A long tube contains air at a pressure of `1 atm` and a temperature of `107^(@) C`. The tube is open at one end and closed at the other by a movable piston. A tuning fork near the open end is vibrating with a frequency of `500 Hz`. Resonance is produced when the piston is at distance `19 , 58.5 and 98 cm` from the open end.
The molar mass of air is `28.8 g//mol`. The ratio of molar heat capacities at constant pressure and constant volume for air at this temperature is nearly

Options

A. `1.66`
B. `1.4`
C. `1.33`
D. `1.5`

Explanation

To solve the problem, we need to find the ratio of molar heat capacities at constant pressure and constant volume (γ) for air at the given temperature. We will follow these steps:

### Step 1: Identify the resonance lengths
The resonance occurs at three positions:
- \( L_1 = 19 \, \text{cm} \)
- \( L_2 = 58.5 \, \text{cm} \)
- \( L_3 = 98 \, \text{cm} \)

### Step 2: Calculate the wavelength (λ)
For a tube open at one end and closed at the other, the resonance lengths can be expressed as:
- First resonance: \( L_1 + e = \frac{\lambda}{4} \)
- Second resonance: \( L_2 + e = \frac{3\lambda}{4} \)

Subtracting these two equations gives:
\[ L_2 - L_1 = \frac{2\lambda}{4} = \frac{\lambda}{2} \]
Thus, we can find the wavelength:
\[ \lambda = 2(L_2 - L_1) = 2(58.5 \, \text{cm} - 19 \, \text{cm}) = 2 \times 39.5 \, \text{cm} = 79 \, \text{cm} = 0.79 \, \text{m} \]

### Step 3: Calculate the speed of sound (v)
Using the frequency (f) and wavelength (λ), the speed of sound can be calculated as:
\[ v = f \lambda = 500 \, \text{Hz} \times 0.79 \, \text{m} = 395 \, \text{m/s} \]

### Step 4: Use the formula for the speed of sound
The speed of sound in a gas is given by the formula:
\[ v = \sqrt{\frac{\gamma R T}{M}} \]
Where:
- \( \gamma \) = ratio of molar heat capacities
- \( R \) = universal gas constant = \( 8.314 \, \text{J/(mol K)} \)
- \( T \) = absolute temperature in Kelvin = \( 107 + 273 = 380 \, \text{K} \)
- \( M \) = molar mass in kg/mol = \( 28.8 \, \text{g/mol} = 0.0288 \, \text{kg/mol} \)

### Step 5: Rearranging the formula to find γ
Rearranging the formula to solve for γ:
\[ \gamma = \frac{v^2 M}{RT} \]
Substituting the values:
\[ \gamma = \frac{(395 \, \text{m/s})^2 \times 0.0288 \, \text{kg/mol}}{8.314 \, \text{J/(mol K)} \times 380 \, \text{K}} \]

### Step 6: Perform the calculations
Calculating the numerator:
\[ (395)^2 = 156025 \, \text{m}^2/\text{s}^2 \]
\[ \text{Numerator} = 156025 \times 0.0288 = 4491.6 \]
Calculating the denominator:
\[ \text{Denominator} = 8.314 \times 380 = 3160.32 \]
Now, calculate γ:
\[ \gamma = \frac{4491.6}{3160.32} \approx 1.42 \]

### Step 7: Conclusion
The ratio of molar heat capacities at constant pressure and constant volume for air at this temperature is approximately \( \gamma \approx 1.4 \).

### Final Answer
The correct option is \( 2 \).

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