A vessel contains 1.6 gm 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 B 11.2
The number of moles of oxygen $= n = \frac{1.6}{32} = 0.05$

Final temperature $= 2 \times 273 = 546$ K.

Firnal presure = 0.2 atm

The volume of new vessel is $V = \frac{n R T}{P}$

$V = \frac{0.05 \times 0.0821 \times 546}{0.2} = 11.2 L$

Hence, the correct option is $B$

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A vessel contains 1.6 gm 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 B 11.2The number of moles of oxygen =n=1.632=0.05Final temperature =2×273=546 K.Firnal presure = 0.2 atmThe volume of new vessel is V=nRTPV=0.05×0.0821×5460.2=11.2LHence, the correct option is B