Calculate the pressure exerted by one mole of methane in a 450 mL container at 25^@C using van der Waals' equation. What pressure will be predicted by ideal gas equation ? (Given : a = 2.253 " atm " L^2mol^(-2), b = 0.0428 L mol^(-2), R =0.0821 L " atm " K^(-1) mol^(_1))

Calculate the pressure exerted by one mole of methane in a 450 mL cont

1.	Calculate the pressure exerted by one mole of methane in a 450 mL container at 25^@C using van der Waals' equation. What pressure will be predicted by ideal gas equation ? (Given : a = 2.253 " atm " L^2mol^(-2), b = 0.0428 L mol^(-2), R =0.0821 L " atm " K^(-1) mol^(_1))
Answer» <P> Solution :Using van der WAALS. equation : For ONE mole of a gas, van der Waals. equation can be written as `(p + a/V_2) (v - b) = RT " or " P= (RT)/(V-b) - a/V_2` Substituting `a = 2.253 " atm " L_2mol^(-2), b = 0.0428 L mol^(-1)`, `V= 450/1000 = 0.450 L, T = 25 + 273 = 298 K` and `R = 0.0821 L " atm " K^(-1) mol^(-1)`, we have `P = (0.0821 xx 298)/(0.450-0.0428)- (2.253)/((0.450)^2)" or " P = 0.0428` atm Using ideal gas equation : `PV = nRT " or " P =(nRT)/V` Substituting, `n = 1, R = 0.0821 L " atm " K^(-1) mol^(-1)`, `T = 298 K " and " V = 0.450 L`, we have `P= (1 xx 0.0821 xx 298)/(0.450) = 54.4` atm Hence, the PRESSURE of methane under given conditions is 48.96 atm. Ideal gas equation predicts the pressure 54.4 atm under similar conditions.