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Pressure and Temperature

2 mol of ammo...

Question

$2 mol$ of ammonia occupied a volume of $5 L$ at $27^{\circ} C$ . Calculate the pressure if the gas obeyed van der Waals equation. $(a = 4.17 {atm L}^{2} {mol}^{- 2}, b = 0.0371 {L mol}^{- 1})$ .

9.33 atm

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18.66 atm

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4.66 atm

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12.32 atm

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Solution

The correct option is A $9.33 atm$
Given:
$n = 2 mol, V = 5 L$
$T = 273 + 27 = 300 K$
$a = 4.17 {atm L}^{2} {mol}^{- 2}, b = 0.0371 {L mol}^{- 1}$ .
$R = 0.082 atm L {mol}^{- 1} K^{- 1}$
Applying van der Waals equation for n moles
$(P + \frac{a n^{2}}{V^{2}}) (V - n b) = n R T$
$\Rightarrow (P + \frac{4.17 \times 2^{2}}{5^{2}}) (5 - 2 \times 0.0371) = 2 \times 0.082 \times 300 P = \frac{2 \times 0.082 \times 300}{4.92} - 0.667$
on solving we get :
$P = 9.33 atm$

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Pressure and Temperature

Standard XII Chemistry

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2 mol of ammonia occupied a volume of 5 L at 27 ∘C. Calculate the pressure if the gas obeyed van der Waals equation. (a=4.17 atm L2 mol−2,b=0.0371 L mol−1).

$2 mol$ of ammonia occupied a volume of $5 L$ at $27^{\circ} C$ . Calculate the pressure if the gas obeyed van der Waals equation. $(a = 4.17 {atm L}^{2} {mol}^{- 2}, b = 0.0371 {L mol}^{- 1})$ .