Question

Pressure exerted by 1 mole of methane in a 0.25L
container at 300K using van der Waals' equation is
(Given a = 2.253 atm L’mol?, b = 0.0428 L mol')
a) 82.82 atm b) 152.51 atm
c) 190.52 atm d ) 70.52 atm​

Answer 1

Find ur answer in the pic

Answer 2

Answer:

$\sf \red{{P = 82.8 \ atm}}$

a). option is correct.

Explanation:

$\sf Given :$

$\sf n = 1 \ mol$

$\sf V = 0.25 \ L$

$\sf a = 2.253 \ L^2 \ atm / mol^2$

$\sf b = 0.0428 \ L / mol$

We have Van der Wall equation for real gases :

$\sf \left(P+\dfrac{an^2}{V^2} \right)\left(V-nb\right) =nRT$

Putting values here we get :

$\sf \left(P+\dfrac{1^2\times(2.253)}{(0.25)^2} \right)\left(0.25-1\times0.0428 \right) =1\times0.082\times300$

$\sf \left(P+\dfrac{2.253}{0.0625} \right)\left(0.25-0.0428 \right) =24.6$

$\sf \left(P+36.048 \right) = \left(\dfrac{24.6}{0.207} \right)$

$\sf \left(P+36.048 \right) = \left(118.84 \right)$

$\sf P = 118.84-36.048$

$\sf P = 82.8 \ atm$

Therefore we get required pressure.