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Byju's Answer

Standard X

Physics

Ideal Gas Equation

28 gm of N2...

Question

$28 g m$ of $N_{2}$ gas is contained in a flask at a pressure of $10 a t m$ and at a temperature of $57^{\circ}$ . It is found that due to leakage in the flask, the pressure is reduced to half and the temperature reduced to $27^{\circ} C$ . The quantity of $N_{2}$ gas that leaked out is:

11 / 20 g m

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20 / 11 g m

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5 / 63 g m

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63 / 5 g m

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Solution

The correct option is C $11 / 20 g m$
The ideal gas law must hold before & after the leakage of gas.
Thus, $\frac{P_{i} V}{n_{i} T_{i}} = \frac{P_{f} V}{n_{f} T_{f}}$
$\Rightarrow n_{f} = \frac{P_{f}}{P_{i}} \frac{T_{i}}{T_{f}} n_{i}$
$\Rightarrow n_{f} = \frac{1}{2} \frac{273 + 57}{273 + 27} n_{i}$
$= \frac{11}{20} n_{i}$
$\Rightarrow m_{f} = \frac{11}{20} m_{i}$
Thus, leaked out mass $= \frac{9}{20} m_{i} = \frac{63}{5} g$

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28 gm of N2 gas is contained in a flask at a pressure of 10 atm and at a temperature of 57∘. It is found that due to leakage in the flask, the pressure is reduced to half and the temperature reduced to 27∘C. The quantity of N2 gas that leaked out is:

The correct option is C 11/20 gmThe ideal gas law must hold before & after the leakage of gas.Thus, PiVniTi=PfVnfTf⇒nf=PfPiTiTfni⇒nf=12273+57273+27ni=1120ni⇒mf=1120miThus, leaked out mass=920mi=635g

$28 g m$ of $N_{2}$ gas is contained in a flask at a pressure of $10 a t m$ and at a temperature of $57^{\circ}$ . It is found that due to leakage in the flask, the pressure is reduced to half and the temperature reduced to $27^{\circ} C$ . The quantity of $N_{2}$ gas that leaked out is: