A vessel contains 1.6 g of oxygen gas at 1 atm and 273 K. The gas is transferred to another container. The temperature of which is double to that of initial temperature. The new pressure of the gas is 0.20 atm. The volume (in litre) of the new vessel is:

22.4

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11.2

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5.6

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None of these

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Solution

The correct option is C 11.2
Ideal gas equation,
$P V = n R T$ ,

From this
$V_{1} = \frac{n R T}{P} = \frac{1}{20} \times 22.4 = 1.12 L$

$(\frac{P_{2}}{P_{1}}) (\frac{V_{2}}{V_{1}}) = (\frac{n_{2}}{n_{1}}) (\frac{T_{2}}{T_{1}})$

$(\frac{0.2}{1}) (\frac{V_{2}}{1.12}) = 2$

$V_{2} = 11.2 L$

Hence, the correct option is $B$

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A vessel contains 1.6 g of oxygen gas at 1 atm and 273 K. The gas is transferred to another container. The temperature of which is double to that of initial temperature. The new pressure of the gas is 0.20 atm. The volume (in litre) of the new vessel is:

The correct option is C 11.2Ideal gas equation,PV=nRT,From thisV1=nRTP=120×22.4=1.12L(P2P1)(V2V1)=(n2n1)(T2T1)(0.21)(V21.12)=2V2=11.2LHence, the correct option is B