A 10 litre flask contains 0.2 mole of methane, 0.3 mole of hydrogen and 0.4 mole of nitrogen at `25^(@)C`. What is the partial pressure of each component and what is the pressure inside the flask ?

Explanation

`P = (nRT)/(V)`
Partial pressure of methane `= (0.2 xx 0.00821 xx 298)/(10) = 0.489 atm`
Partial pressure of hydrogen `= (0.3 xx 0.0821 xx 298)/(10) = 0.734 atm`
Partial pressure of nitrogen `= (0.4 xx 0.0821 xx 298)/(10) = 0.979` atm
Total pressure `= (0.489 + 0.8734 + 0.979) atm`
`= 2.202 atm`
Alternative solution: Total number of moles
`= 0.2 + 0.3 + 0.4 = 0.9`
Let the total pressure be P.
We know that,
`P = ("total number of moles")/(V). RT`
`V = 10` litre, `R = 0.0821 L atm K^(-1), T = (25 + 273) = 298 K`
`+ (0.9)/(10) xx 0.0821 xx 298 = 2.20 atm`
Partial pressure of `CH_(4) =` Mole fraction of methane `xx` total pressure
`= (0.2)/(0.9) xx 2.20 = 0.489 atm`
Partial pressure of `H_(2) =` Mole fraction of `H_(2) xx` total pressure
`= (0.3)/(0.9) xx 2.20 = 0.489 atm`
Partial pressure of `H_(2) =` Mole fraction of `H_(2) xx` total pressure
`= (0.3)/(0.9) xx 2.20 = 0.733` atm
Partial pressure of `N_(2) =` Mole fraction of `N_(2) xx` total pressure
`= (0.3)/(0.9) xx 2.20 = 0.733 atm`
Partial pressure of `N_(2) =` Mole fraction of `N_(2) xx` total pressure
`= (0.4)/(0.9) xx 2.20 = 0.978 atm`