Pressure exerted by 1 mole of methane in a 0.25 litre container at 300 K using van der Waals’ equation (given a = 2.253 atm L2 mol2, b = 0.0428 L mol -1) is

Question

Pressure exerted by 1 mole of methane in a 0.25 litre container at 300 K using van der Waals’ equation (given a = 2.253 atm L2 mol2, b = 0.0428 L mol -1) is (A) 82.82 atm (B) 152.51 atm (C) 190.52 atm (D) 70.52 atm

Tardigrade Admin · Accepted Answer

(P+(n2a/V2)) (V-nb)=nRT (P+(2.253/0.25×0.25))(0.25-0.0428)=0.0821×300 or (P+36.048)(0.2072)=24.63 ⇒ P+36.075=118.87 ⇒ P=82.82 atm

Q. Pressure exerted by $1$ mole of methane in a $0.25$ litre container at $300 K$ using van der Waals’ equation (given $a = 2.253$ atm $L^{2} m o l^{2}, b = 0.0428 L m o l^{- 1})$ is

$82.82 a t m$

$152.51 a t m$

$190.52 a t m$

$70.52 a t m$

$(P + \frac{n ^{2} a}{V ^{2}}) (V - nb) = n RT$
$(P + \frac{2.253}{0.25 \times 0.25}) (0.25 - 0.0428) = 0.0821 \times 300$
or $(P + 36.048) (0.2072) = 24.63$
$\Rightarrow P + 36.075 = 118.87$
$\Rightarrow P = 82.82 a t m$

Q. Pressure exerted by 1 mole of methane in a 0.25 litre container at 300K using van der Waals’ equation (given a=2.253 atm L2mol2,b=0.0428Lmol−1) is

82.82atm

152.51atm

190.52atm

70.52atm