Calculate the pressure exerted by 16 g of methane in a 250 mL container at 300 K using van der Waal's equation. What pressure will be predicted by ideal gas equation ?
`a = 2.253 atm L^(2) mol^(-2), b = 0.0428 L mol^(-1)`
`R = 0.0821 L atm K^(-1) mol^(-1)`

Explanation

Given, `16 g CH_(4) = (16)/(16) = 1` mole
Applying van der Waal's equation
`(P + (n^(2)a)/(V^(2))) (V - nb) = nRT`
`P = (nRT)/((V - nb)) - (n^(2)a)/(V^(2))`
Substituting `n = 1`,
`R = 0.0821 L atm K^(-1) mol^(-1), T = 300 K, V = 0.250 L`,
`a = 2.253 atm L^(2) mol^(-2), b = 0.0428 L mol^(-1)`
`P = (1 xx 0.0821 xx 300)/((0.250 - 1 xx 0.0428)) - (1 xx 2.253)/((0.250)^(2)) = 82.822 atm`
The ideal gas equation predicts that,
`P = (nRT)/(V) = (1 xx 0.0821 xx 300)/(0.250) = 98.52 atm`