Question

A sample of air contains \(\mathrm{N}_{2}, \mathrm{O}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). It is saturated with water vapour and total pressure is 640 torr. The vapour pressure of water is 40 torr and the molar ratio of \(\mathrm{N}_{2}: \mathrm{O}_{2}\) is \(3: 1\). The partial pressure of \(\mathrm{N}_{2}\) in the sample is (1) 480 torr (2) 600 torr (3) 525 torr (4) 450 torr

Short answer

450 torr

Step-by-step solution

- Determine partial pressure of water

Given the total pressure of the sample is 640 torr and the vapour pressure of water is 40 torr. Since the sample is saturated with water vapor, the partial pressure of water vapour is 40 torr.

- Calculate partial pressure of dry air components

Subtract the partial pressure of water vapor from the total pressure: \[ P_{dry air} = P_{total} - P_{H_2O} = 640 - 40 = 600 \text{ torr} \] The pressure of the dry air, which consists of only \ \mathrm{N}_{2} and \ \mathrm{O}_{2}, is 600 torr.

- Find mole fraction of \ \mathrm{N}_{2} and \ \mathrm{O}_{2}

The molar ratio of \ \mathrm{N}_{2} to \ \mathrm{O}_{2} is given as 3:1. Hence, let the mole fraction of \ \mathrm{N}_{2} be \ \mathrm{x} and the mole fraction of \ \mathrm{O}_{2} be \ \mathrm{y}. From the ratio \ \mathrm{N}_{2}: \ \mathrm{O}_{2} = 3:1, we get: \[ x = 3k \] \[ y = k \] where \ k is any positive constant.

- Apply mole fraction normalization

The sum of mole fractions must equal 1: \[ x + y = 1 \ \Rightarrow 3k + k = 1 \ \Rightarrow 4k = 1 \ \Rightarrow k = \frac{1}{4} \] Thus, the mole fraction of \ \mathrm{N}_{2} \left(x\right) and \ \mathrm{O}_{2} \left(y\right) are: \[ x = 3k = 3 \times \frac{1}{4} = \frac{3}{4} \] \[ y = k = \frac{1}{4} \]

- Calculate partial pressure of \ \mathrm{N}_{2}

Using the mole fraction of \ \mathrm{N}_{2} \left(\frac{3}{4}\right) and the dry air pressure (600 torr): \[ P_{N_2} = \left(\frac{3}{4}\right) \times 600 = 450 \text{ torr} \]

Key concepts

Gas Laws

Gas laws help us understand the behavior of gases under different conditions of pressure, volume, and temperature. The main gas laws include:

• Boyle's Law (Pressure-Volume Relationship)
• Charles's Law (Temperature-Volume Relationship)
• Gay-Lussac's Law (Pressure-Temperature Relationship)

For our specific problem, we use Dalton's Law of Partial Pressures. Dalton's Law states that the total pressure exerted by a gas mixture is the sum of the partial pressures of each component gas.
Knowing how to calculate and use these different gas laws is critical for solving problems involving gas mixtures and their behaviors. This is key to understanding our given air sample problem where multiple gases coexist.

Mole Fraction

Mole fraction represents the ratio of the number of moles of one component to the total number of moles in a mixture. It is a way of expressing concentration:

• Mole fraction of component i, \(\text{x}_i\): \[ x_i = \frac{{n_i}}{{n_{\text{total}}}} \]

Here, \(n_i\) is the number of moles of component i and \(n_{\text{total}}\) is the total moles of all components in the mixture. In our problem, the mole ratio of \(N_2\) to \(O_2\) is given as 3:1. Therefore, the total moles proportional for \(N_2\) and \(O_2\) are \(3k\) and \(k\), respectively, ensuring their mole fractions sum up to 1.
This allows us to distribute the pressure correctly according to each gas's mole fraction.

Vapour Pressure

Vapour pressure is the pressure exerted by the vapor of a liquid in dynamic equilibrium with its liquid at a given temperature. It is crucial in determining the saturation point of a liquid in the air.
For a sample of air containing water vapor, the saturation pressure, or the vapor pressure of water, is provided. In this exercise, the vapor pressure of water is given as 40 torr.
Since the sample is saturated with water vapor, the partial pressure of water vapor is 40 torr. This partial pressure needs to be deducted from the total pressure to find out the pressure that belongs to the rest of the components in the air mixture.

Partial Pressure

Partial pressure is the pressure that a single component of a gas mixture would exert if it occupied the entire volume alone at the same temperature. It is calculated using the mole fraction and the total pressure of the gas mixture.
The partial pressure can be determined as:
\[ P_i = x_i \times P_{\text{total}} \]
From the given problem, after deducting the vapor pressure of water, we calculate the partial pressure of the dry air components (other{N_2} and other{O_2}). Given the molar ratio and the calculated mole fraction of other{N_2} as \(3/4\), the partial pressure of other{N_2} is:
\[ P_{N_2} = \frac{3}{4} \times 600 \text{torr} = 450 \text{torr} \]
Understanding partial pressures assists in deciphering the individual pressures contributing to the total pressure in a mixture, like in our given sample of air.