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# Densities of two gases are in the ratio $1:2$ and their temperatures are in the ratio $2:1$, then the ratio of their respective molar mass at a certain pressure is A.$1:1$B.$1:2$C.$2:1$D.$4:1$ Verified
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Hint: The ideal equation says that the pressure and volume are directly proportional to the moles of the gas and the temperature of the system. For an ideal gas to exist, it has to follow the above-mentioned relation. The ideal gas has their particles to be small and there is no considerable repulsion force between them.

It is given that the two gases are in a system. These two gases have a specific ratio between the densities of the two gases. There is also a $2:1$ratio between the temperatures of the two gas.
Let the gas be two ideal gas, we know that the pressure and volume is directly proportional to the number of moles and temperature, also we know that moles are equal to the given mass of the gas divided by the molar mass of the gas, now that means volume and pressure are directly proportional to the given mass of gas and inversely proportional to the gas.
Also, we know that the given mass divide by the volume of the gas gives the density of the gas.
We can tweak the ideal gas into: -
$PV = nRT \\ \Rightarrow PV = \dfrac{{given\,mass}}{{molar\,mass}} \times RT \\ \Rightarrow P \times molar\,mass = \dfrac{{given\,mass}}{V} \times RT \\ \Rightarrow P \times molar\,mass = density \times R \times T \\$ …eq(1)

From the above relation we can say that molar mass is directly proportional to the density and temperature and inversely proportional to the pressure.
We have been given for the two gases, the density is in the ratio $1:2$ and the temperatures are in the ratio $2:1$, rest the pressure and is constant throughout.
Let’s take the density of gas one be $d$and the density of second gas be $2d$
The temperature of gas one be $2T$and the temperature of second gas is $T$
From eq (1)
We can say that the ratio of molar masses will be equal to ratio of density multiplied by the ratio of temperatures as they are directly proportional
$\dfrac{{molar\,mass\,1}}{{molar\,mass\,2}} = \dfrac{{density1}}{{density2}} \times \dfrac{{temperature\,1}}{{temperature\,2}} \\ \Rightarrow \dfrac{{MM\,1}}{{MM\,2}} = \dfrac{d}{{2d}} \times \dfrac{{2T}}{T} \\ \Rightarrow \dfrac{{MM1}}{{MM2}} = \dfrac{1}{1} \\$
Therefore, the ratio between the molar masses of two gases given is just $1:1$.
The ratio of their certain molar masses at a certain pressure is $1:1$.

Note:
For a real gas the pressure is not equal to the actual pressure and the volume is also not ideal because of the attractive forces between molecules and their big size. The attractive forces are so high that it hinders the motion of the particle and lower pressure is observed.