The ratio of speed of sound in neon to that in water vapours at any temperature (when molecular weight of neon is `2.02xx10^(-2)kg mol^(-1)`

Options

A. 1.1
B. 1.7
C. 1.9
D. 1.3

Answer

To find the ratio of the speed of sound in neon to that in water vapors, we can use the formula for the speed of sound in a gas, which is given by:

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

Where:
- \( v \) = speed of sound
- \( \gamma \) = ratio of specific heats (Cp/Cv)
- \( R \) = universal gas constant
- \( T \) = absolute temperature
- \( M \) = molar mass of the gas

### Step 1: Identify the values of \( \gamma \) for both gases

- Neon is a monoatomic gas, so:
\[ \gamma_{\text{Neon}} = \frac{5}{3} \]

- Water vapor (H2O) is a triatomic gas, so:
\[ \gamma_{\text{Water}} = \frac{4}{3} \]

### Step 2: Identify the molar masses of both gases

- Molar mass of neon is given as:
\[ M_{\text{Neon}} = 2.02 \times 10^{-2} \, \text{kg/mol} \]

- Molar mass of water (H2O) is:
\[ M_{\text{Water}} = 18 \, \text{g/mol} = 18 \times 10^{-3} \, \text{kg/mol} \]

### Step 3: Set up the ratio of the speeds of sound

The ratio of the speeds of sound in neon and water vapor can be expressed as:
\[ \frac{v_{\text{Neon}}}{v_{\text{Water}}} = \sqrt{\frac{\gamma_{\text{Neon}} \cdot R \cdot T / M_{\text{Neon}}}{\gamma_{\text{Water}} \cdot R \cdot T / M_{\text{Water}}}} \]

Since \( R \) and \( T \) are constants and cancel out, we have:
\[ \frac{v_{\text{Neon}}}{v_{\text{Water}}} = \sqrt{\frac{\gamma_{\text{Neon}}}{\gamma_{\text{Water}}} \cdot \frac{M_{\text{Water}}}{M_{\text{Neon}}}} \]

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

Substituting the values we found:
\[ \frac{v_{\text{Neon}}}{v_{\text{Water}}} = \sqrt{\frac{\frac{5}{3}}{\frac{4}{3}} \cdot \frac{18 \times 10^{-3}}{2.02 \times 10^{-2}}} \]

This simplifies to:
\[ = \sqrt{\frac{5}{4} \cdot \frac{18 \times 10^{-3}}{2.02 \times 10^{-2}}} \]

### Step 5: Calculate the numerical values

Calculating the fraction:
\[ \frac{18 \times 10^{-3}}{2.02 \times 10^{-2}} = \frac{18}{20.2} = 0.8911 \]

Now substituting this back:
\[ \frac{v_{\text{Neon}}}{v_{\text{Water}}} = \sqrt{\frac{5}{4} \cdot 0.8911} = \sqrt{1.113875} \approx 1.06 \]

### Final Answer

Thus, the ratio of the speed of sound in neon to that in water vapors is approximately:
\[ \frac{v_{\text{Neon}}}{v_{\text{Water}}} \approx 1.06 \]