Calculate the temperature at which rms velocity of a gas molecule is same as that of a molecule of another gas at `47^(@)C`. Molecular weight of first and secon gases are 64 and 32 respectively.

Explanation

To solve the problem, we need to find the temperature at which the root mean square (rms) velocity of a gas with molecular weight \( M_1 = 64 \) g/mol is the same as that of another gas with molecular weight \( M_2 = 32 \) g/mol at a temperature of \( T_1 = 47^\circ C \).

### Step-by-Step Solution:

1. **Convert the temperature from Celsius to Kelvin:**
\[ T_1 = 47^\circ C + 273 = 320 \, K \]
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