 Solution

ID: past-exam-of-the-thermal-physics-course-of-the-university-of-oxford/2016/1/iii/solution

For steady insulated flow through a porous plug, neglect changes of bulk kinetic energy and gravitational potential energy, and allow no shaft work. Each amount of gas brings flow work

p_{1} V_{1}

upstream and delivers

p_{2} V_{2}

downstream. The steady energy conservation balance is

U_{1} + p_{1} V_{1} = U_{2} + p_{2} V_{2}, H_{1} = H_{2} .

Thus enthalpy is conserved in this Joule-Thomson effect. For an ideal gas,

H = H (T)

, so

T_{1} = T_{2}

and

U_{1} = U_{2}

as well. The entropy increases:

Δ S = n_{mol} R ln \frac{p _{1}}{p _{2}} > 0.

The porous-plug process is irreversible despite being slow and insulated; it is not the reversible expansion of the preceding part.

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