Question

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 the total pressure is 1.15 atm, then calculate the ratio of the following 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)

• A. 4.26, 19.3
• B. 2.59, 11.85
• C. 5.46, 17.8
• D. 2.96, 11.2

Hint:

Dalton’s Law of partial pressures allows you to find the partial pressures of gases from their mole fractions. Ensure to use mass percentages and molar masses to first calculate the moles of each gas.

Answer 1

The Correct Option is D

Step 1: Understanding Dalton’s Law of Partial Pressures
Dalton’s Law states that the partial pressure of a gas is proportional to its mole fraction in the mixture. The partial pressure is given by: \[ P_{\text{gas}} = X_{\text{gas}} \times P_{\text{total}} \] Where: \( P_{\text{gas}} \) is the partial pressure of the gas, \( X_{\text{gas}} \) is the mole fraction of the gas, \( P_{\text{total}} \) is the total pressure.
Step 2: Calculate the mole fraction of each gas
We are given the mass percentages of the gases and their molar masses:
Mass percentage of nitrogen (\(N_2\)) = 70.0% Mass percentage of oxygen (\(O_2\)) = 27.0% Mass percentage of argon (\(Ar\)) = 3.0%
The moles of each gas are calculated by dividing the mass by the molar mass of each gas: \[ \text{Moles of } N_2 = \frac{70}{14} = 5 \, \text{moles} \] \[ \text{Moles of } O_2 = \frac{27}{16} = 1.6875 \, \text{moles} \] \[ \text{Moles of } Ar = \frac{3}{40} = 0.075 \, \text{moles} \] Now, the total moles of gases: \[ \text{Total moles} = 5 + 1.6875 + 0.075 = 6.7625 \, \text{moles} \]
Step 3: Calculate the partial pressures
The mole fraction of each gas is given by: \[ X_{N_2} = \frac{5}{6.7625} = 0.7396, \quad X_{O_2} = \frac{1.6875}{6.7625} = 0.2499, \quad X_{Ar} = \frac{0.075}{6.7625} = 0.0111 \] Now, calculate the partial pressures: \[ P_{N_2} = X_{N_2} \times P_{\text{total}} = 0.7396 \times 1.15 = 0.8516 \, \text{atm} \] \[ P_{O_2} = X_{O_2} \times P_{\text{total}} = 0.2499 \times 1.15 = 0.2879 \, \text{atm} \] \[ P_{Ar} = X_{Ar} \times P_{\text{total}} = 0.0111 \times 1.15 = 0.0128 \, \text{atm} \]
Step 4: Calculate the ratios
(i) The ratio of the partial pressures of nitrogen to oxygen is: \[ \frac{P_{N_2}}{P_{O_2}} = \frac{0.8516}{0.2879} = 2.96 \] (ii) The ratio of the partial pressures of oxygen to argon is: \[ \frac{P_{O_2}}{P_{Ar}} = \frac{0.2879}{0.0128} = 11.2 \]
Thus, the correct answer is (4) 2.96, 11.2.

Answer 2

We will apply Dalton's Law of Partial Pressures. According to Dalton's law, the partial pressure of a gas in a mixture is directly proportional to its mole fraction in the mixture. The equation is: \[ \frac{P_{N_2}}{P_{O_2}} = \frac{X_{N_2}}{X_{O_2}} = \frac{n_{N_2}}{n_{O_2}} \] Where: - $P_{N_2}$: Partial pressure of $N_2$ - $P_{O_2}$: Partial pressure of $O_2$ - $X_{N_2}$: Mole fraction of $N_2$ - $X_{O_2}$: Mole fraction of $O_2$ - $n_{N_2}$: Number of moles of $N_2$ - $n_{O_2}$: Number of moles of $O_2$ Now, using the given data: - Partial pressure of $N_2 = 70$ kPa - Partial pressure of $O_2 = 27$ kPa - Mole fraction of $O_2 = \frac{27}{32}$ - Mole fraction of $N_2 = \frac{70}{28}$ Applying Dalton's law and substituting the given values: \[ \frac{P_{N_2}}{P_{O_2}} = \frac{X_{N_2}}{X_{O_2}} = \frac{\frac{70}{28}}{\frac{27}{32}} = 2.96 \] This means the ratio of partial pressures is 2.96, indicating that the partial pressure of $N_2$ is higher than $O_2$ by a factor of 2.96. Next, we calculate the ratio of the partial pressure of $O_2$ to that of $Ar$ (argon): \[ \frac{P_{O_2}}{P_{Ar}} = \frac{n_{O_2}}{n_{Ar}} \] Given: - $n_{O_2} = \frac{27}{32}$ - $n_{Ar} = \frac{3}{40}$ Substituting the values into the equation: \[ \frac{P_{O_2}}{P_{Ar}} = \frac{\frac{27}{32}}{\frac{3}{40}} = 11.25 \] Thus, the ratio of partial pressures of $O_2$ to $Ar$ is 11.25. This indicates that $O_2$ has a significantly higher partial pressure compared to $Ar$. This is the final solution using Dalton's Law of Partial Pressures.