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Q.

A nitrogen cylinder contains 84 grams of N₂ at a pressure of 1013.25 kPa and 300K. On heating the cylinder to 340K, one and half moles of gas was expelled. The pressure of the remaining gas (in atm) in the cylinder is

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a

5.7

b

10.4

c

9.9

d

7.8

answer is B.

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Detailed Solution

Volume of $N_{2} = \frac{nRT}{P} = \frac{84 \times 8 .314 \times 300}{28 \times 1013 .25 \times 10^{3}} = 7 .38 \times 10^{- 3} m^{3}$
Remaining gas after heating $= \frac{84}{28} - 1 .5 = 1 .5 mols$
$∴ P = \frac{1.5 \times 8.314 \times 340}{7.38 \times 10^{- 3}} = 574545 Pa$
= 574.545 kPa = 5.7 atm.

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