A closed organ pipe resonates in its fundamental mode at a frequency of `200 H_(Z)` with `O_(2)` in the pipe at a certain temperature. If the pipe now contains `2` moles of `O_(2)` and `3` moles of ozone, then what will be fundamental frequency of same pipe at same temperature?

Options

A. `268.23 H_(Z)`
B. `175.4 H_(Z)`
C. `149.45 H_(Z)`
D. none of these

Explanation

To solve the problem, we need to determine the fundamental frequency of a closed organ pipe when the gas inside it changes from pure oxygen to a mixture of oxygen and ozone. We will follow these steps:

### Step 1: Determine the molecular masses of the gases involved.
- The molecular mass of O2 (oxygen) is 32 g/mol.
- The molecular mass of O3 (ozone) is 48 g/mol.

### Step 2: Calculate the average molecular mass of the gas mixture.
- We have 2 moles of O2 and 3 moles of O3.
- The average molecular mass (M_avg) can be calculated using the formula:

\[ M_{avg} = \frac{(n_1 \cdot M_1) + (n_2 \cdot M_2)}{n_1 + n_2} \]

Where:
- \( n_1 = 2 \) moles of O2
- \( M_1 = 32 \) g/mol (molecular mass of O2)
- \( n_2 = 3 \) moles of O3
- \( M_2 = 48 \) g/mol (molecular mass of O3)

Plugging in the values:

\[ M_{avg} = \frac{(2 \cdot 32) + (3 \cdot 48)}{2 + 3} = \frac{64 + 144}{5} = \frac{208}{5} = 41.6 \text{ g/mol} \]

### Step 3: Understand the relationship between frequency and molecular mass.
- The frequency of the sound in the pipe is related to the speed of sound, which depends on the molecular mass of the gas. The speed of sound \( v \) in a gas is given by:

\[ v = \sqrt{\frac{\gamma R T}{M}} \]

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

### Step 4: Relate the frequencies before and after the change in gas.
- The fundamental frequency \( f \) of a closed organ pipe is inversely proportional to the square root of the molecular mass of the gas inside it:

\[ f \propto \frac{1}{\sqrt{M}} \]

Thus, we can set up the ratio of the new frequency \( f_2 \) to the old frequency \( f_1 \):

\[ \frac{f_2}{f_1} = \sqrt{\frac{M_1}{M_2}} \]

Where:
- \( f_1 = 200 \, \text{Hz} \) (initial frequency),
- \( M_1 = 32 \, \text{g/mol} \) (molecular mass of O2),
- \( M_2 = 41.6 \, \text{g/mol} \) (average molecular mass of the mixture).

### Step 5: Calculate the new frequency.
Substituting the values into the equation:

\[ \frac{f_2}{200} = \sqrt{\frac{32}{41.6}} \]

Calculating the right side:

\[ \sqrt{\frac{32}{41.6}} \approx \sqrt{0.768} \approx 0.876 \]

Now, we can find \( f_2 \):

\[ f_2 = 200 \times 0.876 \approx 175.2 \, \text{Hz} \]

### Final Answer:
The fundamental frequency of the pipe with the new gas mixture is approximately **175.2 Hz**.

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