Chapter 13, Problem 34A

Interpretation Introduction

Interpretation: -The pressure that each gas would exert if present alone in that space is known as the partial pressure of gas.

Concept Introduction: - The total pressure of a mixture of gases which do not react chemically is the sum of the partial pressure of the component gases.Total pressure of the gaseous mixture = sum of partial pressure of each gas

  $P=P_1+P_2+P_3$

Answer to Problem 34A

1) Partial pressure of ${\mathrm{N}}_2=3800Torr$

2) Partial pressure of ${\mathrm{O}}_2=2508Torr$

3) Partial pressure of ${\mathrm{CO}}_2=1216Torr$

Explanation of Solution

In 1807 John Dalton put forward a generalization dealing with the pressure exerted by the mixture of two or more chemically non − reacting gases enclosed in a same vessel.

This is known as Dalton’s law of partial pressure.

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

Where,

  $\mathrm{P}$ = total pressure.

  ${\mathrm{P}}_1$ = Partial pressure of gas 1

  ${\mathrm{P}}_2$ = Partial pressure of gas 2

  ${\mathrm{P}}_3$ = Partial pressure of gas 3

  ${\mathrm{P}}_n$ = Partial pressure for n gases.

According to Ideal gas equation

  $\begin{array}{l}PV=nRT\\ P_T=\frac{n_{T}RT}{V}\end{array}$

  $\begin{array}{l}\text{n = number of moles}\text{.}\\ \text{P = pressure exerted by gas }\\ \text{V = volume of container}\\ \text{T = Temperature}\\ \text{R = Gas constant}\end{array}$

For calculating the total pressure

  ${\mathrm{P}}_T=\frac{n_{T}RT}{V}$

  $\begin{array}{l}{P}_T=P_1+P_2+P_3\\ n_T=n_1+n_2+n_3\end{array}$

  $\begin{array}{l}{P}_1=\frac{n_{1}RT}{V},P_2=\frac{n_{2}RT}{V},P_3=\frac{n_{3}RT}{V}\\ so,P_T=\frac{(n_1+n_2+n_3)RT}{V}\end{array}$

For calculating partial pressure of each gas individually.

  $\begin{array}{l}\frac{P_1}{P_T}=\frac{n_1\raisebox{1ex}{$RT$}\!\left/ \!\raisebox{-1ex}{$V$}\right.}{n_T\raisebox{1ex}{$RT$}\!\left/ \!\raisebox{-1ex}{$V$}\right.}=\frac{P_1}{P_T}=\frac{n_1}{n_T}\\ P_1=\frac{n_1}{n_T}\times P_T,\\ P_2=\frac{n_2}{n_T}\times P_T,\\ P_3=\frac{n_3}{n_T}\times P_T\end{array}$

Here, the three gases are

1) Nitrogen = 3.0 moles.

2) Oxygen = 2.0 moles

3) Carbon dioxide = 1.0 moles

Total moles ( ${\mathrm{n}}_T$ ) = 3.0 + 2.0 + 1.0 = 6.0 moles

Total Pressure = 10 atm

Temperature = ${25}^{0}C$

1) Partial pressure of Nitrogen

  $\frac{P_1}{P_T}=\frac{n_1}{n_T},P_1=\frac{n_1}{n_T}\times P_T$

  $\begin{array}{l}{P}_1=\frac{3.0}{6.0}\times 10=\frac{30}{6}=5atm\\ P_1=5atm\end{array}$

2) Partial pressure of Oxygen

  $\frac{P_2}{P_T}=\frac{n_2}{n_T},P_2=\frac{n_2}{n_T}\times P_T$

  $\begin{array}{l}{P}_2=\frac{2.0}{6.0}\times 10=\frac{20}{6}=3.3atm\\ P_2=3.3atm\end{array}$

3) Partial pressure of Oxygen

  $\frac{P_3}{P_T}=\frac{n_3}{n_T},P_2=\frac{n_3}{n_T}\times P_T$

  $\begin{array}{l}{P}_3=\frac{1.0}{6.0}\times 10=\frac{10}{6}=1.6atm\\ P_3=1.6atm\end{array}$

Therefore,

  $\begin{array}{l}{P}_1=5atm\\ P_2=3.3atm\\ P_3=1.6atm\end{array}$

Now, we have to covert the values of partial pressure in Torr.

1atm = 760Torr

So, ${\text{P}}_{\text{1}}\text{ = 5atm = 5×760 = 3800Torr}$

  ${\text{P}}_2\text{ = 3}\text{.3atm = 3}\text{.3×760 = 2508Torr}$

  ${\text{P}}_3\text{ = 1}\text{.6atm = 1}\text{.6×760 = 1216Torr}$

Conclusion

1) Partial pressure of ${\mathrm{N}}_2=3800Torr$

2) Partial pressure of ${\mathrm{O}}_2=2508Torr$

3) Partial pressure of ${\mathrm{CO}}_2=1216Torr$