Question

closed container with fine hole contains a mixture of hydrogen and oxygen gases at 2 atm pressure. the mixture contains 4% hydrogen by weight. Calculate the ratio of rates of effusion of two gases.​

Answer 1

Assume 1 mole of hydrogen and 1 mole of oxygen gases are placed in a container with a pin-hole through which both can escape.

n H t O

n O t H = M OM H

21nO

= 322

∵t O

=t H

and n H = 21 n O

= 21 × 41

= 81

The fraction of the oxygen that escapes in the time required for one-half of the hydrogen to escape is 81

Note: Here n O

represents the number of moles of oxygen effused in time t O

and M O

is the molar mass of oxygen.

Similarly, n H

represents the number of moles of hydrogen effused in time t H

and M H

is the molar mass of hydrogen.