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

The Ideal gas equation is given by
`PV=nRT`
...(i)
For oxygen, `p=1` atm.
`V=1L,n=n_(0_2)`
Therefore Eq. (i) becomes Therefore
`1 xx 1`
`:." "1 xx 1=n_(o_2)RT`
`:." "n_(o_2)=1/(RT)`
`:." "n_(o_2)=1/(RT)`
For nitrogen `p=0.5` atm,
`V=2L,n= nN_2` RT
`implies" "nN_2=1/"RT"`
For the mixture of gas
`P_"mix"V_"mix"RT`
Here, `n_"mix"=n_(o_2)+n_(N_2)`
`(P_"mix"V_"mix")/(RT)=1/(RT)+1/(RT)`
`implies P_"mix"V_"mix"=2`
`implies P_"mix"=2 "atm"`