Question

A flask of 2 ${\text{d}}{{\text{m}}^{\text{3}}}$ capacity contains ${O_2}$ at $101.325$ kPa and 300 K. The gas pressure is reduced to $0.1$ Pa. Assuming ideal behavior, answer the following:

1. What will be the volume of gas which is left behind?
2. What amount of ${O_2}$ and the corresponding number of molecules are left behind in the flask?
3. In now $2g$ of ${N_2}$ is introduced, what will be the pressure of the flask

Answer 1

We know that the ideal gas law expresses the quantitative relation between the four variables that describe the state of the gas. And the four variables are pressure, temperature, number of moles, and volume. To solve this question, we must know the relation between all the variables.

Formula used: ${\text{PV = nRT}}$
Here, ${\text{P}}$= pressure
${\text{V}}$ = volume
${\text{n}}$ = number of moles
${\text{R}}$ = Gas constant
${\text{T}}$ = temperature

Complete step by step answer:
Let us solve the first part of the question:
As it is not mentioned if the gas is undergoing compression or expanding, the volume will not change and it will be the same as the volume of the flask.
Given the capacity of the flask is 2 ${\text{d}}{{\text{m}}^{\text{3}}}$
Therefore, the volume of gas which is left behind is ${\text{V}} = 2{\text{d}}{{\text{m}}^{\text{3}}}$
Let us solve the second part of the question:
Here we are asked the number of a molecule of ${O_2}$ if the gas pressure is reduced to $0.1$ Pa pressure:
To find the number of molecules first, we have to find the number of moles of ${O_2}$ left behind:
By modifying ideal gas law we get:
${\text{n = }}\dfrac{{{\text{PV}}}}{{{\text{RT}}}}$
By substituting the value we get:
$n = \dfrac{{0.1 \times {{10}^{ - 3}}\left( {{\text{kPa}}} \right)2\left( {{\text{d}}{{\text{m}}^{\text{3}}}} \right)}}{{8.314\left( {{\text{kPa d}}{{\text{m}}^{\text{3}}}{\text{ }}{{\text{K}}^{{\text{ - 1}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}} \right)300\left( {\text{K}} \right)}}$
By solving we get the number of moles of ${O_2}$ :
$n = 8.019 \times {10^{ - 8}}{\text{mol}}$
Now we know that 1 mole of a substance comprises $6.023 \times {10^{23}}$ molecules.
So, $8.019 \times {10^{ - 8}}$ a mole of ${O_2}$ will comprise $\left( {6.023 \times {{10}^{23}} \times 8.019 \times {{10}^{ - 8}}} \right)$ molecules of ${O_2}$ .
So we get $48.29 \times {10^{15}}$ molecules of ${O_2}$ left behind in the flask.
Now let us solve the third part of the question:
We are given 2 g of ${N_2}$ introduced in the flask.
So the number of moles ${N_2}$ will be:
$n = \dfrac{{2{\text{g}}}}{{14{\text{gmo}}{{\text{l}}^{{\text{ - 1}}}}}} = 0.143{\text{mol}}$
We get the total number of moles left in the flask by adding the number of moles of nitrogen and oxygen:
${n_{total}} = 0.143 + 8.019 \times {10^{ - 8}} \approx 0.143mol$
Here we have very few moles of oxygen so it can be neglected.
By modifying ideal gas law we get:
${\text{P = }}\dfrac{{{\text{nRT}}}}{{\text{V}}}$
By substituting the value we get:
$P = \dfrac{{0.143\left( {{\text{mol}}} \right)8.314\left( {{\text{kPa d}}{{\text{m}}^{\text{3}}}{\text{ }}{{\text{K}}^{{\text{ - 1}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}} \right)300\left( {\text{K}} \right)}}{{2\left( {{\text{d}}{{\text{m}}^{\text{3}}}} \right)}}$
By solving we get the pressure of the flask as:
${\text{P}} = 89.08{\text{ kPa}}$

**Therefore, we can conclude that:

1. Volume of the gas left behind is 2 ${\text{d}}{{\text{m}}^{\text{3}}}$ .
2. The number of molecules ${O_2}$ left behind in the flask is $48.29 \times {10^{15}}$
3. Pressure of the flask when 2 g of ${N_2}$ is introduced is $89.08{\text{ kPa}}$ **

Note: In this type of question we should always focus on the unit conversion like in the second part of the question the pressure reduced is given in pascal and should be converted into kilopascal. And we know that nitrogen and oxygen are non-reacting gaseous mixtures so we can add their number of moles.