A box of 1 L capacity is divided into two equal compartments by a thin partition which are filled with 2g `H_(2)` and 16 g`CH_(4)` respectively. The pressure in each compartment is recorded as P atm. The total pressure when partition is removed will be:

Options

A. P
B. 2P
C. `P//2`
D. `P//4`

Answer

A

Explanation

`{:(500 mL,500 mL,),(2 g H_(2),16 g CH_(4),),(P_(H_(2)),P_(CH_(4)),):}`
`P = (nRT)/(V) = (mRT)/(MV)`
(m = mass of gas, M = Molecular mass)
`P_(H_(2)) = (2RT)/(2 xx 500) = (RT)/(500) = P`(given)
`P_(CH_(4)) = (16RT)/(16 xx 500) = (RT)/(500) = P_(H_(2)) = P`
Partial pressure of `H_(2)`,
`P_(H_(2)) = (2 xx RT)/(2 xx 1000) = (RT)/(1000) = (P)/(2)`
Partial pressure of `CH_(4)`,
`P_(CH_(4)) = (16 xx RT)/(16 xx 1000) = (RT)/(1000) = (P)/(2)`
`P_("Total") = P_(H_(2)) + P_(CH_(4)) = P//2 + P//2 = P`