguation of state (4 points) V-nb Derive an expression for Joule-Thomson coefficient μ for a gas with this equation of state (1.5 points) b RT op乃-下

Answer 1

Isothermal Joule - Thomson coefficient , $\inline \mu_{T}= (\frac{\partial H}{\partial P})_{T}$

a) P (V -nb) = nRT

V - nb = nRT/P

V = nRT/P + nb

$\inline (\frac{\partial V}{\partial T})_{P} = \frac{nR}{P}$  

$\inline T(\frac{\partial V}{\partial T})_{P} = \frac{nRT}{P} = V - nb$

$\inline (\frac{\partial H}{\partial P})_{T} = V- T(\frac{\partial V}{\partial T})_{P} = V - (V - nb)=nb$

Isothermal Joule - Thomson coefficient , $\inline \mu_{T} = nb$

b) PV = nRT

V = nRT/P

$\inline (\frac{\partial V}{\partial T})_{P}= \frac{nR}{P}$

$\inline T(\frac{\partial V}{\partial T})_{P}= \frac{nRT}{P} = V$

$\inline (\frac{\partial H}{\partial P})_{T} = V - T(\frac{\partial V}{\partial T})_{P}= V - V =0$

Isothermal Joule - Thomson coefficient , $\inline \mu_{T}=0$

NOTE: Isothermal Joule - Thomson coefficient , $\inline \mu_{T}= (\frac{\partial H}{\partial P})_{T}$ (asked here)

      Joule - Thomson coefficient , $\inline \mu_{JT}= -\frac{\mu_{T}}{C_{P}}= (\frac{\partial T}{\partial P})_{H}$ ​​​​​​​