Chapter 5, Problem 5.160QP

Interpretation Introduction

Interpretation:

The density of a gas having root-mean-square speed of $\text{5}{\text{.00×10}}^{\text{2}}\text{m/s}$ has a pressure of $\text{2}\text{.50}\text{atm}\text{at}{\text{25}}^{\text{0}}\text{C}$ has to be calculated.

Concept Introduction:

1. Density = an amount of substance per unit volume is known as density.

$\begin{array}{l}\text{ ρ = }\frac{\text{M}}{\text{V}}\\ \text{Where,}\\ \text{ρ=density}\\ \text{M=mass}\\ \text{V=Volume}\end{array}$

2. Ideal gas equation:

At a constant temperature (K) and pressure (P), the volume (v) occupied by the no of moles of any gas is known as ideal gas equation.

Ideal gas equation:

$\begin{array}{l}\text{PV}\text{=}\text{nRT}\\ \text{nRT = PV}\\ \text{n = }\frac{\text{PV}}{\text{RT}}\end{array}$

3. Root mean square value:

${\text{μ}}_{\text{rms}}\text{=}\sqrt{\frac{\text{3RT}}{{\text{M}}_{\text{m}}}}$

Where,

$\begin{array}{l}{\text{μ}}_{\text{rms}}\text{=}\text{root}\text{mean}\text{square}\text{value}\\ \text{R}\text{= ideal}\text{gas}\text{constant = 8}\text{.314}\frac{\text{joule}}{\text{mole}\text{.Kelvin}}\\ \text{K}\text{=}\text{absolute}\text{temperature}\text{in}\text{kelvin}\\ \text{M}\text{=}\text{molar}\text{mass}\text{of}\text{the}\text{molecule}\text{in}\text{kilograms}\end{array}$

Answer to Problem 5.160QP

The density of the gas is $\text{3}\text{.04 g/L}$

Explanation of Solution

Given,

An ideal gas with RMS of a gas is $\text{5}{\text{.00×10}}^{\text{2}}\text{m/s}$ and pressure of $\text{2}\text{.5}\text{atm}$ and temperature ${\text{25}}^{\text{0}}\text{C}$

We can rewrite the ideal gas equation as

  $\begin{array}{l}\text{u}\text{=}{\left(\frac{\text{3RT}}{{\text{M}}_{\text{m}}}\right)}^{\frac{\text{1}}{\text{2}}}\\ \text{rearranging the above equation we get}\\ {\text{u}}^{\text{2}}\text{=}\frac{\text{3RT}}{{\text{M}}_{\text{m}}}\\ {\text{M}}_{\text{m}}\text{=}\frac{\text{3RT}}{{\text{u}}^{\text{2}}}\\ {\text{M}}_{\text{m}}\text{=}\frac{\left(\text{3}\right)\left(\text{8}\text{.31kg}{\text{.m}}^{\text{2}}{\text{/s}}^{\text{2}}\text{.mol}\text{.K}\right)\left(\text{298}\text{.2}\text{K}\right)}{\left(\text{5}{\text{.00×10}}^{\text{2}}{\text{m/s}}^{\text{2}}\right)}\\ \text{=0}\text{.029737}\text{kg/mol}\\ \text{=29}\text{.74}\text{g/mol}\\ \end{array}$

Ideal gas equation and density equation can be rearranged as ${\text{PM}}_{\text{m}}\text{=dRT(where}\text{n}\text{=1)}$

And density can be found using ideal gas equation

$\begin{array}{l}{\text{PM}}_{\text{m}}\text{=}\text{dRT(where}\text{n}\text{=1)}\\ \text{d}\text{=}\frac{{\text{PM}}_{\text{m}}}{\text{RT}}\\ \text{=}\frac{\left(\text{2}\text{.50}\text{atm}\right)\left(\text{29}\text{.74}\text{g/mol}\right)}{\left(\text{0}\text{.08206}\text{L}\text{.atm/K}\text{.mol}\right)\left(\text{298}\text{.2}\text{K}\right)}\\ \text{=3}\text{.038}\text{g/L}\\ \text{=3}\text{.04g/L}\end{array}$

Conclusion

The density of an ideal gas mixture is calculated as $\text{3}\text{.04 g/L}$.