2 non reacting gases G1 (of molar mass M1) and G2 (of molar mass M2) were taken in their weight ratio 2:3 in a closed container at constant temperature. If the pressure of the gas G1 was just double of the gas G2 in the mixture. Then M1/M2 will be

Dear Student,

Weight ratios of G1 and G2 = 2:3
$\frac{m_1}{m_2}=\frac{2}{3}$
Pressure of G1 is double of gas G2.
So, P₁ = 2P_{2
$\frac{P_1}{P_2}=\frac{2}{1}$}
Acording to  ideal  gas equation,
PV = nRT
n = number of moles = mass/molar mass
So, $\frac{n_1}{n_2}=\frac{m_1/M_1}{m_2/M_2}$
pressure is directly proportional to number of moles.
So,$\frac{P_1}{P_2}=\frac{n_1}{n_2}$
$\frac{2}{1}=\frac{m_1/M_1}{m_2/M_2}$
$$\begin{gathered} \frac{2}{1}=\frac{2/M_1}{3/M_2} \\ \frac{2}{1}=\frac{2}{3}\frac{M_2}{M_1} \end{gathered}$$

After solving this,
$\frac{M_1}{M_2}=\frac{1}{3}$

Regards