Chapter 13, Problem 63A

Interpretation Introduction

Interpretation:

The volume of the system at STP is to be determined. Also, the partial pressure of each gas is to be determined.

Concept introduction:

The ideal gas equation for one mole gas is written as

  $\text{PV}\text{=}\text{RT}$

P= pressure of the gas

V= volume of the gas

R= gas constant

T = temperature of the gas

Dalton’s law also says that the entire pressure of the system is equal to the sum of the partial pressure of individual ones.

  ${\text{P}}_{\text{T}}\text{=}{\text{P}}_{\text{1}}\text{+}{\text{P}}_{\text{2}}\text{+}{\text{P}}_{\text{3}}$

Answer to Problem 63A

The volume of the mixture at STP is $\text{32}\text{.4L}$

The partial pressure of gases He, Ar, and Ne are $\text{0}\text{.864}\text{atm}$ , $\text{0}\text{.0173}\text{atm}$ and $\text{0}\text{.129}\text{atm}$ respectively.

Explanation of Solution

Given information:

Standard temperature, (T) = $\text{0°C}$

Initial temperature in K= $\text{0+273}\text{=}\text{273K}$

Standard pressure (P₁) = $\text{1atm}$

Mass of helium gas = $\text{5g}$

The Molar mass of Helium gas = $\text{4}\text{.0026}\text{g/mol}$

Mass of argon gas = $\text{1g}$

The Molar mass of argon gas = $\text{39}\text{.948}\text{g/mol}$

Mass of neon gas = $\text{3}\text{.5g}$

The Molar mass of neon gas = $\text{20}\text{.1797}\text{g/mol}$

Calculation:

To find the number of moles of He, Ar, and Ne atom:

Number of moles of helium is

  $\begin{array}{l}{\text{n}}_{\text{He}}\text{=}\frac{\text{mass}}{\text{molar}\text{mass}\text{of}He}\\ {\text{n}}_{\text{He}}\text{=}\frac{\text{5}}{\text{4}\text{.0026}}\\ \text{=}\text{1}\text{.249}\text{moles}\end{array}$

The number of moles of argon is

  $\begin{array}{l}{\text{n}}_{\text{Ar}}\text{=}\frac{\text{mass}}{\text{molar}\text{mass}\text{of}\text{Ar}}\\ {\text{n}}_{\text{Ar}}\text{=}\frac{\text{1}}{\text{39}\text{.948}}\\ \text{=}\text{0}\text{.025}\text{moles}\end{array}$

The number of moles of neon is

  $\begin{array}{l}{\text{n}}_{\text{Ne}}\text{=}\frac{\text{mass}}{\text{molar}\text{mass}\text{of}\text{Ne}}\\ {\text{n}}_{\text{Ne}}\text{=}\frac{\text{3}\text{.5}}{\text{20}\text{.1797}}\\ \text{=}\text{0}\text{.1734}\text{moles}\end{array}$

To find the volume of the given atoms

  $\begin{array}{l}\text{ccordingtoidealgasequation}\\ \text{V}\text{=}\frac{\text{nRT}}{\text{P}}\\ \text{So,}{\text{V}}_{\text{He}}\text{=}\frac{{\text{n}}_{\text{He}}\text{RT}}{\text{P}}{\text{V}}_{\text{Ar}}\text{=}\frac{{\text{n}}_{\text{Ar}}\text{RT}}{\text{P}}{\text{V}}_{\text{Ne}}\text{=}\frac{{\text{n}}_{\text{Ne}}\text{RT}}{\text{P}}\\ {\text{V}}_{\text{total}}\text{=}{\text{V}}_{\text{He}}\text{+}{\text{V}}_{\text{Ar}}\text{+}{\text{V}}_{\text{Ne}}\\ \text{=}\frac{{\text{n}}_{\text{He}}\text{RT}}{\text{P}}\text{+}\frac{{\text{n}}_{\text{Ar}}\text{RT}}{\text{P}}\text{+}\frac{{\text{n}}_{\text{Ne}}\text{RT}}{\text{P}}\\ \text{=}\left({\text{n}}_{\text{He}}{\text{+n}}_{\text{Ar}}{\text{+n}}_{\text{Ne}}\right)\frac{\text{RT}}{\text{P}}\\ \text{=}\left(\text{1}\text{.249+0}\text{.025+0}\text{.1734}\right)\frac{\text{0}\text{.0821×273}}{\text{1}}\\ {\text{V}}_{\text{total}}\text{=}\text{32}\text{.4L}\end{array}$

To find partial pressure of each gas:

  $\begin{array}{l}{\text{P}}_{\text{He}}\text{=}\frac{{\text{n}}_{\text{He}}\text{RT}}{\text{V}}\\ \text{=}\frac{\text{1}\text{.249×0}\text{.0821×273}}{\text{32}\text{.4}}\\ \text{=}\text{0}\text{.864}\text{atm}\end{array}$

For argon

  $\begin{array}{l}{\text{P}}_{Ar}\text{=}\frac{{\text{n}}_{\text{Ar}}\text{RT}}{\text{V}}\\ \text{=}\frac{\text{0}\text{.025×0}\text{.0821×273}}{\text{32}\text{.4}}\\ \text{=}0.0173\text{atm}\end{array}$

For neon

  $\begin{array}{l}{\text{P}}_{\text{Ne}}\text{=}\frac{{\text{n}}_{\text{Ne}}\text{RT}}{\text{V}}\\ \text{=}\frac{\text{0}\text{.1734×0}\text{.0821×273}}{\text{32}\text{.4}}\\ \text{=}\text{0}\text{.129}\text{atm}\end{array}$

Conclusion

The total volume is $\text{32}\text{.4L}$ and the partial pressure of He is $\text{0}\text{.0173}\text{atm}$ , for Ar is $\text{0}\text{.0173}\text{atm}$ and for neon is $\text{0}\text{.129}\text{atm}$