Question

A container with volume 1.62 L is initially evacuated. Then it is filled with 0.293g of N2. Assume that the pressure of the gas is low enough for the gas to obey the ideal-gas law to a high degree of accuracy. If the root-mean-square speed of the gas molecules is 178 m/s, what is the pressure of the gas?

Answer 1

Concepts and reason

The concept of ideal gas equation and root mean square velocity is required to solve this problem.

Initially write the ideal gas equation. Then convert this equation in terms of root mean square velocity. Finally rearrange equation for the pressure.

Fundamentals

The idea gas equation is as follows:

$pV = nRT$ …… (1)

Here, p is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

The root mean square speed of the gas molecules will be given by the following formula:

${v_{{\rm{rms}}}} = \sqrt {\frac{{3RT}}{M}}$ …… (2)

Here, ${v_{{\rm{rms}}}}$ is the root mean square velocity and M is the mass of the gas molecule.

Rearrange equation (1) for RT.

$RT = \frac{{pV}}{n}$

Substitute $\frac{{pV}}{n}$ for RT in equation (2).

$\begin{array}{c}\\{v_{{\rm{rms}}}} = \sqrt {\frac{{3\frac{{pV}}{n}}}{M}} \\\\{\left( {{v_{{\rm{rms}}}}} \right)^2} = \frac{{3pV}}{{nM}}\\\end{array}$

$p = \frac{{{{\left( {{v_{{\rm{rms}}}}} \right)}^2}nM}}{{3V}}$ …… (3)

The relation between number of moles and mass of a molecule is as follows:

$n = \frac{m}{M}$

Here, m is the total mass of the gas and M is the mass of one mole of gas.

Substitute $\frac{m}{M}$ for n in equation (3).

$\begin{array}{c}\\p = \frac{{{{\left( {{v_{{\rm{rms}}}}} \right)}^2}\left( {\frac{m}{M}} \right)M}}{{3V}}\\\\ = \frac{{{{\left( {{v_{{\rm{rms}}}}} \right)}^2}m}}{{3V}}\\\end{array}$ …… (4)

Substitute $0.293 \times {10^{ - 3}}{\rm{ kg}}$ for m, $178{\rm{ m/s}}$ for ${v_{{\rm{rms}}}}$ , $1.62 \times {10^{ - 3}}{\rm{ }}{{\rm{m}}^3}$ for V in equation (4).

$\begin{array}{c}\\p = \frac{{{{\left( {178{\rm{ m/s}}} \right)}^2}\left( {0.293 \times {{10}^{ - 3}}{\rm{ kg}}} \right)}}{{3\left( {1.62 \times {{10}^{ - 3}}{\rm{ }}{{\rm{m}}^3}} \right)}}\\\\ = 1.91 \times {10^3}{\rm{ Pa}}\\\end{array}$

Ans:

The pressure of the gas is $1.91 \times {10^3}{\rm{ Pa}}$ .