At normal temperature and pressure, the speed of sound in air is `332 ms^(-1)`. What will be the speed of sound in hydrogen (i) at normal temperature and pressure (ii) at `819^@C` temperature and 4 atmospheric pressure. Given air is 16 times heavier than hydrogen.

Answer

To solve the problem, we need to find the speed of sound in hydrogen under two different conditions: (i) at normal temperature and pressure (NTP) and (ii) at 819°C and 4 atmospheric pressure.

### Part (i): Speed of Sound in Hydrogen at Normal Temperature and Pressure

1. **Understanding the Relationship**: The speed of sound in a gas is given by the formula:
\[ v = \sqrt{\frac{P}{\rho}} \]
where \( v \) is the speed of sound, \( P \) is the pressure, and \( \rho \) is the density of the gas.

2. **Setting Up the Ratio**: We can set up a ratio of the speed of sound in air to that in hydrogen:
\[ \frac{v_{\text{air}}}{v_{\text{hydrogen}}} = \sqrt{\frac{\rho_{\text{hydrogen}}}{\rho_{\text{air}}}} \]
Since the pressure is the same for both gases at NTP, it cancels out.

3. **Using the Given Information**: We know that air is 16 times heavier than hydrogen, which means:
\[ \rho_{\text{air}} = 16 \cdot \rho_{\text{hydrogen}} \]

4. **Substituting Densities into the Ratio**:
\[ \frac{v_{\text{air}}}{v_{\text{hydrogen}}} = \sqrt{\frac{\rho_{\text{hydrogen}}}{16 \cdot \rho_{\text{hydrogen}}}} = \sqrt{\frac{1}{16}} = \frac{1}{4} \]

5. **Finding the Speed of Sound in Hydrogen**:
\[ v_{\text{hydrogen}} = 4 \cdot v_{\text{air}} = 4 \cdot 332 \, \text{m/s} = 1328 \, \text{m/s} \]

### Part (ii): Speed of Sound in Hydrogen at 819°C and 4 Atmospheric Pressure

1. **Converting Temperature to Kelvin**:
\[ T_{\text{initial}} = 0°C = 273 \, K \]
\[ T_{\text{final}} = 819°C = 819 + 273 = 1092 \, K \]

2. **Using the Temperature Relationship**: The speed of sound in a gas is also related to the temperature:
\[ v \propto \sqrt{T} \]
Therefore, we can set up the ratio:
\[ \frac{v_{\text{hydrogen, initial}}}{v_{\text{hydrogen, final}}} = \sqrt{\frac{T_{\text{initial}}}{T_{\text{final}}}} = \sqrt{\frac{273}{1092}} \]

3. **Calculating the Ratio**:
\[ \frac{v_{\text{hydrogen, final}}}{v_{\text{hydrogen, initial}}} = \sqrt{\frac{273}{1092}} \approx \sqrt{\frac{1}{4}} = \frac{1}{2} \]

4. **Finding the New Speed of Sound in Hydrogen**:
\[ v_{\text{hydrogen, final}} = 2 \cdot v_{\text{hydrogen, initial}} = 2 \cdot 1328 \, \text{m/s} = 2656 \, \text{m/s} \]

### Final Answers:
- (i) Speed of sound in hydrogen at normal temperature and pressure: **1328 m/s**
- (ii) Speed of sound in hydrogen at 819°C and 4 atmospheric pressure: **2656 m/s**