A 4: 1 molar mixture of He and CH4 is contained in a vessel at 20 bar pressure. Due to a hole in the vessel, the gas mixture leaks out. What is the composition of the mixture effusing out initially?

Explanation

Total pressure of the mixture= 20 bar Since He and `CH_(4)` are present in the molar ratio `4:1`, we have
Partial pressure of He `(P_(He))= (4)/(5) xx 20= 16` bar and Partial pressure of `CH_(4)(P_(CH_(4))= (1)/(5) xx 20 =4` bar
If t is the time taken in effusion, rate of diffusion of He `(r_(He))= (n_(He))/(t)`and rate of diffusion of `CH_(4)(r_(CH_(4)))= (n_(CH_(4)))/(t)`
where `n_(He)a nd n_(CH_(4))` are respectively the number of moles of He and `CH_(4)` effused initially. According to Graham.s law of diffusion, for two gases at different pressures, we have `(r_(He))/(r_(CH_(4)))= sqrt((M_(CH_(4)))/(M_(He)))xx (P_(1))/(P_(2))`
Therefore,
`((n_(He))/(t))/((n_(CH_(4)))/(t))= sqrt((16)/(4)) xx (16)/(4)` or `((n_(He))/(t))/((n_(CH_(4)))/(t))= (8)/(1)`
Hence, the initially effused mixture contains the moles of Helium and Methane in the ratio `8: 1`