`X mL` of `H_(2)` gas effuses through a hole in a container in `5 s`. The time taken for the effusion of the same volume of the gas specified below, under identical conditions, is
XmL of H_(2) gas effuses through a hole in a container is 5 second. The time taken for the effusion of the same volume of the gas specified below under identical conditions is .
The constant motion and high velocities of gas particles lead to some important practical consquences. One such consquences is that gases mix rapidly when they come in contact. Take the stopper off a bottle of perfume, for instance, and the odour will spread rapidly through the room as perfume molecules mix with the molecules in the air. This mixing of different gases by random molecular motion and with frequent collision is called diffusion. A similar process in which gas molecules escape without collision through a tiny hole into a vacuum is called effusion. Both the processes follow Graham's law which is mathematically put as r prop sqrt(1//d) . The average distance travelled by molecules between successive collisions is called mean free path. Answer the following questions on the basis of the above information: X mL H_(2) effuses through a hole in a container in 5 s . The time taken for the effusion of the same volume of the gas specified below under identical conditions is
The time taken for effusion of 64 mL of oxygen will be as the time taken for the effusion of which of the following gases under identical conditions ?
V ml of H_(2) gas diffuses through a small hole in a container in time t_(1). How much time will be required by oxygen gas for the diffusion of same volume?
The time taken for effusion of 32 mL of oxygen will be the same as the time taken for effusion under identical condition of :
