Let

(T_{1}, V_{1})

be the final equilibrium state. The suddenly added weight supplies the constant external pressure

p_{1}

during the displacement. Since the process is adiabatic, the first law gives

\frac{5}{2} N k_{B} (T_{1} - T_{0}) = p_{1} (V_{0} - V_{1}) .

The initial and final ideal-gas equations are

p_{0} V_{0} = N k_{B} T_{0}, p_{1} V_{1} = N k_{B} T_{1} .

Substitution yields

\frac{5}{2} (T_{1} - T_{0}) = \frac{p _{1}}{p _{0}} T_{0} - T_{1},

and hence

T_{1} = T_{0} \frac{5 + 2 p _{1} / p _{0}}{7} .

Therefore

Δ T = T_{1} - T_{0} = \frac{2 T _{0}}{7} (\frac{p _{1}}{p _{0}} - 1) .

ii

Solution

At fixed pressure, the heat supplied equals the enthalpy change:

Q = Δ H = \frac{7}{2} N k_{B} Δ T .

Thus

Δ T = \frac{2 Q}{7 N k _{B}} .

Equivalently, the gas does work

W = p Δ V = N k_{B} Δ T = \frac{2 Q}{7},

while its internal energy increases by

5 Q /7

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