A long tube contains air pressure of `1 atm` and a temperature of `59^(@) 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 distances `16 cm , 49.2 cm and 82.4 cm` from open end. Molar mass of air is `28.8 g//mol`.
Ratio of heat capacities at constant pressure and constant volume for air at `59^(@) C` is

Options

A. `1.4`
B. `1.152`
C. `1.60`
D. `2`

Explanation

To find the ratio of heat capacities at constant pressure and constant volume (denoted as \( \gamma = \frac{C_P}{C_V} \)) for air at \( 59^{\circ} C \), we can follow these steps:

### Step 1: Identify the distances for resonance
The distances at which resonance occurs are given as:
- \( L_1 = 16 \, \text{cm} = 0.16 \, \text{m} \)
- \( L_2 = 49.2 \, \text{cm} = 0.492 \, \text{m} \)
- \( L_3 = 82.4 \, \text{cm} = 0.824 \, \text{m} \)

### Step 2: Calculate the wavelength \( \lambda \)
In a tube open at one end and closed at the other, the resonance conditions can be expressed as:
- For the first resonance (fundamental frequency): \( L_1 + \frac{1}{4} \lambda = \text{constant} \)
- For the second resonance: \( L_2 + \frac{1}{4} \lambda = \text{constant} \)

The difference between the two resonance lengths gives:
\[ L_2 - L_1 = \frac{3}{4} \lambda - \frac{1}{4} \lambda = \frac{2}{4} \lambda = \frac{1}{2} \lambda \]
Thus, we can express \( \lambda \) as:
\[ \lambda = 2(L_2 - L_1) \]

Calculating \( L_2 - L_1 \):
\[ L_2 - L_1 = 0.492 \, \text{m} - 0.16 \, \text{m} = 0.332 \, \text{m} \]
Then,
\[ \lambda = 2 \times 0.332 \, \text{m} = 0.664 \, \text{m} \]

### Step 3: Calculate the speed of sound \( V \)
Using the frequency \( f = 500 \, \text{Hz} \) and the wavelength \( \lambda \):
\[ V = f \lambda = 500 \, \text{Hz} \times 0.664 \, \text{m} = 332 \, \text{m/s} \]

### Step 4: Use the formula for the speed of sound in an ideal gas
The speed of sound in an ideal gas is given by:
\[ V = \sqrt{\gamma \frac{R T}{M}} \]
where:
- \( R = 8.314 \, \text{J/(mol K)} \) (universal gas constant)
- \( T = 59^{\circ} C = 59 + 273 = 332 \, \text{K} \)
- \( M = 28.8 \, \text{g/mol} = 0.0288 \, \text{kg/mol} \) (molar mass of air)

### Step 5: Rearranging to find \( \gamma \)
Squaring both sides:
\[ V^2 = \gamma \frac{R T}{M} \]
Rearranging gives:
\[ \gamma = \frac{V^2 M}{R T} \]
Substituting the values:
\[ \gamma = \frac{(332 \, \text{m/s})^2 \times 0.0288 \, \text{kg/mol}}{8.314 \, \text{J/(mol K)} \times 332 \, \text{K}} \]

### Step 6: Calculate \( \gamma \)
Calculating the numerator:
\[ (332)^2 = 110224 \, \text{m}^2/\text{s}^2 \]
Then:
\[ \gamma = \frac{110224 \times 0.0288}{8.314 \times 332} \]
Calculating the denominator:
\[ 8.314 \times 332 = 2766.088 \, \text{J/(mol K)} \]
Now substituting back:
\[ \gamma = \frac{110224 \times 0.0288}{2766.088} \approx 1.152 \]

### Final Answer
Thus, the ratio of heat capacities at constant pressure and constant volume for air at \( 59^{\circ} C \) is approximately \( 1.152 \).

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