At the sea level, the dry air mass percentage composition is given as nitrogen gas : 70.0, oxygen gas : 27.0 and argon gas : 3.0. If total pressure is 1.15 atm, then calculate the ratio of followings respectively : (i) partial pressure of nitrogen gas to

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Solution

Q.
At the sea level, the dry air mass percentage composition is given as nitrogen gas : 70.0, oxygen gas : 27.0 and argon gas : 3.0. If total pressure is 1.15 atm, then calculate the ratio of followings respectively :

(i) partial pressure of nitrogen gas to partial pressure of oxygen gas
(ii) partial pressure of oxygen gas to partial pressure of argon gas

(Given: Molar mass of N, O and Ar are 14, 16, and 40 g ${mol}^{- 1}$ respectively) [2025]

1 4.26, 19.3

2 2.59, 11.85

3 5.46, 17.8

4 2.96, 11.2

Ans.
(4)

As per Dalton's law of partial pressure, partial pressure of a gas = mole fraction of that gas $\times$ total pressure.

So:

$\frac{P_{N_{2}}}{P_{O_{2}}} = \frac{x_{N_{2}} P_{Total}}{x_{O_{2}} P_{Total}} = \frac{x_{N_{2}}}{x_{O_{2}}} = \frac{n_{N_{2}}}{n_{O_{2}}} = \frac{W_{N_{2}} / M_{N_{2}}}{W_{O_{2}} / M_{O_{2}}} = \frac{70 / 28}{27 / 32} = 2.96$

$\frac{P_{O_{2}}}{P_{A r}} = \frac{x_{O_{2}} P_{Total}}{x_{A r} P_{Total}} = \frac{x_{O_{2}}}{x_{A r}} = \frac{n_{O_{2}}}{n_{A r}} = \frac{W_{O_{2}} / M_{O_{2}}}{W_{A r} / M_{A r}} = \frac{27 / 32}{3 / 40} = 11.25$

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