- #1
mef51
- 23
- 0
Homework Statement
One method for cooling a gas is adiabatic throttling (Joule-Thomson Experiment). Another method is a reversible adiabatic expansion. Show that if the initial and final pressures are the same, the difference in temperature obtained by the second method is always higher.
Hint: Express ##\frac{\partial T}{\partial P}## in each process as a function of the expansivity ##\beta##, the isothermal compressibility ##\kappa## and ##c_P## and compare the results.
Homework Equations
The question has to do with thermodynamic potentials.
$$
\beta=\frac{1}{V}\Big(\frac{\partial V}{\partial T}\Big)_{P} \\
\kappa=-\frac{1}{V}\Big(\frac{\partial V}{\partial P}\Big)_{T} \\
c_{P}=\frac{T}{n}\Big(\frac{\partial S}{\partial T}\Big)_{P}
$$
Cyclic Relation:
$$\Big(\frac{\partial T}{\partial P}\Big)_{S}\Big(\frac{\partial P}{\partial S}\Big)_{T}\Big(\frac{\partial S}{\partial T}\Big)_{P}=-1$$
Reciprocal Relation
$$
\Big(\frac{\partial x}{\partial y}\Big)_{z}=\Big(\frac{\partial y}{\partial x}\Big)_{z}^{-1}
$$
The temperature is a function of ##S## and ##P## so
$$T(P,S) \implies dT=\Big(\frac{\partial T}{\partial S}\Big)_{P}dS+\Big(\frac{\partial T}{\partial P}\Big)_{S}dP$$
The Maxwell Relations are also relevant.
The Attempt at a Solution
Using the cyclic and reciprocal relation and the definitions of ##\beta##, ##\kappa## and ##c_P## I can get that, for the second process
$$
\Big(\frac{\partial T}{\partial P}\Big)_{S}=\frac{\beta VT}{nc_{p}}
$$
To get the change in temperature I integrate the differential. Since the second process is reversible ##dS=0## so
$$
\Delta T=T_{f}-T_{i}=\int_{P_{i}}^{P_{f}}\Big(\frac{\partial T}{\partial P}\Big)_{S}dP
$$
But since the final and initial pressures are the same this change in temperature is always 0! How can it ever be greater than the temperature difference in the first method? What am I missing here?