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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. |
Answer» 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`. |
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