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 = 2P2
$\frac{{P}_{1}}{{P}_{2}}=\frac{2}{1}$

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,$\frac{{P}_{1}}{{P}_{2}}=\frac{{n}_{1}}{{n}_{2}}$
$\frac{2}{1}=\frac{{m}_{1}/{M}_{1}}{{m}_{2}/{M}_{2}}$
$\frac{2}{1}=\frac{2/{M}_{1}}{3/{M}_{2}}\phantom{\rule{0ex}{0ex}}\frac{2}{1}=\frac{2}{3}\frac{{M}_{2}}{{M}_{1}}$

After solving this,
$\frac{{M}_{1}}{{M}_{2}}=\frac{1}{3}$

Regards

• 9
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