Question

When $2gms$ of gas is introduced into an evacuated flask kept at $25{}^\circ C$ the pressure is found to be one atmosphere. If $3gms$ of another gas is added to the same flask the pressure becomes $1.5$ atmospheres. The ratio of the molecular weight of these gases will be
A. $1:3$
B. $3:1$
C. $2:3$
D. $3:2$

Answer 1

Every ideal gas obeys the ideal gas equation. An ideal gas equation gives a relation between pressure, volume, number of molecules, gas constant, and temperature. This equation is constant for all the gasses. The no of molecules in a gas can be given as the ratio of the given weight of the gas to the molecular weight of the gas.

As per the given data,
The weight of gas molecules is ${{W}_{1}}=2gms$
The pressure is found to be $1atm$
The weight of other gas molecules is ${{W}_{2}}=3gms$
Total pressure after the addition of both the gasses is $1.5atm$

Formula used-
$PV=nRT$
No of moles, $n=\dfrac{W}{M}$

Complete answer:
The ideal gas equation is given as,
$PV=nRT$
The number of moles is given as the ratio of the given weight of the gas to the molecular weight of the gas.
$n=\dfrac{W}{M}$
Let ${{M}_{1}}$ and ${{M}_{2}}$ be the molecular weight of the gasses 1 and 2.
By putting the value of no of moles in the ideal gas equation and rearranging it,
The Molecular weight of $2grms$ gas will be,
$\begin{aligned} & {{M}_{1}}=\dfrac{WRT}{PV} \\\ & \Rightarrow {{M}_{1}}=\dfrac{2RT}{V}\quad ...\left( 1 \right) \\\ \end{aligned}$
The Molecular weight of $3grms$ gas will be,
${{M}_{2}}=\dfrac{3RT}{0.5V}\quad \ldots ...\left( 2 \right)$
By taking the ratio of the equations (1) and (2),
$\begin{aligned} & {{M}_{1}}:{{M}_{2}}=\dfrac{2RT}{V}\times \dfrac{0.5V}{3RT} \\\ & \Rightarrow {{M}_{1}}:{{M}_{2}}=\dfrac{2(0.5)}{3} \\\ & \therefore {{M}_{1}}:{{M}_{2}}=1:3 \\\ \end{aligned}$
Thus, the ratio of the molecular weight of the two gases is $1:3$.

And the correct option which satisfies the question is Option A.

Note:
The molecular weight of a gas is the average mass of the molecule which is compared to $\dfrac{1}{12}$ the mass of the carbon 12. It is calculated as the sum of the atomic weight values of the atoms in molecules. This is used in physical chemistry to study the properties of the gas.