Question

A gaseous mixture was prepared by taking equal moles of $CO$ and $N_2.$ If the total pressure of the mixture was found $1$ atmosphere, the partial pressure of the nitrogen $(N_2)$ in the mixture is

• A. 0.8 atm
• B. 0.9 atm
• C. 1 atm
• D. 0.5 atm

Answer 1

The Correct Option is D

To solve the problem of finding the partial pressure of nitrogen \( (N_2) \) in a gaseous mixture, we can use Dalton's Law of Partial Pressures. Dalton's law states that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of individual gases.

Given that the gaseous mixture was prepared by taking equal moles of carbon monoxide \( (CO) \) and nitrogen \( (N_2) \), let's explore this concept mathematically:

1. Understanding the composition of the mixture:

• Since the moles of \( CO \) and \( N_2 \) are equal, we can denote the moles of \( CO \) as \( n \) and the moles of \( N_2 \) as \( n \).

2. Applying Dalton's Law:

• According to Dalton's Law, the total pressure \( P \) of the mixture is the sum of partial pressures of \( CO \) and \( N_2 \).
• Total pressure, \( P_{\text{total}} = P_{CO} + P_{N_2} = 1 \) atm
• Since both gases have equal moles, the partial pressures will depend on their molar ratios, which are equal in this case.

3. Calculating partial pressures:

• If \( n_{CO} = n_{N_2} \), then \( \text{mole fraction of } CO = \text{mole fraction of } N_2 = \frac{1}{2} \).
• Partial pressure of \( N_2 , P_{N_2} = \left(\text{mole fraction of } N_2\right) \times P_{\text{total}} = \left(\frac{1}{2}\right) \times 1 \text{ atm} = 0.5 \text{ atm}.

Thus, the partial pressure of nitrogen \( (N_2) \) in the mixture is \(\mathbf{0.5\, \text{atm}}\).

By following the step-by-step reasoning and applying Dalton's Law, we can conclude that the partial pressure of \( N_2 \) is indeed 0.5 atm, which matches the given correct answer.