A $2.24 L$ cylinder of oxygen at $1 a t m$ and $273 K$ is found to develop a leakage. When the leakage was plugged the pressure dropped to $570 m m$ of Hg. The number of moles of gas that escaped will be:

0.025

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0.050

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0.075

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0.09

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Solution

The correct option is D $0.025$
Original pressure, $(P_{i}) = 1 atm = 760 mmHg$
Final pressure, $(P_{f}) = 570 mmHg$

$∴$ Drop in pressure, $(P) = P_{i} - P_{f} = (760 - 570) mmHg = 190 mmHg = \frac{190}{760} atm$
Volume, $(V) = 2.24 L$
$R = 0.0821 L atm K^{- 1} {m o l}^{- 1}$
Temperature, $(T) = 273 K$

According to ideal gas equation,
$P V = n R T$
$\Rightarrow n = \frac{P V}{R T}$
$\Rightarrow n = \frac{(\frac{190}{760}) \times 2.24}{0.0821 \times 273} = 0.02498 = 0.025 moles$

Hence the number of moles of gas that escaped will be $0.025 moles$ .

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A 2.24L cylinder of oxygen at 1atm and 273K is found to develop a leakage. When the leakage was plugged the pressure dropped to 570mm of Hg. The number of moles of gas that escaped will be:

A $2.24 L$ cylinder of oxygen at $1 a t m$ and $273 K$ is found to develop a leakage. When the leakage was plugged the pressure dropped to $570 m m$ of Hg. The number of moles of gas that escaped will be: