Chapter 9, Problem 132QP

Interpretation Introduction

Interpretation:

The total pressure of mixture of gas and partial pressure of argon gas and nitrogen gas in the tank at $45°\text{C}$ are to be determined.

Concept Introduction:

The number of moles $\left(n\right)$ present in the gas container is equal to the ratio of amount of gas $\left(m\right)$ present in the container to the molar mass of the gas $\left(M\right)$ .

$n=\frac{m}{M}$ … (1)

Ideal gas law is followed by ideal gases. It states that pressure $\left(P\right)$ and volume $\left(V\right)$ of the gas are directly proportional to the number of moles $\left(n\right)$ present in the gas and the temperature $\left(T\right)$ of the gas.

$\begin{array}{c}PV\propto nT\\ PV=nRT\end{array}$ … (2)

Where, $R$ is universal gas constant.

Dalton’s law of partial pressure states that the total pressure of the mixture of gases is equal to the sum of partial pressure of gases present in the mixture of gases.

$P_{Total}=P_1+P_2+P_3+\mathrm{..........}+P_n$

Answer to Problem 132QP

Solution:

The partial pressure of argon gas and nitrogen gas in the tank are $21.5\text{ atm}$ and $33.6\text{ atm}$ , respectively. The total pressure of the mixture of gases in the tank is $55.1\text{ atm}$ .

Explanation of Solution

Given Information: The volume of the tank is $2.00\text{ L}$ . Tank contains $72.0{\text{ g N}}_2$ and $66.0\text{g Ar}$ .

The tank contains a mixture of gases $72.0{\text{ g N}}_2$ and $66.0\text{g Ar}$ . The moles of each gas present in the tank can be calculated by using equation (1). The molar mass of $\text{Ar}$ gas is $39.95{\text{ g mol}}^{-1}$ . Substitute $m_{\text{Ar}}$ as $66.0\text{g}$ and $M_{\text{Ar}}$ as $39.95{\text{ g mol}}^{-1}$ in equation (1) to determine the no. of moles of argon gas.

$\begin{array}{c}{n}_{\text{Ar}}=\frac{m_{\text{Ar}}}{M_{\text{Ar}}}\\ =\frac{66.0\text{ g}}{\text{39}{\text{.95 g mol}}^{-1}}\\ =1.65\text{ mol}\end{array}$

The molar mass of ${\text{N}}_2$ gas is $28{\text{ g mol}}^{-1}$ . Substitute $m_{{\text{N}}_2}$ as $72.0\text{ g}$ and $M_{{\text{N}}_2}$ as $28{\text{ g mol}}^{-1}$ in equation (1) to determine the number of moles of nitrogen gas.

$\begin{array}{c}{n}_{{\text{N}}_2}=\frac{m_{{\text{N}}_2}}{M_{{\text{N}}_2}}\\ =\frac{72.0\text{ g}}{28{\text{ g mol}}^{-1}}\\ =2.57\text{mol}\end{array}$

The moles of argon gas are $1.65\text{ mol}$ and the moles of nitrogen gas are $2.57\text{mol}$ in the tank. The temperature of the mixture of gas is $45.0°\text{C}$ and the volume of the gas is $2\text{ L}$ . Pressure exerted by moles of argon gas is determined by equation (2).

Convert temperature units from Celsius to Kelvin.

$T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15$

For $45.0°\text{C}$ ,

$\begin{array}{c}T=45.0°\text{C}+273.15\\ =318.2\text{ K}\end{array}$

Substitute $T$ as $318.2\text{ K}$ , $n$ as $1.65\text{ mol}$ , $V$ as $2\text{L}$ , and $R$ as $0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}$ in equation (2) to determine the partial pressure of argon gas.

$\begin{array}{c}{P}_{Ar}\times 2\text{ L}=1.65\text{ mol}\times 0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}\times 318.2\text{ K}\\ P_{Ar}=\frac{1.65\text{ mol}\times 0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}\times 318.2\text{ K}}{\text{2 L}}\\ =21.5\text{ atm}\end{array}$

Therefore, the partial pressure of argon gas in the tank is $21.5\text{ atm}$ . Now, to determine the partial pressure of nitrogen gas, substitute $T$ as $318.2\text{ K}$ , $n$ as $2.57\text{mol}$ , $V$ as $2\text{L}$ , and $R$ as $0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}$ in equation (2).

$\begin{array}{c}{P}_{{\text{N}}_2}\times 2\text{ L}=2.57\text{ mol}\times 0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}\times 318.2\text{ K}\\ P_{{\text{N}}_2}=\frac{2.57\text{ mol}\times 0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}\times 318.2\text{ K}}{2\text{ L}}\\ =33.6\text{ atm}\end{array}$

Therefore, the partial pressure of nitrogen gas in the tank is $33.6\text{ atm}$ .

The total pressure of the mixture of gases in the tank is the sum of the partial pressure of argon gas and nitrogen gas according to Dalton’s law. Therefore, the total pressure of gases is,

$\begin{array}{c}{P}_T=P_{\text{Ar}}+P_{{\text{N}}_2}\\ =21.5\text{ atm}+33.6\text{ atm}\\ =55.1\text{ atm}\end{array}$

Therefore, the total pressure of the mixture of gases in the tank is $55.1\text{ atm}$ .

Conclusion

The partial pressure of argon gas and nitrogen gas in the tank are $21.5\text{ atm}$ and $33.6\text{ atm}$ . The total pressure of the mixture of gases in the tank is $55.1\text{ atm}$ .