The ratio of speed of sound in monomatomic gas to that in water vapours at any temperature is. (when molecular weight of gas is `40 gm//mol` and for water vapours is `18 gm//mol`)

Options

A. `0.75`
B. `0.73`
C. `0.68`
D. None of these

Explanation

To find the ratio of the speed of sound in a monatomic gas to that in water vapors, we can use the formula for the speed of sound in a gas:

\[ V = \sqrt{\frac{\gamma RT}{M}} \]

Where:
- \( V \) is the speed of sound,
- \( \gamma \) is the adiabatic index (ratio of specific heats),
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature,
- \( M \) is the molar mass of the gas.

### Step 1: Write the ratio of the speeds of sound

Let \( V_1 \) be the speed of sound in the monatomic gas and \( V_2 \) be the speed of sound in water vapors. The ratio can be expressed as:

\[ \frac{V_1}{V_2} = \sqrt{\frac{\gamma_1 RT}{M_1}} \div \sqrt{\frac{\gamma_2 RT}{M_2}} = \sqrt{\frac{\gamma_1}{\gamma_2} \cdot \frac{M_2}{M_1}} \]

### Step 2: Identify the values of \( \gamma \)

For a monatomic gas, \( \gamma_1 = \frac{5}{3} \).

For water vapors (considered as a polyatomic gas), \( \gamma_2 = \frac{4}{3} \).

### Step 3: Identify the molar masses

Given:
- Molar mass of the monatomic gas, \( M_1 = 40 \, \text{g/mol} \)
- Molar mass of water vapors, \( M_2 = 18 \, \text{g/mol} \)

### Step 4: Substitute the values into the ratio

Now substituting the values into the ratio:

\[ \frac{V_1}{V_2} = \sqrt{\frac{\frac{5}{3}}{\frac{4}{3}} \cdot \frac{18}{40}} \]

### Step 5: Simplify the expression

This simplifies to:

\[ \frac{V_1}{V_2} = \sqrt{\frac{5}{4} \cdot \frac{18}{40}} = \sqrt{\frac{5 \cdot 18}{4 \cdot 40}} = \sqrt{\frac{90}{160}} = \sqrt{\frac{9}{16}} = \frac{3}{4} \]

### Step 6: Final result

Thus, the ratio of the speed of sound in the monatomic gas to that in water vapors is:

\[ \frac{V_1}{V_2} = 0.75 \]

### Conclusion

The final answer is:

**The ratio of the speed of sound in monatomic gas to that in water vapors is 0.75.**

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