Question

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

Answer 1

Answer:

Explanation:

PV

Answer 2

Answer:

The quantity of N₂ gas that leaked out is 12.6 g.

Explanation:

Given that,

Weight of gas = 28 g

Pressure = 10 atm

Temperature = 57° C

28 g of N₂ mean 1 g- mol.

If the pressure is reduced to half and the temperature reduced to 27 degree Celsius.

We need to calculate the quantity of N₂ gas that leaked out

Using formula of ideal gas

$PV=nRT$

$n = \dfrac{PV}{RT}$

So, $n\propto\dfrac{P}{T}$

The ratio of initial and final quantity of gas is

$\dfrac{n_{f}}{n_{i}}=\dfrac{P_{f}}{P_{i}}\tinmes\dfrac{T_{i}}{T_{f}}$

Put the value into the formula

$\dfrac{n_{f}}{n_{i}}=\dfrac{1}{2}\times\dfrac{273+57}{273+27}$

$\dfrac{n_{f}}{n_{i}}=\dfrac{11}{20}$

$n_{f}=\dfrac{11}{20}n_{i}$

$n_{f}=\dfrac{11}{20}\ mol$

We need to calculate the change in quantity

$\Delta n=n_{i}-n_{f}$

Put the value into the formula

$\Delta n=1-\dfrac{11}{20}$

$\Delta n=\dfrac{9}{20}\ g-mol$

We need to calculate the moles

$\Delta m=\Delta n\times M$

Where, M = molar mass

n = moles of the gas

m = given mass of gas

$\Delta m=\dfrac{9}{20}\times28$

$\Delta m=12.6\ g$

Hence, The quantity of N₂ gas that leaked out is 12.6 g.