Showing That the Change in State Variables between Fixed Initial and Final States Is Independent of the Path Followed It is possible to go from a given initial equilibrium state of a system to a given final equilibrium state by a number of different paths, involving different intermediate states an

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Showing That the Change in State Variables between Fixed Initial and Final States Is Independent of the Path Followed

It is possible to go from a given initial equilibrium state of a system to a given final equilibrium state by a number of different paths, involving different intermediate states and different amounts of heat and work. Since the internal energy of a system is a state property, its change between any two states must be independent of the path chosen (see Sec. 1.3). The heat and work flows are, however, path-dependent quantities and can differ on different paths between given initial and final states. This assertion is established here by example. One mole of a gas at a temperature of 25°C and a pressure of 1 bar (the initial state) is to be heated and compressed in a frictionless piston and cylinder to 300°C and 10 bar (the final state). Compute the heat and work required along each of the following paths.

Path A. Isothermal (constant temperature) compression to 10 bar, and then isobaric (constant pressure) heating to 300°C

Path B. Isobaric heating to 300°C followed by isothermal compression to 10 bar

Path C. A compression in which [latex]P \underline{V}^{\gamma}= ext { constant, where } \gamma=C_{ P }^{*} / C_{ V }^{*}[/latex] , followed by an isobaric cooling or heating, if necessary, to 300°C.

For simplicity, the gas is assumed to be ideal with [latex]C_{ P }^{*}[/latex] = 38 J/(mol K).

Accepted Answer

The 1-mol sample of gas will be taken as the thermodynamic system. The difference form of the mass balance for this closed, deforming volume of gas is
N = constant = 1 mol
and the difference form of the energy balance is

[latex]\Delta U=Q-\int P d V=Q+W[/latex]

Path A
i. Isothermal compression

[latex]\begin{aligned}W_{ i } &=-\int_{\underline{V}_{1}}^{\underline{V}_{2}} P d V=-\int_{\underline{V}_{1}}^{\underline{V}_{2}} R T \frac{d V}{\underline{V}}=-R T \int_{\underline{V}_{1}}^{\underline{V}_{2}} \frac{d V}{\underline{V}}=-R T \ln \frac{\underline{V}_{2}}{\underline{V}_{1}}=R T \ln \frac{P_{2}}{P_{1}} \&=8.314 J /( mol K ) imes 298.15 K imes \ln \frac{10}{1}=5707.7 J / mol\end{aligned}[/latex]

Since

[latex]\Delta \underline{U}=\int_{T_{1}}^{T_{2}} C_{ V }^{*} d T=C_{ V }^{*}\left(T_{2}-T_{1} ight) \quad ext { and } \quad T_{2}=T_{1}=25^{\circ} C[/latex]

we have

[latex]\Delta \underline{U}=0 \quad ext { and } \quad Q_{ i }=-W_{ i }=-5707.7 J / mol[/latex]

ii. Isobaric heating

[latex]\begin{gathered}W_{ ii }=-\int_{\underline{V}_{2}}^{\underline{V}_{3}} P_{2} d V=-P_{2} \int_{\underline{V}_{2}}^{\underline{V}_{3}} d V=-P_{2}\left(\underline{V}_{3}-\underline{V}_{2} ight)=-R\left(T_{3}-T_{2} ight) \\Delta \underline{U}=\int_{T_{2}}^{T_{3}} C_{ V }^{*} d T=C_{ V }^{*}\left(T_{3}-T_{2} ight)\end{gathered}[/latex]

and

[latex]\begin{aligned}Q_{ ii } &=\Delta \underline{U}-W_{ ii }=C_{ V }^{*}\left(T_{3}-T_{2} ight)+R\left(T_{3}-T_{2} ight)=\left(C_{ V }^{*}+R ight)\left(T_{3}-T_{2} ight) \&=C_{ P }^{*}\left(T_{3}-T_{2} ight)\end{aligned}[/latex]

[This is, in fact, a special case of the general result that at constant pressure for a closed system, [latex]Q=\int C_{ P }^{*}[/latex] dT. This is easily proved by starting with

[latex]\dot{Q}=\frac{d \underline{U}}{d t}+P \frac{d \underline{V}}{d t}[/latex]

and using the fact that P is constant to obtain

[latex]\dot{Q}=\frac{d \underline{U}}{d t}+\frac{d}{d t}(P \underline{V})=\frac{d}{d t}(\underline{U}+P \underline{V})=\frac{d \underline{H}}{d t}=C_{ P }^{*} \frac{d T}{d t}[/latex]

Now setting [latex]Q=\int \dot{Q} d t ext { yields } Q=\int C_{ P }^{*} d T ext {.] }[/latex]
Therefore,

[latex]\begin{aligned}W_{ ii } &=-8.314 J /( mol K ) imes 275 K =-2286.3 J / mol \Q_{ ii } &=38 J /( mol K ) imes 275 K =10450 J / mol \Q &=Q_{ i }+Q_{ ii }=-5707.7+10450=4742.3 J / mol \W &=W_{ i }+W_{ ii }=5707.7-2286.3=3421.4 J / mol\end{aligned}[/latex]

Path B
i. Isobaric heating

[latex]\begin{aligned}Q_{ i } &=C_{ P }^{*}\left(T_{2}-T_{1} ight)=10450 J / mol \W_{ i } &=-R\left(T_{2}-T_{1} ight)=-2286.3 J / mol\end{aligned}[/latex]

ii. Isothermal compression

[latex]\begin{aligned}W_{ ii } &=R T \ln \frac{P_{2}}{P_{1}}=8.314 imes 573.15 \ln \left(\frac{10}{1} ight)=10972.2 J / mol \Q_{ ii } &=-W_{ ii }=-10972.2 J / mol \Q &=10450-10972.2=-522.2 J / mol \W &=-2286.3+10972.2=8685.9 J / mol\end{aligned}[/latex]

Path C
i. Compression with [latex]P \underline{V}^{\gamma}[/latex] = constant

[latex]\begin{aligned}W_{ i } &=-\int_{\underline{V}_{1}}^{\underline{V}_{2}} P d \underline{V}=-\int_{\underline{V}_{1}}^{\underline{V}_{2}} \frac{ ext { constant }}{\underline{V}^{\gamma}} d \underline{V}=-\frac{ ext { constant }}{1-\gamma}\left(\underline{V}_{2}^{1-\gamma}-\underline{V}_{1}^{1-\gamma} ight) \&=-\frac{1}{1-\gamma}\left(P_{2} \underline{V}_{2}-P_{1} \underline{V}_{1} ight)=\frac{-R\left(T_{2}-T_{1} ight)}{1-\gamma}=\frac{-R\left(T_{2}-T_{1} ight)}{1-\left(C_{ P }^{*} / C_{ V }^{*} ight)}=C_{ V }^{*}\left(T_{2}-T_{1} ight)\end{aligned}[/latex]

where [latex]T_{2}[/latex] can be computed from

[latex]P_{1} \underline{V}_{1}^{\gamma}=P_{1}\left(\frac{R T_{1}}{P_{1}} ight)^{\gamma}=P_{2} \underline{V}_{2}^{\gamma}=P_{2}\left(\frac{R T_{2}}{P_{2}} ight)^{\gamma}[/latex]

or

[latex]\frac{T_{2}}{T_{1}}=\left(\frac{P_{2}}{P_{1}} ight)^{(\gamma-1) / \gamma}[/latex]

Now

[latex]\gamma=\frac{C_{ P }^{*}}{C_{ V }^{*}}=\frac{38}{38-8.314}=1.280[/latex]

so that

[latex]T_{2}=298.15 K (10)^{0.280 / 1.280}=493.38 K[/latex]

and

[latex]\begin{aligned}W_{ i } &=C_{ V }^{*}\left(T_{2}-T_{1} ight)=(38-8.314) J /( mol K ) imes(493.38-298.15) K \&=5795.6 J / mol \\Delta U_{ i } &=C_{ V }^{*}\left(T_{2}-T_{1} ight)=5795.6 J / mol \Q_{ i } &=\Delta U_{ i }-W_{ i }=0\end{aligned}[/latex]

ii. Isobaric heating

[latex]\begin{aligned}&Q_{ ii }=C_{ P }^{*}\left(T_{3}-T_{2} ight)=38 J /( mol K ) imes(573.15-493.38) K =3031.3 J / mol \&W_{ ii }=-R\left(T_{3}-T_{2} ight)=-8.314 J /( mol K ) imes(573.15-493.38) K =-663.2 J / mol\end{aligned}[/latex]

and
Q = 0 + 3031.3 = 3031.3 J/mol
W = 5795.6 - 663.2 = 5132.4 J/mol

[latex]\begin{array}{cccc}\hline ext { Path } & Q( J / mol ) & W( J / mol ) & Q+W=\Delta U( J / mol ) \\hline ext { A } & 4742.3 & 3421.4 & 8163.7 \ ext { B } & -522.2 & 8685.9 & 8163.7 \ ext { C } & 3031.3 & 5132.4 & 8163.7 \\hline\end{array}[/latex]

Comment
Notice that along each of the three paths considered (and, in fact, any other path between the initial and final states), the sum of Q and W, which is equal to ΔU, is 8163.7 J/mol, even though Q and W separately are different along the different paths. This illustrates that whereas the internal energy is a state property and is path independent (i.e., its change in going from state 1 to state 2 depends only on these states and not on the path between them), the heat and work flows depend on the path and are therefore path functions.

Question 3.4-6: Showing That the Change in State Variables between Fixed Ini...

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