A flask contains hydrogen and helium in the ratio `2:1` by mass. The temperature of the mixture is `27^(@)C`. Obtin the ratio of rms speeds of the molecules of the two gases. Atomic mass of hydrogen =2, molecular mass of helium =4

Explanation

To find the ratio of the root mean square (rms) speeds of hydrogen and helium in a mixture, we will follow these steps:

### Step 1: Understand the formula for rms speed
The root mean square speed (Vrms) of a gas is given by the formula:
\[ V_{rms} = \sqrt{\frac{3RT}{M}} \]
where:
- \(R\) is the universal gas constant,
- \(T\) is the absolute temperature in Kelvin,
- \(M\) is the molar mass of the gas.

### Step 2: Convert temperature to Kelvin
The temperature given is \(27^\circ C\). To convert this to Kelvin:
\[ T(K) = 27 + 273 = 300 \, K \]

### Step 3: Identify the molar masses
From the problem statement:
- The molar mass of hydrogen (\(M_H\)) = 2 g/mol
- The molar mass of helium (\(M_{He}\)) = 4 g/mol

### Step 4: Write the expressions for Vrms for both gases
Using the formula for rms speed, we can write:
- For hydrogen:
\[ V_{rms, H} = \sqrt{\frac{3RT}{M_H}} = \sqrt{\frac{3R \cdot 300}{2}} \]
- For helium:
\[ V_{rms, He} = \sqrt{\frac{3RT}{M_{He}}} = \sqrt{\frac{3R \cdot 300}{4}} \]

### Step 5: Find the ratio of Vrms speeds
Now, we take the ratio of the rms speeds of hydrogen to helium:
\[ \frac{V_{rms, H}}{V_{rms, He}} = \frac{\sqrt{\frac{3R \cdot 300}{2}}}{\sqrt{\frac{3R \cdot 300}{4}}} \]

### Step 6: Simplify the ratio
The \(3R \cdot 300\) terms cancel out:
\[ \frac{V_{rms, H}}{V_{rms, He}} = \frac{\sqrt{\frac{1}{2}}}{\sqrt{\frac{1}{4}}} = \frac{\sqrt{4}}{\sqrt{2}} = \frac{2}{1} \]

### Step 7: Final result
Thus, the ratio of the rms speeds of hydrogen to helium is:
\[ V_{rms, H} : V_{rms, He} = \sqrt{2} : 1 \]