Calculate the pressure exerted by $16 g$ of methane in a $250 m L$ container at $300 K$ using van der Waals' equation. What pressure will be predicted by ideal gas equation?
$a = 2.253 a t m L^{2} m o l^{- 2}, b = 0.0428 L m o l^{- 1}$ ,
$R - 0.0821 L a t m K^{- 1} m o l^{- 1}$ .

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Solution

Given, $16 g C H_{3} = \frac{16}{16} = 1 m o l e$
Applying van der Waal's equation,
$(P + \frac{n^{2} a}{V^{2}}) (V - n b) = n R T$
$P = \frac{n R T}{(V - n b)} - \frac{n^{2} a}{V^{2}}$
Substituting $n = 1$ ,
$R = 0.0821 L a t m K^{- 1} m o l^{- 1}; T = 300 K; V = 0.250 L$ ;
$a = 2.253 a t m L^{- 2} m o l^{- 2}; b = 0.0428 L m o l^{- 1}$
$P = \frac{1 \times 0.0821 \times 300}{(0.250 - 1 \times 0.0428)} - \frac{1 \times 2, 253}{(0.250)^{2}} = 82.822 a t m$
The ideal gas equation predicts that,
$P = \frac{n R T}{V} = \frac{1 \times 0.0821 \times 300}{0.250} = 98.52 a t m$

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b = 0.05 L/mol

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0.5 d m^{3}

at 300 K using:

(a) Ideal gas law and
(b) Van der Waals equation

[Given:

a = 360 k P a d m^{6} m o l^{- 2}

and

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Calculate the pressure exerted by 16 g of methane in a 250 mL container at 300 K using van der Waals' equation. What pressure will be predicted by ideal gas equation?a=2.253 atm L2mol−2,b=0.0428 L mol−1,R−0.0821 L atm K−1mol−1.

Calculate the pressure exerted by $16 g$ of methane in a $250 m L$ container at $300 K$ using van der Waals' equation. What pressure will be predicted by ideal gas equation?
$a = 2.253 a t m L^{2} m o l^{- 2}, b = 0.0428 L m o l^{- 1}$ ,
$R - 0.0821 L a t m K^{- 1} m o l^{- 1}$ .