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Two identical vessels contain two differ...

Two identical vessels contain two different ideal gases at the same temperature. If the average speed of gas molecules in the first vessel is euqal to the most probable speed of molecules in the second vessel, then the ratio of the mass of gas molecules in the first vessel to that in the second vessel is

A

`4/(pi)`

B

`8/(pi)`

C

`2/(pi)`

D

`(pi)/2`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to establish the relationship between the average speed of gas molecules in the first vessel and the most probable speed of gas molecules in the second vessel. ### Step-by-Step Solution: 1. **Define the Variables:** - Let the molar mass of gas in the first vessel (gas A) be \( M_A \). - Let the molar mass of gas in the second vessel (gas B) be \( M_B \). - Both gases are at the same temperature \( T \). 2. **Average Speed of Gas Molecules:** The average speed \( v_{\text{avg}} \) of gas molecules is given by the formula: \[ v_{\text{avg}} = \sqrt{\frac{8RT}{\pi M_A}} \] where \( R \) is the universal gas constant. 3. **Most Probable Speed of Gas Molecules:** The most probable speed \( v_{\text{mp}} \) of gas molecules is given by the formula: \[ v_{\text{mp}} = \sqrt{\frac{2RT}{M_B}} \] 4. **Equating the Speeds:** According to the problem, the average speed of gas A is equal to the most probable speed of gas B: \[ \sqrt{\frac{8RT}{\pi M_A}} = \sqrt{\frac{2RT}{M_B}} \] 5. **Squaring Both Sides:** To eliminate the square roots, we square both sides: \[ \frac{8RT}{\pi M_A} = \frac{2RT}{M_B} \] 6. **Canceling Common Terms:** Since \( RT \) is common on both sides, we can cancel it out (assuming \( RT \neq 0 \)): \[ \frac{8}{\pi M_A} = \frac{2}{M_B} \] 7. **Cross Multiplying:** Cross-multiplying gives us: \[ 8M_B = 2\pi M_A \] 8. **Rearranging for the Mass Ratio:** Rearranging the equation to find the ratio of the masses: \[ \frac{M_A}{M_B} = \frac{8}{2\pi} = \frac{4}{\pi} \] ### Final Answer: The ratio of the mass of gas molecules in the first vessel to that in the second vessel is: \[ \frac{M_A}{M_B} = \frac{4}{\pi} \]
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