One litre of oxygen at a pressure of 1 atm and two litres of nitrogen at a pressure of 0.5 atm are introduced into a vessel of volume 1 litre. If there is no change in temperature, the final pressure of the mixture of gas (in atm) is

Options

A. 1.5
B. 2.5
C. 2
D. 4

Answer

C

Explanation

Ideal gas equation is given by:
pV = nRT.........(i)
For oxygen, p = 1 atm, V = 1 L, `n = n_(o_(2))`.
Therefore, Eq. (i) becomes
`therefore 1 xx 1 = n_(o_(2))RT`
`rArr n_(o_(2)) = 1/(RT)`
For nitrogen `p = 0.5 atm`, V = 2L, `n = n_(N_(2))`
`therefore 0.5 xx 2 = n_(N_(2)RT)`
`rArr n_(N_(2)) = 1/(RT)`
For mixture of gas
`p_("mix")V_("mix") = n_("mix") RT`
Here, `n_("mix") = n_(O_(2)) + n_(N_(2))`
`therefore (p_("mix")V_("mix"))/(RT) = 1/(RT) + 1/(RT)`
`rArr p_("mix") V_("mix") = 2(V_("mix") =1)`