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Q.

A vessel containing 1 gm of oxygen at a pressure of 10 atm and a temperature of 47°C. The pressure drops to  of its original value and temperature falls to 27°C due to leakage of gas. Then mass of oxygen leaked out will be

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a

0.22 g

b

0.82 g

c

0.432 g

d

0.33 g

answer is D.

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Detailed Solution

PV=nRT−10×V=132R×320V = R litreAfter leakage10×58R=n×R×320n=10×58×300=148 mass of gas =3248=23gmMass of gas leaked out 1 – 2/3  gm = 1/3 gm = 0.33gmMore than one correct answer.
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