A vessel contains equal masses of `H_2` and `CH_4`. The fraction of the partial pressure of methane in the total pressure is
(1) 1/9
(2) 1/2
(3) 8/9
(4) 3/4

Same mass of CH_(4) and H_(2) is taken in container. The partial pressure caused by H_(2) is

1 mole of methane ( CH_4 ) contains

A vessel of 4.00L capacity contains 4.00 g of methane and 1.00 g of hydrogen at 27^(@)C . Calculate the partial pressure of each gas and also the total pressure in the container.

A gaseous mixture contains 56 g of N_2 , 44 g CO_2 and 16 g of CH_4 The total pressure of the mixture is 720 mm Hg. The partial pressure of CH_4 is

A closed vessel contains equal number of oxygen and hydrogen molecules at a total pressure of 740 mm. If oxygen is removed from the system, the pressure: 1)becomes half of 740 mm 2)becomes 1/3 rd of 740 mm 3)becomes double of 740 mm 4)remains unchanged.

If the ration of the masses of SO_(3) and O_(2) gases confined in a vessel is 1 : 1 , then the ratio of their partial pressure would be

If the ration of the masses of SO_(3) and O_(2) gases confined in a vessel is 1 : 1 , then the ratio of their partial pressure would be

Equal mass of H_(2) , He and CH_(4) are mixed in empty container at 300 K, when total pressure is 2.6 atm The partial pressure of H_(2) in the mixture is

A cylinder contains 0.3 moles of N_(2) , 0.1 mole of O_(2) and 0.1 mole of helium. If the total pressure is one atm, what is the partial pressure of oxygen in mm Hg:-

A gaseous mixture contains 1 g of H_(2) , 4 g of He , 7 g of N_(2) and 8 g of O_(2) . The gas having the highest partial pressure is :-