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Standard XII

Chemistry

First Law of Thermodynamics

56 g of nitro...

Question

$56 g$ of nitrogen are confined to a $6$ litre flask at $37^{\circ} C$ . Calculate its pressure using van der Waals' equation for nitrogen. $a = 4.17 a t m l i t r e^{2} m o l^{- 2}$ and $b = 0.037 l i t r e m o l^{- 1}$ .

9.1175 a t m

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10.1175 a t m

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8.1175 a t m

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7.1175 a t m

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Solution

The correct option is C $8.1175 a t m$ .
van der Waal's equation-
$(P + \frac{a n^{2}}{V^{2}}) (V - n b) = n R T$

n = $\frac{g i v e n w e i g h t}{m o l e c u l a r w e i g h t} = \frac{56}{28} = 2$

on putting values we have-

$(P + \frac{4.17 \times 4}{36}) (6 - 2 \times 0.037) = 2 \times 0.0821 \times 310$

$P = 8.1175 a t m$

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56 g of nitrogen are confined to a 6 litre flask at 37∘C. Calculate its pressure using van der Waals' equation for nitrogen. a=4.17 atm litre2mol−2 and b=0.037 litre mol−1.

The correct option is C 8.1175 atm.van der Waal's equation-(P+an2V2)(V−nb)=nRTn = given weightmolecular weight=5628=2on putting values we have-(P+4.17×436)(6−2×0.037)=2×0.0821×310P=8.1175atm

$56 g$ of nitrogen are confined to a $6$ litre flask at $37^{\circ} C$ . Calculate its pressure using van der Waals' equation for nitrogen. $a = 4.17 a t m l i t r e^{2} m o l^{- 2}$ and $b = 0.037 l i t r e m o l^{- 1}$ .