A vessel of `4.00L` capacity contains `4.00 g` of methane and `1.00 g` of hydrogen at `27^(@)C`. Calculate the partial pressure of each gas and also the total pressure in the container.

Explanation

Let the partial pressure of hydrogen be `P_(H_(12))` and the partial pressure of methane be `P_(CH_(4))`.
The number of moles of hydrogen is
`n_(1)=(1)/(2)=0.5 mol`
The number of moles of methane is
`n_(2)=(4)/(16)=0.25 mol`
Now applying ideal gas equation for each gas, we get
`P_(H_(2))=(n_(1)RT)/(V)=(0.5xx0.0821xx300)/(4)=3.07 atm`
Similarly, `P_(CH_(4))V=n_(2)RT`
`P_(CH_(4))=(n_(2)RT)/(V)=(0.25xx0.0821xx300)/(4)=15.4 atm`
Total pressure of the gaseous mixture is
`P_(H_(2))+P_(CH_(4))=3.07+1.54=4.61 atm`.