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For a gas van...

Question

For a gas vander waal's constants are
$a = 2.51 L^{2} a t m {m o l e}^{- 2}$
$b = 0.032 L {m o l e}^{- 1}$
Calculate the pressure of 1 mole gas in a 2l flask at 300 K, using ideal gas equation and vander waal's equation.

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Solution

Given,

$a = 2.51 L^{2} a t m m o l^{- 1}$

$b = 0.032 L m o l^{- 1}$

$M a s s = 1 m o l e$

So van der Waals equation is:

$(p + \frac{a}{V^{2}}) (V - b) = R T$

Using $R = 0.082, T = 300 k, V = 21$

$(P + \frac{2.51}{21}) (21 + 0.032) = 0.082 \times 300$

$(P + 0.12) 21.032 = 24.4$

$21.032 P + 2.51 = 24.41$

$P = 1.04$

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