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. `2/3`
D. `4`

Answer

C

Explanation

Ideal gas equation is given by
`pV=nRT`…….
For oxygen `p=1` atm `V=1l,n=n_(O_(2))`
Therefore Eq I becomes
`1xx1=n_(O_(2))RT`
`impliesn_(O_(2))=1/(RT)`
for nitrogen
`p=0.5atm, V=2L,n=n_(N_(2))`
`:.0.5xx2=-n_(N_(2))RT`
`impliesn_(N_(2))=1/(RT)`
For mixture of gasses
`p_("mix")V_("mix")=n_("mix")RT`
Here `n_("mix")=n_(O_(2))+n_(N_(2))`
`:.(P_("mix")V_("mix"))/(RT)=1/(RT)+1/(RT)`
`impliesP_("mix")V_("mix")=2`
`P_("mix")=2/3`