What is the pressure of $2$ mole of $N H_{3}$ at $27^{o} C$ when its volume is $5$ litre in vander Waal's equation $? (a = 4.17, b = 0.03711)$

10.33

atm

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9.33

atm

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9.74

atm

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9.2

atm

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Solution

The correct option is A $10.33$ atm
$(P + \frac{a n^{2}}{v^{2}}) (v - n b) = n R T$
$(P + \frac{4.17 \times 2^{2}}{5^{2}}) (5 - 2 \times 0.0371) = 2 \times 0.0821 \times 300$
$(P + 0.66) (4.258) = 49.26$
$P = 10.9 a t m$

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Similar questions

The pressure of 2 moles of ammonia at 27⁰ when its volume is 5 L according to van der WaaL's equation is (Given, α=4.17, b=0.03711)

a t m l^{2} m o l^{- 2}

m o l^{- 1}

2 mol

5 L

27^{\circ} C

(a = 4.17 {atm L}^{2} {mol}^{- 2}, b = 0.0371 {L mol}^{- 1})

1 mole

0.25 litre

300 K

a = 2.25 {atm - L}^{2} / {mol}^{2}

b = 0.05 L/mol

The pressure of 2 moles of ammonia at 27⁰ when its volume is 5 L according to van der WaaL's equation is (Given, α=4.17, b=0.03711)

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