Chapter 12.6, Problem 73P

Determine the enthalpy change and the entropy change of oxygen per unit mole as it undergoes a change of state from 220 K and 5 MPa to 300 K and 10 MPa (a) by assuming ideal-gas behavior and (b) by accounting for the deviation from ideal-gas behavior.

To determine

The change in specific enthalpy and entropy of oxygen per unit mole by assuming ideal-gas behavior.

Answer to Problem 73P

The change in specific enthalpy and entropy of oxygen per unit mole by assuming ideal-gas behavior is $2332\text{kJ/kmol}$ and $3.28\text{kJ/kmol}\cdot \text{K}$ respectively.

Explanation of Solution

At ideal gas state, the enthalpy is the function of temperature only.

Write the formula for difference in molar specific enthalpy of oxygen at ideal gas state.

${\left(\overline{h_2}-\overline{h_1}\right)}_{\text{ideal}}={\overline{h_2}}_{\text{ideal}}\left(T_2\right)-{\overline{h_1}}_{\text{ideal}}\left(T_1\right)$ (I)

Here, the molar enthalpy at ideal gas state corresponding to the temperature is $\overline{h}\left(T\right)$ and the subscripts 1 and 2 indicates initial and final states.

Write the formula for change in molar specific entropy.

${\left(\overline{s_2}-\overline{s_1}\right)}_{\text{ideal}}=\overline{s_2^{\circ }}-\overline{s_1^{\circ }}-R_u\ln \frac{P_2}{P_1}$ (II)

Here, the molar specific entropy at reference sate is $\overline{s°}$, the universal gas constant is $R_u$, the pressure is $P$, and the subscripts 1 and 2 indicates initial and final states.

Refer Table A-19, “Ideal-gas properties of oxygen, ${\text{O}}_{\text{2}}$”.

Obtain the initial properties corresponding to the temperature of $220\text{K}$.

$\begin{array}{l}\overline{h_1}=6404\text{kJ/kmol}\\ \overline{s_{{}_1}^{\circ }}=196.171\text{kJ/kmol}\cdot \text{K}\end{array}$

Obtain the final properties corresponding to the temperature of $300\text{K}$.

$\begin{array}{l}\overline{h_2}=8736\text{kJ/kmol}\\ \overline{s_{{}_2}^{\circ }}=205.213\text{kJ/kmol}\cdot \text{K}\end{array}$

The universal gas constant $\left(R_u\right)$ is $8.314\text{kJ/kmol}\cdot \text{K}$.

Conclusion:

Substitute $8736\text{kJ/kmol}$ for ${\overline{h_2}}_{\text{ideal}}\left(T_2\right)$ and $6404\text{kJ/kmol}$ for ${\overline{h_1}}_{\text{ideal}}\left(T_1\right)$ in Equation (I).

$\begin{array}{c}{\left(\overline{h_2}-\overline{h_1}\right)}_{\text{ideal}}=8736\text{kJ/kmol}-6404\text{kJ/kmol}\\ =2332\text{kJ/kmol}\end{array}$

Substitute $205.213\text{kJ/kmol}\cdot \text{K}$ for $\overline{s_{{}_2}^{\circ }}$, $196.171\text{kJ/kmol}\cdot \text{K}$ for $\overline{s_{{}_1}^{\circ }}$, $8.314\text{kJ/kmol}\cdot \text{K}$ for $R_u$, $10\text{MPa}$ for $P_2$, and $5\text{MPa}$ for $P_1$ in Equation (I).

$\begin{array}{c}{\left(\overline{s_2}-\overline{s_1}\right)}_{\text{ideal}}=\left[\begin{array}{l}205.213\text{kJ/kmol}\cdot \text{K}-196.171\text{kJ/kmol}\cdot \text{K}\\ -\left(8.314\text{kJ/kmol}\cdot \text{K}\right)\ln \frac{10\text{MPa}}{5\text{MPa}}\end{array}\right]\\ =9.042\text{kJ/kmol}\cdot \text{K}-5.7628\text{kJ/kmol}\cdot \text{K}\\ =3.2792\text{kJ/kmol}\cdot \text{K}\\ \simeq 3.28\text{kJ/kmol}\cdot \text{K}\end{array}$

Thus, the change in specific enthalpy and entropy of oxygen per unit mole by assuming ideal-gas behavior is $2332\text{kJ/kmol}$ and $3.28\text{kJ/kmol}\cdot \text{K}$ respectively.

To determine

The change in specific enthalpy and entropy of oxygen per unit mole by accounting for the deviation from ideal-gas behavior.

Answer to Problem 73P

The change in specific enthalpy and entropy of oxygen per unit mole by accounting for the deviation from ideal-gas behavior is $2396\text{kJ/kmol}$ and $3.70\text{kJ/kmol}\cdot \text{K}$ respectively.

Explanation of Solution

Write formula for enthalpy departure factor $\left(Z_h\right)$ on molar basis.

$Z_h=\frac{{\left({\overline{h}}_{\text{ideal}}-\overline{h}\right)}_{T,P}}{R_u{T}_{\text{cr}}}$ (III)

Here, the molar enthalpy at ideal gas state is ${\overline{h}}_{\text{ideal}}$, the molar enthalpy and normal state is $\overline{h}$, the universal gas constant is $R_u$, and the critical temperature is $T_{\text{cr}}$; The subscripts $T,P$ indicates the correspondence of given temperature and pressure.

Rearrange the Equation (III) to obtain $\overline{h}$.

$\overline{h}={\overline{h}}_{\text{ideal}}-Z_h{R}_u{T}_{\text{cr}}$ (IV)

Refer Equation (IV) express as two states of enthalpy difference (final – initial).

${\overline{h}}_2-{\overline{h}}_1={\left({\overline{h}}_2-{\overline{h}}_1\right)}_{\text{ideal}}-\left(Z_{h2}-Z_h{}_1\right)R_u{T}_{\text{cr}}$ (V)

Write formula for entropy departure factor $\left(Z_s\right)$ on molar basis.

$Z_s=\frac{{\left({\overline{s}}_{\text{ideal}}-\overline{s}\right)}_{T,P}}{R_u}$ (VI)

Here, the molar entropy at ideal gas state is ${\overline{s}}_{\text{ideal}}$, the molar entropy and normal state is $\overline{s}$, the universal gas constant is $R_u$, and the critical temperature is $T_{\text{cr}}$; The subscripts $T,P$ indicates the correspondence of given temperature and pressure.

Rearrange the Equation (VI) to obtain $\overline{s}$.

$\overline{s}={\overline{s}}_{\text{ideal}}-Z_s{R}_u$ (VII)

Refer Equation (VII) express as two states of entropy difference (final – initial).

${\overline{s}}_2-{\overline{s}}_1={\left({\overline{s}}_2-{\overline{s}}_1\right)}_{\text{ideal}}-\left(Z_{s2}-Z_s{}_1\right)R_u$ (VIII)

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”.

The critical temperature and pressure of oxygen gas is as follows.

$\begin{array}{l}{T}_{\text{cr}}=154.8\text{K}\\ P_{\text{cr}}=5.08\text{MPa}\end{array}$

The reduced pressure $\left(P_{R1}\right)$ and temperature $\left(T_{R1}\right)$ at initial state is expressed as follows.

$\begin{array}{c}{T}_{R1}=\frac{T_1}{T_{\text{cr}}}\\ =\frac{220\text{K}}{154.8\text{K}}\\ =1.42\end{array}$

$\begin{array}{c}{P}_{R1}=\frac{P_1}{P_{\text{cr}}}\\ =\frac{5\text{MPa}}{5.08\text{MPa}}\\ =0.98\end{array}$

The reduced pressure $\left(P_{R2}\right)$ and temperature $\left(T_{R2}\right)$ at final state is expressed as follows.

$\begin{array}{c}{T}_{R2}=\frac{T_2}{T_{\text{cr}}}\\ =\frac{300\text{K}}{154.8\text{K}}\\ =1.94\end{array}$

$\begin{array}{c}{P}_{R2}=\frac{P_2}{P_{\text{cr}}}\\ =\frac{10\text{MPa}}{5.08\text{MPa}}\\ =1.97\end{array}$

At initial:

Refer Figure A-29, “Generalized enthalpy departure chart”.

The enthalpy departure factor $\left(Z_{h1}\right)$ corresponding the reduced pressure $\left(P_{R1}\right)$ and reduced temperature $\left(T_{R1}\right)$ is $0.53$.

Refer Figure A-30, “Generalized entropy departure chart”.

The entropy departure factor $\left(Z_{s1}\right)$ corresponding the reduced pressure $\left(P_{R1}\right)$ and reduced temperature $\left(T_{R1}\right)$ is $0.25$.

At final:

Refer Figure A-29, “Generalized enthalpy departure chart”.

The enthalpy departure factor $\left(Z_{h2}\right)$ corresponding the reduced pressure $\left(P_{R2}\right)$ and reduced temperature $\left(T_{R2}\right)$ is $0.48$.

Refer Figure A-30, “Generalized entropy departure chart”.

The entropy departure factor $\left(Z_{s2}\right)$ corresponding the reduced pressure $\left(P_{R2}\right)$ and reduced temperature $\left(T_{R2}\right)$ is $0.20$.

Conclusion:

Substitute $2332\text{kJ/kmol}$ for ${\left({\overline{h}}_2-{\overline{h}}_1\right)}_{\text{ideal}}$, $0.48$ for $Z_{h2}$, $0.53$ for $Z_{h1}$, $8.314\text{kJ/kmol}\cdot \text{K}$ for $R_u$, and $154.8\text{K}$ for $T_{\text{cr}}$ in Equation (V).

$\begin{array}{c}{\overline{h}}_2-{\overline{h}}_1=\left\{\begin{array}{l}2332\text{kJ/kmol}\\ -\left[\left(0.48-0.53\right)\left(8.314\text{kJ/kmol}\cdot \text{K}\right)\left(154.8\text{K}\right)\right]\end{array}\right\}\\ =2332\text{kJ/kmol}+64.3504\text{kJ/kmol}\\ =2396.3504\text{kJ/kmol}\\ \simeq 2396\text{kJ/kmol}\end{array}$

Substitute $3.28\text{kJ/kmol}\cdot \text{K}$ for ${\left({\overline{s}}_2-{\overline{s}}_1\right)}_{\text{ideal}}$, $0.20$ for $Z_{s2}$, $0.25$ for $Z_{s1}$, and $8.314\text{kJ/kmol}\cdot \text{K}$ for $R_u$, in Equation (VIII).

$\begin{array}{c}{\overline{s}}_2-{\overline{s}}_1=\left\{\begin{array}{l}3.28\text{kJ/kmol}\cdot \text{K}\\ -\left[\left(0.20-0.25\right)\left(8.314\text{kJ/kmol}\cdot \text{K}\right)\right]\end{array}\right\}\\ =3.28\text{kJ/kmol}\cdot \text{K}+1.8785\text{kJ/kmol}\cdot \text{K}\\ =3.6957\text{kJ/kmol}\cdot \text{K}\\ \simeq 3.70\text{kJ/kmol}\cdot \text{K}\end{array}$

Thus, the change in specific enthalpy and entropy of oxygen per unit mole by accounting for the deviation from ideal-gas behavior is $2396\text{kJ/kmol}$ and $3.70\text{kJ/kmol}\cdot \text{K}$ respectively.