A 15.0 L vessel containing 5.6g of `N_(2)` is connected to a 5.0 L vessel containing 8.0g of `O_(2)` by means of a valve. After the valve is opened and the gases are allowed to mix, what will be the partial pressure of each gas in the mixture at `27^(@)C`?

Explanation

5.6g `N_(2)=(5.6)/(28)=0.2` mol `N_(2),8.0g` `O_(2)=(8)/(32)=0.25` mol `O_(2)`
and T=(273+27)K=300K
After opening of the valve, total volume of the gas mixture,
V=(15+5)L=20L
If the partial pressure of `N_(2)` gas in the mixture is `p_(N_(2)),` then `p_(N_(2)) xx V=n_(N_(2))RT`
or, `p_(N_(2))xx20=0.2xx0.0821xx300" "therefore p_(N_(2))=0.2463atm`
if the partial pressure of `O_(2)` gas in the mixture is `p_(O_(2))`,
then `p_(O_(2))xxV=n_(O_(2))RT`
or, `p_(O_(2))xx20=0.25xx0.0821xx300" "thereforep_(O_(2))=0.3078atm`