Chapter 9, Problem 130QP

Interpretation Introduction

Interpretation:

The volume occupied by argon gas in the new system is to be determined.

Concept Introduction:

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

$PV=nRT$ … (1)

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

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

Answer to Problem 130QP

Solution:

The volume occupied by argon gas in new system is $2.88\text{ L}$ .

Explanation of Solution

Given Information: The volume of argon gas collected over water is $3\text{ L}$ at $27.0°\text{C}$ and $765\text{ torr}$ . The vapor pressure of water at $15°\text{C}$ is $13\text{ torr}$ .

A sample of $3\text{ L}$ argon gas is bubbled through and collected over water at $27.0°\text{C}$ and $765\text{ torr}$ . Thus, the number of moles present in the argon gas can be calculated using ideal gas law.

Convert temperature units from Celsius to Kelvin as,

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

For $27.0°\text{C}$ ,

$\begin{array}{c}{T}_1=27.0°\text{C}+273.15\\ =300.15\text{ K}\end{array}$

Covert pressure units from $\text{torr}$ to $\text{atm}$ as,

$\begin{array}{l}1\text{ atm}=760\text{ torr}\\ \text{1 torr}=\frac{1}{760}\text{ atm}\end{array}$

For $765\text{ torr}$ ,

$\begin{array}{c}\text{765 torr}=\frac{1\text{ atm}}{760\text{ torr}}\times \text{765 torr}\\ =1.01\text{ atm}\end{array}$

Substitute $V$ as $3\text{ L}$ , $T$ as $300.15\text{ K}$ , $P$ as $1.01\text{ atm}$ and $R$ as $0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}$ in equation (1).

$\begin{array}{c}1.01\text{ atm}\times 3\text{ L}=n\times 0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}\times 300.15\text{ K}\\ n=\frac{1.01\text{ atm}\times 3\text{ L}}{0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}\times 300.15\text{ K}}\\ =0.123\text{ mol}\end{array}$

Thus, $0.123\text{ mol}$ of argon gas is collected over water. The temperature and pressure of the new system are $15°\text{C}$ and $778\text{ torr}$ . The total pressure of the system and the partial pressure of water is known. So, the partial pressure of argon gas in the new system can be determined using Dalton’s law.

$P_{\text{Total}}=P_{\text{Argon}}+P_{\text{Water}}$ …(2)

Substitute $P_{\text{Water}}$ as $13\text{ torr}$ and $P_{\text{Total}}$ as $778\text{ torr}$ in equation (2).

$\begin{array}{c}778\text{ torr}=P_{\text{Argon}}+13\text{ torr}\\ P_{\text{Argon}}=778\text{ torr}-13\text{ torr}\\ =765\text{ torr}\end{array}$

The new pressure of argon gas in new system is $765\text{ torr}$ . Thus, the volume occupied by argon gas in new system is determined using ideal gas law.

Covert pressure units from $\text{torr}$ to $\text{atm}$ as,

$\begin{array}{l}1\text{ atm}=760\text{ torr}\\ \text{1 torr}=\frac{1}{760}\text{ atm}\end{array}$

For $765\text{ torr}$ ,

$\begin{array}{c}\text{765 torr}=\frac{1\text{ atm}}{760\text{ torr}}\times \text{765 torr}\\ =1.01\text{ atm}\end{array}$

Convert temperature units from Celsius to Kelvin as,

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

For $15.0°\text{C}$ ,

$\begin{array}{c}{T}_1=15.0°\text{C}+273.15\\ =298.15\text{ K}\end{array}$

Substitute $P$ as $1.01\text{ atm}$ , $T$ as $298.15\text{ K}$ , $n$ as $0.123\text{ mol}$ and $R$ as $0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}$ in equation (1).

$\begin{array}{c}1.01\text{ atm}\times V=0.123\text{ mol}\times 0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}\times 298.15\text{ K}\\ V=\frac{0.123\text{ mol}\times 0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}\times 298.15\text{ K}}{1.01\text{ atm}}\\ =2.88\text{ L}\end{array}$

Conclusion

The volume occupied by argon gas in the new system is $2.88\text{ L}$ .