Question

Pressure of $1g$ of an ideal gas A at ${27^\circ }C$ is found to be $2bar$. When $2g$ of another ideal gas B is introduced in the same flask at same temperature, the pressure becomes $3bar$ . Find a relationship between their molecular masses?

Answer 1

To solve this question, first we will suppose the molecular masses of both the gases. And then we will find their number of moles with the help of molecular masses. And finally we can relate all the given and found values.

Complete step-by-step answer: Let the molecular masses of A and B are ${M_A}$ and ${M_B}$ $g.mo{l^{ - 1}}$ respectively.
Then, their number of moles will be:
${n_A} = \dfrac{1}{{{M_A}}}$
here, ${n_A}$ is the number of moles of gas A.
Again,
${n_B} = \dfrac{2}{{{M_B}}}$
here, ${n_B}$ is the number of moles of gas B.
The given pressure for gas A:
${P_A} = 2bar$
And, according to the question: the sum of the pressure of both the ideal gases:
${P_A} + {P_B} = 3bar$
i.e.. ${P_B} = 1bar$
Pressure for gas B is $1bar$ .
So, we will apply the relation between all the given information about this question:
$PV = nRT$
Whereas,
${P_A}V = {n_A}RT$ , the relation for gas A
and,
${P_B}V = {n_B}RT$
So, we have:
$\therefore \dfrac{{{P_A}}}{{{P_B}}} = \dfrac{{{n_A}}}{{{n_B}}} = \dfrac{{1/{M_A}}}{{2/{M_B}}} = \dfrac{{{M_B}}}{{2{M_A}}}$
or, $\dfrac{{{M_B}}}{{{M_A}}} = 2 \times \dfrac{{{P_A}}}{{{P_B}}} = 2 \times \dfrac{2}{1} = 4$
or, $\Rightarrow {M_B} = 4{M_A}$

Hence, the above conclusion is the relationship between the molar masses of both the gases A and B.

Note: Basically, this question is solved with the help of ideal gas law, that describes the behavior of an ideal sample of gas, and how that behavior is related to the pressure, temperature, volume, and molarity of the gas sample.