Question 4.5.4: Illustration 3.4-6 Continued, Showing That Entropy Is a Stat...

Since the piston-and-cylinder device is frictionless (see Illustration 3.4-6), each of the expansion processes will be reversible (see also Illustration 4.5-8). Thus, the entropy balance for the gas within the piston and cylinder reduces to

\frac{d S}{d t}=\frac{\dot{Q}}{T}.

Path A

i. Isothermal compression. Since T is constant,

\Delta S_{A}=\frac{Q_{A}}{T}=-\frac{5707.7 J / mol }{298.15 K }=-19.14 J /( mol K ).

ii. Isobaric heating

\dot{Q}=C_{ P }^{*} \frac{d T}{d t}.

so

\frac{d \underline{S}}{d t}=\frac{\dot{Q}}{T}=\frac{C_{ P }^{*}}{T} \frac{d T}{d t}.

and

\Delta \underline{S}_{B}=C_{ P }^{*} \ln \frac{T_{2}}{T_{1}}=38 \frac{ J }{ mol K } \times \ln \frac{573.15}{298.15}=24.83 J /( mol K ).

\Delta \underline{S}=\Delta \underline{S}{ }_{A}+\Delta \underline{S}_{B}=-19.14+24.83=5.69 J /( mol K ).

Path B
i. Isobaric heating

\Delta \underline{S}_{A}=C_{ P }^{*} \ln \frac{T_{2}}{T_{1}}=24.83 J /( mol K ).

ii. Isothermal compression

\Delta\underline{S}_{B}=\frac{Q}{T}=-\frac{10972.2}{573.15}=-19.14 J /( mol K ).

\Delta \underline{S}=24.83-19.14=5.69 J /( mol K ).

Path C

\text { i. Compression with } P V^{\gamma}=\text { constant }.

\Delta \underline{S}_{A}=\frac{Q}{T}=0.

ii. Isobaric heating

\Delta \underline{S}_{B}=C_{ P }^{*} \ln \frac{T_{3}}{T_{2}}=38 \frac{ J }{ mol K } \times \ln \frac{573.15}{493.38}=5.69 J /( mol K ).

\Delta \underline{S}=5.69 J /(\operatorname{mol~K}).

Comment

This example verifies, at least for the paths considered here, that the entropy is a state function.
For reversible processes in closed systems, the rate of change of entropy and the ratio \dot{Q} / T are equal. Thus, for reversible changes, \dot{Q} / T is also a state function, even though the total heat flow \dot{Q}is a path function.