Question

A tank of volume $V$ contains a homogeneous mixture of two ideal gases, A and B, at temperature $T$ and pressure $P$. The mixture contains $n_A$ moles of gas A and $n_B$ moles of gas B. If $P_A$ and $P_B$ are the partial pressures of gas A and B, respectively, then:

• A. $P_A = \dfrac{n_A}{n_A+n_B} P, P_B = \dfrac{n_B}{n_A+n_B} P$
• B. $P_A = \dfrac{n_B}{n_A} P, P_B = \dfrac{n_A}{n_B} P$
• C. $P_A = \dfrac{n_A}{n_B} P, P_B = \dfrac{n_B}{n_A} P$
• D. $P_A = \dfrac{n_B}{n_A+n_B} P, P_B = \dfrac{n_A}{n_A+n_B} P$

Hint:

In mixtures of ideal gases, partial pressure = mole fraction $\times$ total pressure.

Answer 1

The Correct Option is A

Step 1: Dalton’s Law of Partial Pressures.
In an ideal gas mixture, the partial pressure of a component is proportional to its mole fraction: \[ P_i = y_i P \] where $y_i = \dfrac{n_i}{n_A + n_B}$.
Step 2: Apply to each gas.
For gas A: \[ P_A = \frac{n_A}{n_A+n_B} P \] For gas B: \[ P_B = \frac{n_B}{n_A+n_B} P \]
Step 3: Verification.
Clearly, (A) is the correct relation. Final Answer: \[ \boxed{P_A = \frac{n_A}{n_A+n_B} P, \; P_B = \frac{n_B}{n_A+n_B} P} \]