Chapter 3.8, Problem 97P

Nitrogen at 150 K has a specific volume of 0.041884 m³/kg. Determine the pressure of the nitrogen using (a) the ideal-gas equation and (b) the Beattie-Bridgeman equation. Compare your results to the experimental value of 1000 kPa.

To determine

The pressure of the nitrogen using the ideal gas equation.

Answer to Problem 97P

The pressure of the nitrogen using the ideal gas equation is $\underset{\_}{1063\text{kPa}}$.

Explanation of Solution

Determine the pressure of the tank using the ideal gas equation.

$P=\frac{RT}{v}$ (I)

Here, the universal gas constant is $R$, the specific volume of the tank is $v$, and the temperature of the tank is $T$.

Conclusion:

Refer to Table A-1 to find the gas constant and molar mass of nitrogen as $0.2968\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}$ and $28.013\text{kg}/\text{kmol}$.

Substitue $0.2968\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}$ for $R$, $150\text{K}$ for $T$, and $0.041884{\text{m}}^{\text{3}}/\text{kg}$ for $v$ in Equation (I).

$\begin{array}{c}P=\frac{\left(0.2968\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(150\text{K}\right)}{\left(0.041884{\text{m}}^{\text{3}}/\text{kg}\right)}\\ =\frac{\left(44.52\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\right)}{\left(0.041884{\text{m}}^{\text{3}}/\text{kg}\right)}\\ =1062.9\text{kPa}\\ \cong 1063\text{kPa}\end{array}$

Therefore, the error compression with the actual value is 6.3%.

Thus, the pressure of the nitrogen using the ideal gas equation is $\underset{\_}{1063\text{kPa}}$.

To determine

The pressure of the nitrogen using the Beattie-Bridgeman equation.

Answer to Problem 97P

The pressure of the nitrogen using the Beattie-Bridgeman equation is $\underset{\_}{1000.4\text{kPa}}$.

Explanation of Solution

Determine the pressure in the tank using the Beattie-Bridgeman equation.

$\begin{array}{c}P=\frac{R_{u}T}{{\overline{v}}^2}\left(1-\frac{c}{\overline{v}{T}^3}\right)\left(\overline{v}+\left(B_{\text{0}}\left(1-\frac{b}{\overline{v}}\right)\right)\right)-\frac{\left(A_0\left(1-\frac{a}{\overline{v}}\right)\right)}{{\overline{v}}^2}\\ =\frac{R_{u}T}{{\left(Mv\right)}^2}\left(1-\frac{c}{\left(Mv\right)T^3}\right)\left(\left(Mv\right)+\left(B_{\text{0}}\left(1-\frac{b}{\left(Mv\right)}\right)\right)\right)-\frac{\left(A_0\left(1-\frac{a}{\left(Mv\right)}\right)\right)}{{\left(Mv\right)}^2}\end{array}$ (II)

Here, the five constant are $A_0,a,B_0,b,\text{and}c$.

Conclusion:

From the Table 3-4, “Constant that appear in the Beattie-Bridgeman and the Benedict-Webb-Rubin equation of state” to obtain of value of the five constant are $A_0,a,B_0,b,\text{and}c$ as 136.2315, 0.02617, 0.05046, -0.00691, and $4.20\times {10}^4$.

Substitute 136.2315 for $A_0$, 0.02617 for $a$, 0.05046 for $B_0$, -0.00691 for $b$, $4.20\times {10}^4{\text{m}}^{\text{3}}\cdot {\text{K}}^{\text{3}}/\text{kmol}$ for $c$, $0.041884{\text{m}}^{\text{3}}/\text{kg}$ for $v$, $28.013\text{kg}/\text{kmol}$ for $M$ $150\text{K}$ for $T$, and $8.314\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kmol}\cdot \text{K}$ for $R$ in Eqaution (IV).

$\begin{array}{c}P=\left[\begin{array}{l}\left(\frac{\left(8.314\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kmol}\cdot \text{K}\right)\left(150\text{K}\right)}{{\left(\left(28.013\text{kg}/\text{kmol}\right)\times \left(0.041884{\text{m}}^{\text{3}}/\text{kg}\right)\right)}^2}\right)\\ \times \left(1-\frac{\left(4.20\times {10}^4{\text{m}}^{\text{3}}\cdot {\text{K}}^{\text{3}}/\text{kmol}\right)}{\left(\left(28.013\text{kg}/\text{kmol}\right)\times \left(0.041884{\text{m}}^{\text{3}}/\text{kg}\right)\right){\left(150\text{K}\right)}^3}\right)\times \\ \left(\begin{array}{l}\left(\left(28.013\text{kg}/\text{kmol}\right)\times \left(0.041884{\text{m}}^{\text{3}}/\text{kg}\right)\right)\\ +\left(\left(0.05046\right)\left(1-\frac{\left(-0.00691\right)}{\left(\left(28.013\text{kg}/\text{kmol}\right)\times \left(0.041884{\text{m}}^{\text{3}}/\text{kg}\right)\right)}\right)\right)\end{array}\right)\\ -\frac{\left(\left(136.2315\right)\left(1-\frac{\left(0.02617\right)}{\left(\left(28.013\text{kg}/\text{kmol}\right)\times \left(0.041884{\text{m}}^{\text{3}}/\text{kg}\right)\right)}\right)\right)}{{\left(\left(28.013\text{kg}/\text{kmol}\right)\times \left(0.041884{\text{m}}^{\text{3}}/\text{kg}\right)\right)}^2}\end{array}\right]\\ =\left[\begin{array}{l}\left(\frac{\left(1247.1\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kmol}\right)}{{\left(1.1733{\text{m}}^{\text{3}}/\text{kmol}\right)}^2}\right)\times \left(1-\frac{\left(4.20\times {10}^4{\text{m}}^{\text{3}}\cdot {\text{K}}^{\text{3}}/\text{kmol}\right)}{\left(1.1733{\text{m}}^{\text{3}}/\text{kmol}\right)\left(3375000{\text{K}}^3\right)}\right)\times \\ \left(\left(1.1733{\text{m}}^{\text{3}}/\text{kmol}\right)+\left(0.05076{\text{m}}^{\text{3}}/\text{kmol}\right)\right)-\frac{\left(133.193{\text{m}}^{\text{3}}/\text{kmol}\right)}{{\left(1.1733{\text{m}}^{\text{3}}/\text{kmol}\right)}^2}\end{array}\right]\\ =\left[\begin{array}{l}\left(\frac{\left(1247.1\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kmol}\right)}{\left(1.615203{{\text{m}}^{\text{6}}/\text{kmol}}^2\right)}\right)\times \left(1-\frac{\left(4.20\times {10}^4{\text{m}}^{\text{3}}\cdot {\text{K}}^{\text{3}}/\text{kmol}\right)}{\left(1.1733{\text{m}}^{\text{3}}/\text{kmol}\right)\left(3375000{\text{K}}^3\right)}\right)\times \\ \left(\left(1.1733{\text{m}}^{\text{3}}/\text{kmol}\right)+\left(0.05076{\text{m}}^{\text{3}}/\text{kmol}\right)\right)-\frac{\left(133.193{\text{m}}^{\text{3}}/\text{kmol}\right)}{\left(1.615203{{\text{m}}^{\text{6}}/\text{kmol}}^2\right)}\end{array}\right]\\ =1000.4\text{kPa}\end{array}$

Therefore, the error compression with the actual value is negligible error.

Thus, the pressure of the nitrogen using the Beattie-Bridgeman equation is $\underset{\_}{1000.4\text{kPa}}$.