Chapter 10, Problem 149AP

A 5.00-mol sample of ${\text{NH}}_{\text{3}}$ gas is kept in a 1.92-L container at 300 K. If the van der Waals equation is assumed to give the correct answer for the pressure of the gas. calculate the percent error made in using the ideal gas equation to calculate the pressure.

Interpretation Introduction

Interpretation:

The percent error in pressure value from the ideal gas equation as compared to the van der Waal’s equation for ${\text{NH}}_3$ gas is to be calculated.

Concept introduction:

The ideal gas equation elaborates the physical properties of gases by relating the pressure, volume, temperature and number of moles linked with each other with the help of gas laws.

This can be shown by:

$P=\frac{nRT}{V}$

Here, R represents the universal gas constant, $n$ is the number of moles, $T$ is the temperature, $P$ is the pressure, and $V$ is the volume.

The percentage error is equal to the percent of difference between exact and experimental values, that is, if this difference is divided by the exact value and multiplied by 100 to make percent, then the percent error is obtained.

$\text{PE=}\frac{\text{|experimental value-exact value|}}{\text{|exact value|}}\text{×100}$

The deviation of real gas from the ideal gas can be calculated with the help of van der walls equation as follows:

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

Here, a and b represent the van der wall constants, which have different values for different gases, R represents the gas constant, $n$ is the number of moles, $T$ is the temperature, $P$ is the pressure, and $V$ is the volume.

Answer to Problem 149AP

Solution: $50\%$

Explanation of Solution

Given information:

Number of moles $n=5\text{ mol}$

Volume $V=1.92\text{ L}$

Temperature $T=300\text{ K}$

The equation for an ideal gas is as follows:

$P=\frac{nRT}{V}$

Substitute $5\text{ mol}$ for $n$, $300\text{ K}$ for $T$, $1.92\text{ L}$ for $V$, and $0.08206\text{ L}\text{.atm}/\text{K}\text{.mol}$ for $R$ in the above equation

$\begin{array}{c}P=\frac{\left(\text{5 }\cancel{\text{mol}}\right)\left(0.08206\cancel{\text{L}}\text{.atm}/\cancel{\text{K}}\text{.}\cancel{\text{mol}}\right)\left(300\cancel{\text{K}}\right)}{\text{1}\text{.92 }\cancel{\text{L}}}\\ =\frac{123.09}{\text{1}\text{.92}}\text{ atm}\\ =64.1\text{ atm}\end{array}$ $\begin{array}{c}P=\frac{\left(\text{5 mol}\right)\left(0.08206\text{ L}\text{.atm}/\text{K}\text{.mol}\right)\left(300\text{ K}\right)}{\text{1}\text{.92 L}}\\ =\frac{123.09}{\text{1}\text{.92}}\text{ atm}\\ =64.1\text{ atm}\end{array}$

The value of van der walls constant $a$ for ${\text{NH}}_3$ is $4.17\text{ atm}{\text{.L}}^2/mo{l}^2$.

The value of van der walls constant $b$ for ${\text{NH}}_3$ is $0.0371\text{ L}/mol$.

Now, the corrected value in both pressure and volume can be calculated as follows:

For pressure,

$\begin{array}{c}\frac{a{n}^2}{V^2}=\frac{\left(4.17\text{ atm}{\text{.L}}^2/mo{l}^2\right){\left(5\text{ mol}\right)}^2}{{\left(1.92\text{ L}\right)}^2}\\ =\frac{\left(4.17\text{ atm}\text{.}\cancel{{\text{L}}^2}/\cancel{mo{l}^2}\right)\left(25\cancel{{\text{mol}}^2}\right)}{3.69\cancel{{\text{L}}^2}}\\ =\frac{104.25}{3.69}\text{ atm}\\ =28.25\text{ atm}\end{array}$ $\begin{array}{c}\frac{a{n}^2}{V^2}=\frac{\left(4.17\text{ atm}{\text{.L}}^2/mo{l}^2\right){\left(5\text{ mol}\right)}^2}{{\left(1.92\text{ L}\right)}^2}\\ =\frac{\left(4.17\text{ atm}{\text{.L}}^2/mo{l}^2\right)\left(25{\text{ mol}}^2\right)}{3.69{\text{ L}}^2}\\ =\frac{104.25}{3.69}\text{ atm}\\ =28.25\text{ atm}\end{array}$

For volume,

$\begin{array}{c}nb=\left(5\text{ mol}\right)\left(0.0371\text{ L}/mol\right)\\ =0.1855\text{ L}\end{array}$

The van der walls equation for a gas is as follows:

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

Substitute $28.25\text{ atm}$ for $a{n}^2/V^2$, $0.1855\text{ L}$ for $nb$, $5\text{ mol}$ for $n$, $300\text{ K}$ for $T$, $1.92\text{ L}$ for $V$, and $0.08206\text{ L}\text{.atm}/\text{K}\text{.mol}$ for $R$ in the van der walls equation

$\begin{array}{c}\left(P+28.25\text{ atm}\right)\left(1.92\text{ L}-0.1855\text{ L}\right)=\left(0.08206\text{ L}\text{.atm}/\text{K}\text{.mol}\right)\left(\text{5 mol}\right)\left(300\text{ K}\right)\\ \left(P+28.25\text{ atm}\right)\left(1.7345\text{ L}\right)=123.09\text{ L}\text{.atm}\\ \left(P+28.25\text{ atm}\right)=\frac{123.09\cancel{\text{L}}\text{.atm}}{1.73\cancel{\text{L}}}\\ =70.94\text{ atm}\end{array}$ $\begin{array}{c}\left(P+28.25\text{ atm}\right)\left(1.92\text{ L}-0.1855\text{ L}\right)=\left(0.08206\text{ L}\text{.atm}/\text{K}\text{.mol}\right)\left(\text{5 mol}\right)\left(300\text{ K}\right)\\ \left(P+28.25\text{ atm}\right)\left(1.7345\text{ L}\right)=123.09\text{ L}\text{.atm}\\ \left(P+28.25\text{ atm}\right)=\frac{123.09\text{ L}\text{.atm}}{1.73\text{ L}}\\ =70.94\text{ atm}\end{array}$

The value of pressure can be solved as

$\begin{array}{c}P=70.94\text{ atm}-28.25\text{ atm}\\ =\text{42}\text{.7 atm}\end{array}$

The error in pressure value from the ideal gas equation can be calculated as

$\text{Percent error}=\frac{\left|P_{\text{vanderwall}}-P_{\text{ideal}}\right|}{P_{\text{vanderwall}}}\times 100\%$

Substitute $42.7\text{ atm}$ for $P_{\text{vanderwall}}$ and $64.1\text{ atm}$ for $P_{\text{ideal}}$ in the above equation

$\begin{array}{c}\text{Percent error}=\frac{\left|42.7-64.1\right|}{42.7}\times 100\%\\ =\frac{21.4}{42.7}\times 100\%\\ =0.50\times 100\%\\ =50\%\end{array}$

Conclusion

Hence, the percent error in pressure value from the ideal gas equation is $50\%$.