Chapter 20, Problem 43PQ

To determine

The pressure for which Van der Waals equation of state result in a volume 2% larger than that predicted by the ideal gas law for 1 mol of nitrogen at a temperature of 0°C.

Answer to Problem 43PQ

The pressure is $\underset{\_}{P_0=2.26\times {10}^7\text{Pa}\text{or}\text{-1}\text{.96}\times {\text{10}}^6\text{Pa}}$.

Explanation of Solution

Let the ideal volume be, $V_{\text{ideal}}=V_0$.

The Van der Waals volume is 2% larger than gas volume.

  $\begin{array}{c}{V}_{\text{vdW}}=V_0+\frac{2}{100}{V}_0\\ =1.02V_0\end{array}$

Write the mathematical expression of ideal gas law.

  $P_0{V}_0=nRT$                                                                                                                (I)

Here, $P_0$ is the new pressure, $V_0$ is the ideal gas volume, $n$ is the number of moles, $R$ is the gas constant, and $T$ is the temperature.

Write the Van der Waals equation.

  $\left(P+a\frac{n^2}{V^2}\right)\left(V-nb\right)=nRT$                                                                                      (II)

Here, $a$ and $b$ are constants.

Conclusion:

Substitute, $1.00\text{mol}$ for $n$, $83.14\text{J/mol}\text{.K}$ for $R$, and $273.15\text{K}$ for $T$ in equation (I).

  $\begin{array}{c}{P}_0{V}_0=\left(1.00\text{mol}\right)\left(83.14\text{J/mol}\text{.K}\right)\left(273.15\text{K}\right)\\ =2271\text{J}\end{array}$                                                         (II)

Rearrange equation (II) to obtain an expression for $V_0$.

  $V_0=\frac{2271\text{J}}{P_0}$

Substitute, $0.139\text{Pa}\cdot {\text{m}}^6/{\text{mol}}^{\text{2}}$ for $a$, $1$ for $n$, $1.02V_0$ for $V$, $3.91\times {10}^{-5}{\text{m}}^3/\text{mol}$, $1.00\text{mol}$ for $n$, $83.14\text{J/mol}\text{.K}$ for $R$, and $273.15\text{K}$ for $T$ in equation (III).

  $\begin{array}{c}\left(P_0+\left(0.139\text{Pa}\cdot {\text{m}}^6/{\text{mol}}^{\text{2}}\right)\frac{1^2}{{\left(1.02V_0\right)}^2}\right)\left(1.02V_0-3.91\times {10}^{-5}{\text{m}}^3/\text{mol}\right)=\left(1.00\text{mol}\right)\left(83.14\text{J/mol}\text{.K}\right)\left(273.15\text{K}\right)\\ \left(P_0+\frac{0.1336}{{\left(V_0\right)}^2}\right)\left(1.02V_0-3.91\times {10}^{-5}\right)=2271\end{array}$    (III)

Substitute, $\frac{2271\text{J}}{P_0}$ for $V_0$ in above equation.

  $\begin{array}{c}\left(P_0+\frac{0.1336}{{\left(\frac{2271}{P_0}\right)}^2}\right)\left(1.02\left(\frac{2271}{P_0}\right)-3.91\times {10}^{-5}\right)=2271\\ -1.013\times {10}^{-12}{P}_0^2+2.091\times {10}^{-5}{P}_0+45=0\end{array}$                                           (IV)

Rearrange equation (IV) to obtain values for $P_0$.

  $\begin{array}{c}{P}_0=\frac{-2.091\times {10}^{-5}\pm \sqrt{{\left(2.091\times {10}^{-5}\right)}^2-4\left(-1.013\times {10}^{-12}\right)\left(45\right)}}{2\left(-1.013\times {10}^{-12}\right)}\\ =2.26\times {10}^7\text{Pa or -1}\text{.96}\times {\text{10}}^6\text{Pa}\end{array}$

Therefore, the pressure for which Van der Waals equation of state result in a volume 2% larger than that predicted by the ideal gas law for 1 mol of nitrogen at a temperature of 0°C is $\underset{\_}{P_0=2.26\times {10}^7\text{Pa}\text{or}\text{-1}\text{.96}\times {\text{10}}^6\text{Pa}}$.