The speed of sound in hydrogen gas at N.T.P is `1328ms^(-1)`. If the density of hydrogen is `1//16^("th")` of that of air, then the speed of sound in air in at N.T.P is

Options

A. `340ms^(-1)`
B. `332ms^(-1)`
C. `320ms^(-1)`
D. `280ms^(-1)`

Answer

To find the speed of sound in air at N.T.P given the speed of sound in hydrogen and the relationship between the densities of hydrogen and air, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between speed of sound, pressure, and density:**
The speed of sound (v) in a gas can be expressed using the formula:
\[ v = \sqrt{\frac{\gamma P}{\rho}} \]
where \( \gamma \) is the adiabatic index (ratio of specific heats), \( P \) is the pressure, and \( \rho \) is the density of the gas.

2. **Set up the ratio of speeds of sound in hydrogen and air:**
Since the pressure is the same for both gases at N.T.P, we can write:
\[ \frac{v_H}{v_A} = \sqrt{\frac{P_H}{P_A} \cdot \frac{\rho_A}{\rho_H}} = \sqrt{\frac{\rho_A}{\rho_H}} \]
Here, \( v_H \) is the speed of sound in hydrogen, and \( v_A \) is the speed of sound in air.

3. **Use the given information:**
We know:
- \( v_H = 1328 \, \text{m/s} \)
- The density of hydrogen \( \rho_H \) is \( \frac{1}{16} \) of the density of air \( \rho_A \):
\[ \rho_H = \frac{\rho_A}{16} \]

4. **Substituting the density relationship into the equation:**
Substitute \( \rho_H \) into the ratio:
\[ \frac{v_H}{v_A} = \sqrt{\frac{\rho_A}{\frac{\rho_A}{16}}} = \sqrt{16} = 4 \]

5. **Rearranging to find \( v_A \):**
Now, we can rearrange the equation to find the speed of sound in air:
\[ v_A = \frac{v_H}{4} \]
Substituting the value of \( v_H \):
\[ v_A = \frac{1328}{4} = 332 \, \text{m/s} \]

6. **Final answer:**
Therefore, the speed of sound in air at N.T.P is:
\[ \boxed{332 \, \text{m/s}} \]