Question

One litre of Nitrogen gas and 7 / 8 litres of Oxygen gas under identical conditions of pressure and temperature are mixed the amount of gases present admixture show



A WN2=3WO2



B WN2 = 8Wo2



C WN2 = wo2



D WN2= 16wo2

Answer 1

D WN2=16wo2 is the answer

Answer 2

Answer:

The correct option is C.

Explanation:

Let the pressure and temperature at which both gases are mixed be P and T.

Volume of the nitrogen gas = 1 L

Moles of nitrogen gas = $n_{N_2}$

$PV=nRT$ (Ideal gas equation)

$n_{N_2}=\frac{P\times 1L}{RT}=\frac{w_{N_2}}{28 g/mol}$

Volume of the oxygen gas = 7/8 L

Moles of oxygen gas = $n_{O_2}$

$n_{O_2}=\frac{P\times \frac{7}{8} L}{RT}=\frac{w_{O_2}}{32 g/mol}$

On diving number of moles of both gases:

$\frac{n_{N_2}}{n_{O_2}}=\frac{\frac{P\times 1L}{RT}}{\frac{P\times \frac{7}{8} L}{RT}}=\frac{\frac{w_{N_2}}{28 g/mol}}{\frac{w_{O_2}}{32 g/mol}}$

$\frac{n_{N_2}}{n_{O_2}}=\frac{1}{1}=\frac{w_{N_2}}{w_{O_2}}$

$\frac{1}{1}=\frac{w_{N_2}}{w_{O_2}}$

$w_{N_2}=w_{O_2}$

Hence, the correct option is C.