What is the maximum work that can be obtained in a steady-state flow process from 1 mol of nitrogen in its ideal-gas state at 800 K and 50 bar? Take the temperature and pressure of the surroundings as 300 K and 1.0133 bar.

Question

What is the maximum work that can be obtained in a steady-state flow process from 1 mol of nitrogen in its ideal-gas state at 800 K and 50 bar? Take the temperature and pressure of the surroundings as 300 K and 1.0133 bar

Accepted Answer

The maximum possible work is obtained from any completely reversible process that reduces the nitrogen to the temperature and pressure of the surroundings, i.e., to 300 K and 1.0133 bar. Required here is the calculation of [latex]W_{ideal}[/latex] by Eq. (5.22),

[latex] W_{ ext {ideal }}=\Delta H - T_\sigma \Delta S [/latex] (5.22)

in which ΔS and ΔH are the molar entropy and enthalpy changes of the nitrogen as its state changes from 800 K and 50 bar to 300 K and 1.0133 bar. For the ideal-gas state, enthalpy is independent of pressure, and its change is given by:

[latex] \Delta H^{i g}=\int_{T_1}^{T_2} C_P^{i g} d T [/latex]

The value of this integral is found from Eq. (4.8), and is represented by:

[latex] \int_{T_0}^T \frac{C_P}{R} d T=A\left(T - T_0 ight)+\frac{B}{2}\left(T^2 - T_0^2 ight)+\frac{C}{3}\left(T^3 - T_0^3 ight)+D\left(\frac{T - T_0}{T T_0} ight) [/latex] (4.8)

[latex] 8.314 imes ICPH \left(800,300 ; 3.280,0.593 imes 10^{-3}, 0.0,0.040 imes 10^{-5} ight)=-15,060 J \cdot mol ^{-1} [/latex]

where the parameters for nitrogen come from Table C.1 of App. C.

Similarly, the entropy change is found from Eq. (5.10), here written:

[latex] \frac{\Delta S^{i g}}{R}=\int_{T_0}^T \frac{C_P^{i g}}{R} \frac{d T}{T}-\ln \frac{P}{P_0} [/latex] (5.10)

The value of the integral, found from Eq. (5.11), is represented by:

[latex] x_i \gamma_i^l=z_i \gamma_i^s \psi_i [/latex] (all i ) (15.11)

[latex] 8.314 imes \operatorname{ICPS}\left(800,300 ; 3.280,0.593 imes 10^{-3}, 0.0,0.040 imes 10^{-5} ight)=-29.373 J \cdot mol ^{-1} \cdot K ^{-1} [/latex]

Whence, [latex] \Delta S^{i g}=-29.373 - 8.314 \ln \frac{1.0133}{50}=3.042 J \cdot mol ^{-1} \cdot K ^{-1} [/latex]

With these values of [latex] \Delta H^{i g} ext { and } \Delta S^{i g} ext {, } [/latex], Eq. (5.22) becomes:

[latex] W_{ ext {ideal }}=-15,060 - (300)(3.042)=-15,973 J \cdot mol ^{-1} [/latex]

Question 5.7: What is the maximum work that can be obtained in a steady-st...

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