Chapter 3.8, Problem 95P

a)

To determine

The final pressure of the methane treating it as an ideal gas.

Answer to Problem 95P

The final pressure of the methane treating it as an ideal gas is $156.5\text{ kPa}$.

Explanation of Solution

Write the ideal gas equation.

$P_2=P_1\left(\frac{T_2}{T_1}\right)$ (I)

Here, pressure at state 2 is $P_2$, pressure at state 1 is $P_1$, temperature at state 2 is $T_2$ and temperature at state 1 is $T_1$.

Conclusion:

Substitute 80 kPa for $P_1$, $20°\text{C}$ for $T_1$ and $300°\text{C}$ for $T_2$ in Equation (I).

$\begin{array}{c}{P}_2=\left(80\text{ kPa}\right)\left(\frac{300°\text{C}+273\text{ K}}{20°\text{C}+273\text{ K}}\right)\\ =156.5\text{ kPa}\end{array}$

Thus, the final pressure of the methane treating it as an ideal gas is $156.5\text{ kPa}$.

b)

To determine

The final pressure of the methane treating it as a Benedict-Webb-Rubin gas.

Answer to Problem 95P

The final pressure of the methane treating it as a Benedict-Webb-Rubin gas is $156.5\text{ kPa}$.

Explanation of Solution

Write the expression to obtain the specific molar volume of the methane $\left(\overline{v}\right)$.

${\overline{v}}_1={\overline{v}}_2=\overline{v}=\frac{R_u{T}_1}{P_1}$ (II)

Here, universal gas constant is $R_u$.

Write the Benedict-Webb-Rubin equation of state for state 2.

$P_2=\left\{\begin{array}{l}\frac{R_u{T}_2}{{\overline{v}}_2}+\left(B_0{R}_u{T}_2-A_0-\frac{C_0}{T_2^2}\right)\frac{1}{{\overline{v}}^2}+\\ \frac{b{R}_u{T}_2-a}{{\overline{v}}^3}+\frac{a\alpha }{{\overline{v}}^6}+\frac{c}{{\overline{v}}^3{T}_2^2}\left(1+\frac{\gamma }{{\overline{v}}^2}\right)\exp \left(\frac{-\gamma }{{\overline{v}}^2}\right)\end{array}\right\}$ (III)

Here, Constants that appear in the Benedict-Webb-Rubin equation of state are $a,A_0,b,B_0,c,C_0,\alpha \text{ and }\gamma$.

Conclusion:

Substitute $8.314\text{kPa}\cdot {\text{m}}^3/\text{kmol}\cdot \text{K}$ for $R_u$, $20°\text{C}$ for $T_1$ and 80 kPa for $P_1$ in Equation (II).

$\begin{array}{c}{\overline{v}}_2=\frac{\left(8.314\text{kPa}\cdot {\text{m}}^3/\text{kmol}\cdot \text{K}\right)\left(20°\text{C}+273\right)\text{K}}{80\text{ kPa}}\\ =30.45{\text{m}}^3/\text{kmol}\end{array}$

Refer the Table 3-4, “Constants that appear in the Benedict-Webb-Rubin equation of state”, obtain the following properties for methane ${\text{CH}}_4$.

$\begin{array}{l}a=5\\ A_0=187.91\\ b=0.003380\\ B_0=0.04260\\ c=2.578\times {10}^5\\ C_0=2.286\times {10}^6\\ \alpha =1.244\times {10}^{-4}\\ \gamma =0.0060\end{array}$

Substitute $8.314\text{kPa}\cdot {\text{m}}^3/\text{kmol}\cdot \text{K}$ for $R_u$, $300°\text{C}$ for $T_2$, $30.45{\text{m}}^3/\text{kmol}$ for $\overline{v}$, $0.04260$ for $B_0$, $187.91$ for $A_0$, $2.286\times {10}^6$ for $C_0$, $0.003380$ for $b$, $5$ for a and $0.0060$ for $\gamma$ in Equation (III).

$\begin{array}{c}{P}_2=\left\{\begin{array}{l}\frac{\left(8.314\text{kPa}\cdot {\text{m}}^3/\text{kmol}\cdot \text{K}\right)\left(300°\text{C}+273\right)\text{K}}{30.45{\text{m}}^3/\text{kmol}}+\\ \left(\begin{array}{l}\left(0.04260\right)\left(8.314\text{kPa}\cdot {\text{m}}^3/\text{kmol}\cdot \text{K}\right)\\ \left(300°\text{C}+273\right)\text{K}-187.91\\ -\frac{2.286\times {10}^6}{{\left(\left(300°\text{C}+273\right)\text{K}\right)}^2}\end{array}\right)\frac{1}{{\left(30.45{\text{m}}^3/\text{kmol}\right)}^2}+\\ \frac{\left(0.003380\right)\left(8.314\text{kPa}\cdot {\text{m}}^3/\text{kmol}\cdot \text{K}\right)\left(300°\text{C}+273\right)\text{K}-5}{{\left(30.45{\text{m}}^3/\text{kmol}\right)}^3}+\\ \frac{5\times 1.244\times {10}^{-4}}{{\left(30.45{\text{m}}^3/\text{kmol}\right)}^6}+\\ \left[\begin{array}{l}\frac{2.578\times {10}^5}{{\left(30.45{\text{m}}^3/\text{kmol}\right)}^3{T}_2^2}\\ \left(1+\frac{0.0060}{{\left(30.45{\text{m}}^3/\text{kmol}\right)}^2}\right)\exp \left(\frac{-0.0060}{{\left(30.45{\text{m}}^3/\text{kmol}\right)}^2}\right)\end{array}\right]\end{array}\right\}\\ =156.45+\left(8.7\times {10}^{-3}\right)+\left(3.93\times {10}^{-4}\right)+\left(7.8\times {10}^{-13}\right)+\left(2.78\times {10}^{-5}\right)\\ =156.5\text{kPa}\end{array}$

Thus, the final pressure of the methane treating it as a Benedict-Webb-Rubin gas is $156.5\text{ kPa}$.