A gas is enclosed in a vessel at a constant temperature at a pressure of 5 atmospheres and volume 4 litre. Due to leakage in the vessel, after some time, the pressure is reduced to 4 atmosphere. As a result, the:

volume of the gas decreased by 20%

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average KE of gas molecule decreases by 20%

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20% of the gas escaped due to the leakage

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25% of the gas escaped due to the leakage

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Solution

The correct option is C 20% of the gas escaped due to the leakage
$n^{'} = \frac{P^{'}}{P} \times n = \frac{4}{5} \times n = 0.8 n$
Thus amount of gas escaped is $n - 0.8 n = 0.2 n$ or 20%

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A gas is enclosed in a vessel at a constant temperature at a pressure of 5 atmospheres and volume 4 litre. Due to leakage in the vessel, after some time, the pressure is reduced to 4 atmosphere. As a result, the:

The correct option is C 20% of the gas escaped due to the leakagen′=P′P×n=45×n=0.8nThus amount of gas escaped is n−0.8n=0.2n or 20%

The correct option is C 20% of the gas escaped due to the leakage
$n^{'} = \frac{P^{'}}{P} \times n = \frac{4}{5} \times n = 0.8 n$
Thus amount of gas escaped is $n - 0.8 n = 0.2 n$ or 20%