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Standard X

Physics

Ideal Gas Equation

28 G of N2 ga...

Question

28 G of N2 gas is contained in a flask at a pressure of 10 atm and at a temperature of $57^{\circ} C$ . It is found that due to leakage in the flask, the pressure is reduced to half and the temperature to $27^{\circ} C$ . The quantity of $N_{2}$ gas that leaked out is

11/20 g

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20/11 g

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5/63 g

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63/5 g

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Solution

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

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

The correct option is D 63/5 gThe ideal gas law must hold both before and after the leakage of the gas.Thus PiVniTi=PfVnfTf⟹nf=PfPiTiTfni⟹nf=12273+57273+27ni=1120ni⟹mf=1120miThus leaked out mass=920mi=635g

28 G of N2 gas is contained in a flask at a pressure of 10 atm and at a temperature of $57^{\circ} C$ . It is found that due to leakage in the flask, the pressure is reduced to half and the temperature to $27^{\circ} C$ . The quantity of $N_{2}$ gas that leaked out is