Question

A jar is filled with two monoatomic non-interacting gases A and B with total masses $M_A$ and $M_B$, respectively. The molar mass of A is double the molar mass of B. If the jar is kept at temperature $T$, what is the ratio of the total pressure of the combined gas to the partial pressure due to the gas A?

• A. $1 + 2 \frac{M_B}{M_A}$
• B. $1 + \frac{1}{2} \frac{M_B}{M_A}$
• C. $1 + \frac{1}{2} \frac{M_A}{M_B}$
• D. $1 + 2 \frac{M_A}{M_B}$

Hint:

For ideal gas mixtures, the ratio of partial pressures is simply the ratio of their mole counts ($P_i \propto n_i$).
Convert masses to moles first to simplify any mixture pressure calculations.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
By Dalton's Law of Partial Pressures, the ratio of pressures is the same as the ratio of the number of moles of the gases.

Step 2: Detailed Explanation:
1. Mole Ratios: Let molar mass of B be \( m \). Then molar mass of A is \( 2m \).
Number of moles of A, \( n_{A} = M_{A} / 2m \).
Number of moles of B, \( n_{B} = M_{B} / m \).
2. Pressure Ratio:
\[ \frac{P_{\text{total}}}{P_{A}} = \frac{n_{A} + n_{B}}{n_{A}} = 1 + \frac{n_{B}}{n_{A}} \]
3. Calculation:
\[ \frac{n_{B}}{n_{A}} = \frac{M_{B}/m}{M_{A}/2m} = \frac{M_{B}}{m} \times \frac{2m}{M_{A}} = 2\frac{M_{B}}{M_{A}} \]
\[ \frac{P_{\text{total}}}{P_{A}} = 1 + 2 \frac{M_{B}}{M_{A}} \]

Step 3: Final Answer:
The ratio is \( 1 + 2 M_{B}/M_{A} \).