Chapter 8, Problem 54QRT

(a)

Interpretation Introduction

Interpretation:

Final pressure inside the system has to be determined.

Concept Introduction:

Dalton’s law of partial pressure:

According to this law, the total pressure exerted by each gas in a mixture is equal to the sum of the individual partial pressure of the gases.

Boyle’s law:

At fixed temperature and number of molecules, the volume of a fixed amount of gas is inversely proportional to the pressure exerted by the gas.

  $\begin{array}{l}\text{P}\propto \frac{\text{1}}{\text{V}}\left(\text{n}\text{,}\text{T}\text{will}\text{be}\text{constant}\right)\\ \\ \text{PV}\text{=}\text{constant}\\ \\ {\text{P}}_{\text{1}}{\text{V}}_{\text{1}}\text{=}{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}\end{array}$

Answer to Problem 54QRT

Final pressure inside the system is $\text{1}\text{.98 atm}\text{.}$

Explanation of Solution

Given data is shown below,

  $\begin{array}{c}{P}_{O_2}\text{ = 1}\text{.46 atm}\\ \\ P_{N_2}\text{ = 0}\text{.908 atm}\\ \\ P_{Ar}\text{ = 2}\text{.71 atm}\\ \\ V_{O_2}\text{ = 3}\text{.0 L}\\ \\ V_{N_2}\text{ = 2}\text{.0 L}\\ \\ V_{Ar}\text{ = 5}\text{.0 L}\\ \\ {\text{V}}_{total}\text{ = }{V}_{O_2}+V_{N_2}\text{+ }{V}_{Ar}\\ \\ \text{ = 3}\text{.0 L}+2.0L\text{ + 5}\text{.0 L = 10}\text{.0 L}\end{array}$

Final pressure of $O_2,{\text{ N}}_2\text{ and Ar}$ is calculated using Boyle’s law as follows,

  $\begin{array}{c}{P}_i{V}_i{\text{ = P}}_f{V}_f\\ \\ {\text{P}}_f\text{ = }\frac{P_i{V}_i}{V_f}\end{array}$

Substitute the values to determine the final pressure,

  $\begin{array}{l}{P}_{f,O_2}\text{ = }\frac{\left(1.46\text{ atm}\right)\left(3.0\text{ L}\right)}{10.0\text{ L}}=\text{ 0}\text{.438 atm}\\ \\ P_{f,N_2}\text{ = }\frac{\left(0.908\text{ atm}\right)\left(2.0\text{ L}\right)}{10.0\text{ L}}=\text{ 0}\text{.182 atm}\\ \\ P_{f,Ar}\text{ = }\frac{\left(2.71\text{ atm}\right)\left(5.0\text{ L}\right)}{10.0\text{ L}}=\text{ 1}\text{.36 atm}\end{array}$

According to Dalton’s law, the total pressure of gases in a mixture of gases is the sum of the pressure of each gas in the mixture. Hence, final pressure inside the system can be determined as follows,

  $\begin{array}{c}{\text{P}}_{total}\text{ = }{P}_{f,O_2}\text{+ }{P}_{f,N_2}+P_{f,Ar}\\ \\ =\text{0}\text{.438 atm + 0}\text{.182 atm + 1}\text{.36 atm}\\ \\ \text{= 1}\text{.98 atm}\end{array}$

Final pressure inside the system is $1.98atm.$

(b)

Interpretation Introduction

Interpretation:

Partial pressure of $O_2,{\text{ N}}_2\text{ and Ar}$ has to be determined.

Concept Introduction:

Refer to (a).

Answer to Problem 54QRT

Partial pressure of $O_2$ is $0.438\text{ atm}\text{.}$

Partial pressure of $N_2$ is $0.182\text{ atm}\text{.}$

Partial pressure of $Ar$ is $1.36\text{ atm}\text{.}$

Explanation of Solution

Given data is shown below,

  $\begin{array}{c}{P}_{O_2}\text{ = 1}\text{.46 atm}\\ \\ P_{N_2}\text{ = 0}\text{.908 atm}\\ \\ P_{Ar}\text{ = 2}\text{.71 atm}\\ \\ V_{O_2}\text{ = 3}\text{.0 L}\\ \\ V_{N_2}\text{ = 2}\text{.0 L}\\ \\ V_{Ar}\text{ = 5}\text{.0 L}\\ \\ {\text{V}}_{total}\text{ = }{V}_{O_2}+V_{N_2}\text{+ }{V}_{Ar}\\ \\ \text{ = 3}\text{.0 L}+2.0L\text{ + 5}\text{.0 L = 10}\text{.0 L}\end{array}$

Final pressure of $O_2,{\text{ N}}_2\text{ and Ar}$ is calculated using Boyle’s law as follows,

  $\begin{array}{c}{P}_i{V}_i{\text{ = P}}_f{V}_f\\ \\ {\text{P}}_f\text{ = }\frac{P_i{V}_i}{V_f}\end{array}$

Substitute the values to determine the final pressure,

  $\begin{array}{l}{P}_{f,O_2}\text{ = }\frac{\left(1.46\text{ atm}\right)\left(3.0\text{ L}\right)}{10.0\text{ L}}=\text{ 0}\text{.438 atm}\\ \\ P_{f,N_2}\text{ = }\frac{\left(0.908\text{ atm}\right)\left(2.0\text{ L}\right)}{10.0\text{ L}}=\text{ 0}\text{.182 atm}\\ \\ P_{f,Ar}\text{ = }\frac{\left(2.71\text{ atm}\right)\left(5.0\text{ L}\right)}{10.0\text{ L}}=\text{ 1}\text{.36 atm}\end{array}$

Partial pressure of the gas is the pressure caused by each gas if it was alone inside the container.

  $\begin{array}{c}{P}_{O_2}\text{ = }{P}_{f,O_2}\text{ = 0}\text{.438 atm}\\ \\ P_{N_2}\text{ = }{P}_{f,N_2}\text{ = 0}\text{.182 atm}\\ \\ P_{Ar}\text{ = }{P}_{f,Ar}\text{ = 1}\text{.36 atm}\end{array}$

Partial pressure of $O_2$ is $0.438atm.$

Partial pressure of $N_2$ is $0.182atm.$

Partial pressure of $Ar$ is $1.36atm.$