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?
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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.
Step 1: Understanding the Question:
We are given a mixture of two monoatomic gases, A and B, in a container.
The total masses are $M_A$ and $M_B$, and the molar mass of gas A is twice that of gas B.
We need to find the ratio of the total pressure to the partial pressure of gas A. Step 2: Key Formula or Approach:
According to Dalton's law of partial pressures, the total pressure is the sum of the partial pressures:
\[ P_{total} = P_A + P_B \]
The ratio of total pressure to the partial pressure of A is:
\[ \frac{P_{total}}{P_A} = \frac{P_A + P_B}{P_A} = 1 + \frac{P_B}{P_A} \]
The partial pressure of an ideal gas is proportional to its number of moles ($n$), so:
\[ \frac{P_B}{P_A} = \frac{n_B}{n_A} \]
Step 3: Detailed Explanation:
• Let the molar mass of gas B be $M_{0B}$.
• The molar mass of gas A is $M_{0A} = 2 M_{0B}$.
• The number of moles of gas A is:
\[ n_A = \frac{M_A}{M_{0A}} = \frac{M_A}{2 M_{0B}} \]
• The number of moles of gas B is:
\[ n_B = \frac{M_B}{M_{0B}} \]
• Let us compute the ratio of the number of moles of B to A:
\[ \frac{n_B}{n_A} = \frac{\frac{M_B}{M_{0B}}}{\frac{M_A}{2 M_{0B}}} = 2 \frac{M_B}{M_A} \]
• Substituting this back into the pressure ratio formula:
\[ \frac{P_{total}}{P_A} = 1 + \frac{P_B}{P_A} = 1 + \frac{n_B}{n_A} = 1 + 2 \frac{M_B}{M_A} \]
Step 4: Final Answer:
The ratio of the total pressure to the partial pressure of gas A is $1 + 2 \frac{M_B}{M_A}$.
