A sample of oxygen at NTP has volume V and a sample of hydrogen at NTP has the volume 4V. Both the gases are mixed. If the speed of sound in hydrogen at NTP is 1270 m/s, that in mixture is

Options

A. 317 m/s
B. 635 m/s
C. 830 m/s
D. 950 m/s

Answer

To solve the problem of finding the speed of sound in a mixture of oxygen and hydrogen gases, we can follow these steps:

### Step 1: Understand the Given Information
- Volume of oxygen (O₂) at NTP = V
- Volume of hydrogen (H₂) at NTP = 4V
- Speed of sound in hydrogen (H₂) = 1270 m/s

### Step 2: Calculate the Densities of the Gases
- Molar mass of oxygen (O₂) = 32 g/mol
- Molar mass of hydrogen (H₂) = 2 g/mol

Using the relationship between density, mass, and volume:
\[ \text{Density} (\rho) = \frac{\text{Mass}}{\text{Volume}} \]

For oxygen:
\[ \rho_{O_2} = \frac{M_{O_2}}{V} = \frac{32 \text{ g/mol}}{V} \]

For hydrogen:
\[ \rho_{H_2} = \frac{M_{H_2}}{4V} = \frac{2 \text{ g/mol}}{4V} = \frac{0.5 \text{ g/mol}}{V} \]

### Step 3: Establish the Relationship Between Densities
Since the density of oxygen is 16 times that of hydrogen:
\[ \rho_{O_2} = 16 \cdot \rho_{H_2} \]

### Step 4: Calculate the Density of the Mixture
The total mass of the mixture is:
\[ \text{Mass}_{\text{mixture}} = \text{Mass}_{O_2} + \text{Mass}_{H_2} = \rho_{O_2} \cdot V + \rho_{H_2} \cdot 4V \]

Substituting the densities:
\[ \text{Mass}_{\text{mixture}} = (16 \cdot \rho_{H_2} \cdot V) + (\rho_{H_2} \cdot 4V) = (16\rho_{H_2} + 4\rho_{H_2})V = 20\rho_{H_2}V \]

The total volume of the mixture is:
\[ V_{\text{total}} = V + 4V = 5V \]

Now, the density of the mixture is:
\[ \rho_{\text{mixture}} = \frac{\text{Mass}_{\text{mixture}}}{V_{\text{total}}} = \frac{20\rho_{H_2}V}{5V} = 4\rho_{H_2} \]

### Step 5: Relate the Speed of Sound in the Mixture to the Speed of Sound in Hydrogen
The speed of sound in a gas is given by 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.

Since the pressure and \( \gamma \) remain constant for both gases, we can establish:
\[ \frac{v_{H_2}}{v_{\text{mixture}}} = \sqrt{\frac{\rho_{\text{mixture}}}{\rho_{H_2}}} \]

Substituting the densities:
\[ \frac{1270 \, \text{m/s}}{v_{\text{mixture}}} = \sqrt{\frac{4\rho_{H_2}}{\rho_{H_2}}} = \sqrt{4} = 2 \]

### Step 6: Solve for the Speed of Sound in the Mixture
Rearranging gives:
\[ v_{\text{mixture}} = \frac{1270 \, \text{m/s}}{2} = 635 \, \text{m/s} \]

### Final Answer
The speed of sound in the mixture is **635 m/s**.

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