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Standard XII

Physics

P-T Relationship

28 g of N...

Question

$28$ $g$ of $N_{2}$ gas is contained in a flask at a pressure $10$ $a t m$ and at a temperature of $57^{0}$ $C$ . It is found that due to leakage in the flask, the pressure is reduced to half and the temperature reduced to $27^{0}$ $C$ .The quantity of $N_{2}$ gas that leaked out is:

\frac{11}{20}

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\frac{80}{11}

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\frac{5}{63}

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\frac{63}{5}

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Solution

The correct option is D $\frac{63}{5}$ g
$\frac{P}{n T} =$ constant
( $28$ $g$ of $N_{2} = 1$ $m o l$ )
$\frac{10}{1 \times (273 + 57)} = \frac{5}{x \times (273 + 27)}$
$x = \frac{330}{300} \times \frac{5}{10} = \frac{33}{60} = \frac{11}{20}$
$∴ x$ moles weighs $= \frac{33}{60} \times 28 = \frac{77}{5}$
$∴ N_{2}$ leaked out $= 28 + \frac{77}{5} = \frac{63}{5} g$

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28 g of N2 gas is contained in a flask at a pressure 10 atm and at a temperature of 570 C. It is found that due to leakage in the flask, the pressure is reduced to half and the temperature reduced to 270 C.The quantity of N2 gas that leaked out is:

The correct option is D 635gPnT= constant (28 g of N2=1 mol)101×(273+57)=5x×(273+27)x=330300×510=3360=1120∴x moles weighs =3360×28=775∴N2 leaked out =28+775=635g

$28$ $g$ of $N_{2}$ gas is contained in a flask at a pressure $10$ $a t m$ and at a temperature of $57^{0}$ $C$ . It is found that due to leakage in the flask, the pressure is reduced to half and the temperature reduced to $27^{0}$ $C$ .The quantity of $N_{2}$ gas that leaked out is: