34D (Statistical Physics)

For a fixed amount of simple compressible material,

d U = T, d S - p, d V

and

G = U - TS + p V

gives

d G = - S, d T + V, d p

G

therefore yields the Maxwell relation

(\frac{\partial S}{\partial p})_{T} = - (\frac{\partial V}{\partial T})_{p} .

In the insulated throttling process, piston work into the gas is

p_{1} V_{1}

and work out is

p_{2} V_{2}

. Neglecting changes in macroscopic kinetic and gravitational energy, the first law gives

U_{2} - U_{1} = p_{1} V_{1} - p_{2} V_{2}

. Thus $U_{1} + p_{1} V_{1} = U_{2} + p_{2} V_{2}$ , so enthalpy is conserved. The irreversible process is not assumed isentropic.

Since

d H = T, d S + V, d p

and

d S = (C_{p} / T) d T - (\partial V / \partial T)_{p} d p

, one has

d H = C_{p}, d T + [V - T (\partial V / \partial T)_{p}] d p

μ_{J T} = (\frac{\partial T}{\partial p})_{H} = \frac{T ( \partial V / \partial T ) _{p} - V}{C _{p}} = \frac{V ( α T - 1 )}{C _{p}} .

V \propto T

at fixed pressure and

α = 1/ T

, so $μ_{J T} = 0$ . A pressure drop has

d p < 0

, so cooling,

d T < 0

, requires $μ_{J T} > 0$ in the operating range.

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