Chapter 17.7, Problem 128RP

To determine

To derive an expression for the speed of sound based on van der Walls’s equation of state $P=\frac{RT}{v-b}-\frac{a}{v^2}$ and using this relation determine the speed of sound in carbon dioxide at $80°\text{C}$ and $320\text{kPa}$, and compare the result to that obtained by assuming ideal-gas behavior.

Answer to Problem 128RP

The speed of sound in carbon dioxide by using the relation obtained from the van der Walls equation is $289.43\text{m/s}$.

The speed of sound in carbon dioxide by assuming as ideal gas is $292.04\text{m/s}$.

Explanation of Solution

Write the given equation of state.

$P=\frac{RT}{v-b}-\frac{a}{v^2}$ (I)

Here, the pressure is $P$, the gas constant is $R$, the temperature is $T$, the specific volume is $v$, and the van der Walls constants are $a,b$.

Partially differentiate the Equation (I) with respect to specific volume $v$ by the keeping the temperature $T$ as constant.

$\begin{array}{c}{\left(\frac{\partial P}{\partial v}\right)}_T=\frac{\partial }{\partial v}{\left(\frac{RT}{v-b}-\frac{a}{v^2}\right)}_T\\ =\frac{\partial }{\partial v}{\left(\frac{RT}{v-b}\right)}_T-\frac{\partial }{\partial v}{\left(\frac{a}{v^2}\right)}_T\\ =RT\frac{\partial }{\partial v}\left(\frac{1}{v-b}\right)-a\frac{\partial }{\partial v}\left(\frac{1}{v^2}\right)\\ =RT\left[\frac{-1}{{\left(v-b\right)}^2}\right]-a\left(\frac{-2}{v^3}\right)\end{array}$

$=-\frac{RT}{{\left(v-b\right)}^2}+\frac{2a}{v^3}$ (II)

Write the relation of speed of the sound.

$c^2=k{\left(\frac{\partial P}{\partial \rho }\right)}_T$ (III)

Here, the specific heat ratio is $k$, the pressure is $P$, the density is $\rho$, and the symbol $\partial$ indicates the partial derivative of variables.

Write the relation between density and specific volume.

$\rho =\frac{1}{v}$

Partially differentiate the $\rho$ with respect to specific volume $v$.

$\begin{array}{l}\frac{\partial \rho }{\partial v}=\frac{\partial }{\partial v}\left(\frac{1}{v}\right)\\ \frac{\partial \rho }{\partial v}=\frac{-1}{v^2}\\ \partial \rho =\partial v\left(\frac{-1}{v^2}\right)\\ \partial \rho =-\frac{\partial v}{v^2}\end{array}$

Substitute $-\frac{\partial v}{v^2}$ for $\partial \rho$ in Equation (III).

$\begin{array}{c}{c}^2=k{\left(\frac{\partial P}{-\frac{\partial v}{v^2}}\right)}_T\\ =-k{v}^2{\left(\frac{\partial P}{\partial v}\right)}_T\end{array}$ (IV)

Substitute $-\frac{RT}{{\left(v-b\right)}^2}+\frac{2a}{v^3}$ for ${\left(\frac{\partial P}{\partial v}\right)}_T$ in Equation (IV).

$\begin{array}{c}{c}^2=-k{v}^2\left[-\frac{RT}{{\left(v-b\right)}^2}+\frac{2a}{v^3}\right]\\ =-\frac{\left(-k{v}^2\right)RT}{{\left(v-b\right)}^2}+\frac{\left(-k{v}^2\right)2a}{v^3}\\ =\frac{v^{2}kRT}{{\left(v-b\right)}^2}-\frac{2ak}{v}\end{array}$ (V)

Write the formula for velocity of sound at the given conditions of ${\text{CO}}_2$ by assuming it as ideal gas.

$c=\sqrt{kRT}$ (VI)

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

The molar mass $\left(M\right)$ of carbon dioxide is $44.01\text{kg/kmol}\simeq 44\text{kg/kmol}$.

The gas constant $\left(R\right)$ of carbon dioxide is $0.1889\text{kJ/kg}\cdot \text{K}$ or $0.1889\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}$.

Refer Table A-2, “Ideal-gas specific heats of various common gases”.

The specific heat ratio $\left(k\right)$ of carbon dioxide is $1.289$.

Conclusion:

Express the van der Walls constant $a$ and $b$ per unit mass as follows.

$\begin{array}{c}a=364.3\text{kPa}\cdot {\text{m}}^6{\text{/kmol}}^2\times \frac{1}{{\left(44\text{kg/kmol}\right)}^2}\\ =\frac{364.3\text{kPa}\cdot {\text{m}}^6{\text{/kmol}}^2}{1936{\text{kg}}^2{\text{/kmol}}^2}\\ =0.1882\text{kPa}\cdot {\text{m}}^6{\text{/kg}}^2\end{array}$

$\begin{array}{c}b=0.0427{\text{m}}^3\text{/kmol}\times \frac{1}{44\text{kg/kmol}}\\ =9.70\times {10}^{-4}{\text{m}}^3\text{/kg}\end{array}$

Substitute $320\text{kPa}$ for $P$, $0.1889\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}$ for $R$, $80°\text{C}$ for $T$, $9.70\times {10}^{-4}{\text{m}}^3\text{/kg}$ for $b$, and $0.1882\text{kPa}\cdot {\text{m}}^6{\text{/kg}}^2$ for $a$ in Equation (I).

$\begin{array}{l}320\text{kPa}=\frac{\left(0.1889\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(80°\text{C}\right)}{v-9.70\times {10}^{-4}{\text{m}}^3\text{/kg}}-\frac{0.1882\text{kPa}\cdot {\text{m}}^6{\text{/kg}}^2}{v^2}\\ 320\text{kPa}=\frac{\left(0.1889\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(80+273\right)\text{K}}{v-0.00097{\text{m}}^3\text{/kg}}-\frac{0.1882\text{kPa}\cdot {\text{m}}^6{\text{/kg}}^2}{v^2}\\ 320\text{kPa}=\frac{66.6817\text{kPa}\cdot {\text{m}}^3\text{/kg}}{v-0.00097{\text{m}}^3\text{/kg}}-\frac{0.1882\text{kPa}\cdot {\text{m}}^6{\text{/kg}}^2}{v^2}\end{array}$ (VII)

By using Equation solver or online calculator solve the Equation (VII) and the value of $v$ is obtained as follows.

$v=0.2065{\text{m}}^3\text{/kg}$

Substitute $0.2065{\text{m}}^3\text{/kg}$ for $v$, $1.279$ for $k$, $0.1889\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}$ for $R$, $80°\text{C}$ for $T$, $9.70\times {10}^{-4}{\text{m}}^3\text{/kg}$ for $b$, and $0.1882\text{kPa}\cdot {\text{m}}^6{\text{/kg}}^2$ for $a$ in Equation (V).

$\begin{array}{l}{c}^2=\left[\begin{array}{l}\frac{{\left(0.2065{\text{m}}^3\text{/kg}\right)}^2\left(1.279\right)\left(0.1889\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(80°\text{C}\right)}{{\left(0.2065{\text{m}}^3\text{/kg}-9.70\times {10}^{-4}{\text{m}}^3\text{/kg}\right)}^2}\\ -\frac{2\left(0.1882\text{kPa}\cdot {\text{m}}^6{\text{/kg}}^2\right)\left(1.279\right)}{0.2065{\text{m}}^3\text{/kg}}\end{array}\right]\\ c^2=\left[\begin{array}{l}\frac{\left(0.05454{\text{m}}^6{\text{/kg}}^2\right)\left(0.1889\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(80+273\right)\text{K}}{0.04224{\text{m}}^6{\text{/kg}}^2}\\ -2.3313\text{kPa}\cdot {\text{m}}^3\text{/kg}\end{array}\right]\\ c^2=\left[\begin{array}{l}\frac{\left(0.05454{\text{m}}^6{\text{/kg}}^2\right)\left(66.6817\text{kPa}\cdot {\text{m}}^3\text{/kg}\right)}{0.04224{\text{m}}^6{\text{/kg}}^2}\\ -2.3313\text{kPa}\cdot {\text{m}}^3\text{/kg}\end{array}\right]\\ c^2=86.0989\text{kPa}\cdot {\text{m}}^3\text{/kg}-2.3313\text{kPa}\cdot {\text{m}}^3\text{/kg}\end{array}$

$\begin{array}{l}{c}^2=83.7676\text{kPa}\cdot {\text{m}}^3\text{/kg}\times \frac{1000{\text{m}}^2{\text{/s}}^2}{1\text{kPa}\cdot {\text{m}}^3\text{/kg}}\\ c=\sqrt{83767.6{\text{m}}^2{\text{/s}}^2}\\ c=289.4263\text{m/s}\\ \simeq 289.43\text{m/s}\end{array}$

Thus, the speed of sound in carbon dioxide by using the relation obtained from the van der Walls equation is $289.43\text{m/s}$.

When the carbon dioxide is assumed as ideal gas, the velocity of sound is determined as follows.

Substitute $0.1889\text{kJ/kg}\cdot \text{K}$ for R, $1.279$ for k, and $80°\text{C}$ for $T$ in Equation (VI).

$\begin{array}{c}c=\sqrt{1.279\times 0.1889\text{kJ/kg}\cdot \text{K}\times 80°\text{C}}\\ =\sqrt{1.279\times 0.1889\text{kJ/kg}\cdot \text{K}\times \left(80+273\right)\text{K}}\\ =\sqrt{85.2859\text{kJ/kg}\times \frac{1000{\text{m}}^2{\text{/s}}^2}{1\text{kJ/kg}}}\\ =292.0375\text{m}/\text{s}\end{array}$

$\simeq 292.04\text{m}/\text{s}$

Thus, the speed of sound in carbon dioxide by assuming as ideal gas is $292.04\text{m/s}$.