Question

Two non-reactive monatomic ideal gases have their atomic masses in the ratio 2:3. The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature, is 4:3. The ratio of their densities is:
A. 1:4
B. 1:2
C. 6:9
D. 8:9

Answer 1

Hint: In kinetic theory gases, we assume several data regarding gases to make their behavior close to that of ideal gas in order to analyze their behavior properly and reducing the calculation part. An ideal gas is assumed to have no interaction with the corresponding particles of gases.
Formula used:
${\displaystyle \frac{P_1{M}_1}{{\rho }_1}}={\displaystyle \frac{P_2{M}_2}{{\rho }_2}}$

Complete answer:
Partial pressure of a gas is the pressure exerted by individual gas in a container having more than one gases. We are given the ratio of molecular masses as 2:3 i.e.
${\displaystyle \frac{M_1}{M_2}}={\displaystyle \frac{2}{3}}$and the ratio of their partial pressure as 4:3 i.e. ${\displaystyle \frac{P_1}{P_2}}={\displaystyle \frac{4}{3}}$.
Thus, using the equation of state ${\displaystyle \frac{P_1{M}_1}{{\rho }_1}}={\displaystyle \frac{P_2{M}_2}{{\rho }_2}}$, we have;
${\displaystyle \frac{{\rho }_2}{{\rho }_1}}={\displaystyle \frac{P_2}{P_1}}\times {\displaystyle \frac{M_2}{M_1}}={\displaystyle \frac{3}{4}}\times {\displaystyle \frac{3}{2}}={\displaystyle \frac{9}{8}}$
So, ${\displaystyle \frac{{\rho }_1}{{\rho }_2}}={\displaystyle \frac{8}{9}}$.

So, the correct answer is “Option D”.

Additional Information:
 Equation of ideal gas – The most fundamental equation of an ideal gas is $PV=nRT$. Here, ‘P’ is the partial pressure of a gas contained in a chamber of volume ‘V’ having temperature ‘T’ and quantity of ‘n’ moles. Here ‘R’ is the universal gas constant and is equal to $R=8.314J{K}^{-1}mo{l}^{-1}$. But for getting the equation in terms of density and molecular mass, we put $n={\displaystyle \frac{m}{M}}$. Putting it in $PV=nRT$, we get:
$PV={\displaystyle \frac{m}{M}}RT⟹PM={\displaystyle \frac{m}{V}}RT$
And since ${\displaystyle \frac{m}{V}}=\rho =densityofgas$, thus
We get the final equation as $PM=\rho RT$, where ‘M’ is the molecular mass of gas.

Note:
Many times, we need the relation between different parameters of a gas like, temperature, pressure, volume, moles, density, etc. All these parameters could be found using the basic ideal gas equation. Just we need little modification and we can get the desired results. Students are advised to practice as much as problems based on the ideal gas equation.