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. 1
C. 2
D. 4

Answer

C

Explanation

Ideal gas equation is given by
`PV=nRT`…….i
For oxygen
`P=1` atm `V=1l,n=n_(O_(2))`
Therefore eq I becomes
`:.1xx1=n_(O_(2))RT`
`n_(O_(2))=1/(RT)`
For nitrogen
`P=0.5` atm `V=2l,n=n_(N_(2))`
`:.0.5xx2=n_(N_(2))RT`
`impliesn_(N_(2))=1/(RT)`
For mixture of gas
`P_("mix")V("mix")=n_("mix")RT`
`:.(P_("mix")V_("mix"))/(RT)=1/(RT)+1/(RT)`
`P_("mix")V_("mix")=2(V_("mix")=1)`