We have two vessels of equal volume, one filled with hydrogen and the other with an equal mass of Helium. The common temperature is `27^(@)C`
a. What is the relative number of molecules in the two vessel?
b. If pressure of Hydrogen veseel is 2 atm, what is the pressure of Helium?
c. IF the temperatrue of Helium is kept at `27^(@)C` and that of hydrogen is changed, at what temperature will its pressure become equal to that of helium? THe molceular weights of hydrogen and helium are 2 and 4 respectively.

Explanation

a.The masses of hydrogen and helium gases in the vessels are equal. This means that the product of the number of molecules and the mass of a molecule must be same for `H_(2)` and He gases. Since molecular masses of `H_(2)` and He are in the ratio 1:2, their number of molecules `n_(H)` and `n_(He)` in the vessels must be in the reverse ratio, that is `(n_(H))/(n_(He))=2/1`
b. The equation of state for one mole of gas is
`pV=RT=NkT`
Where N is Avogadro.s number (no. of molecules in one mole) and k is Boltzmann.s constant. If a gas has n molecules, the equation of state will be
`pV=nKT`
For a given volume and a given temperatue we have
`p prop n.`
Since `H_(2)` and He have the same volume and the same temperature `(27^(@)C)` we have
`(p_(H))/(p_(He))=(n_(H))/(n_(He))=2/1`
Here `p_(H)=2` atm
`p_(He)=1` atm
c. Again we have
`pV=nkT`
`H_(2)` and He have equal volumes. Since pressure are equal we must have
`n_(H)T_(H)=n_(He)T_(He)`
or `(T_(He))/(T_(H))=(n_(H))/(n_(He))=2`
Here `T_(He)=27+273=300K`
`:.T_(H)=1/2T_(He)=150K=(150-273)^(@)C=-123^(@)C`