Figure-2.28 shows a cylindrical container which is divided in two equal parts by a clamped diathermic piston. Different ideal gases are filled in the two parts. It is found that the rms speed of molecules in the lower part is equal to the mean speed of molecules in the upper part. Find the ratio ofmass of molecule of gas in lower part to that of the gas in upper part.

Explanation

As piston is diathermic the two gases must be in thermal equilibrium. If `M_(1)` and `M_(2)` be the molecular masses of the gases filled in lower and upper part of the container, the rms speed of gas molecules in lower part is
`v_("rms")=sqrt((3RT)/(M_(1)))" "...(2.61)`
The mean speed of gas molecules in upper part is
`v_("mean")=sqrt((8RT)/(pi M_(2)))" "...(2.62)`
From equation-(2.61) and (2.62), according to the given situation, we have
`v_("rms")=v_("mean")`
or `" "sqrt((3RT)/(M_(1)))=sqrt((8RT)/(pi M_(2)))`
or `" "(M_(1))/(M_(2))=(3pi)/(8)=1.178`
As ratio of masses of molecules and that of molecular masses is same, it is 1.178.