Both neon `[ M_(Ne) = 20 xx 10^(-3) kg]` and helium `[M_(He) = 4 xx 10^(-3) kg]` are monoatomic gases and can be assumed to be ideal gases. The fundamental frequency of a tube (open at both ends ) of neon is `300 Hz` at `270 K (R = ( 25//3) J//K mol)`
The length of the tube is

Options

A. `(5)/(12) m`
B. `(sqrt(3))/(12) m`
C. `( 5 sqrt(3))/(12) m`
D. ` ( 5 sqrt(3)) m`

Explanation

To find the length of the tube open at both ends for neon gas, we can use the formula for the fundamental frequency of a tube, which is given by:

\[ f = \frac{1}{2L} v \]

where:
- \( f \) is the fundamental frequency,
- \( L \) is the length of the tube,
- \( v \) is the velocity of sound in the gas.

The velocity of sound in an ideal gas can be calculated using the formula:

\[ v = \sqrt{\frac{\gamma RT}{M}} \]

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

### Step 1: Identify the values
- For neon:
- \( f = 300 \, \text{Hz} \)
- \( T = 270 \, \text{K} \)
- \( R = \frac{25}{3} \, \text{J/K mol} \)
- \( M = 20 \times 10^{-3} \, \text{kg} \)
- \( \gamma = \frac{5}{3} \)

### Step 2: Calculate the velocity \( v \)
Using the formula for velocity:

\[ v = \sqrt{\frac{\gamma RT}{M}} \]

Substituting the values:

\[ v = \sqrt{\frac{\frac{5}{3} \cdot \frac{25}{3} \cdot 270}{20 \times 10^{-3}}} \]

### Step 3: Simplify the expression for \( v \)
Calculating the numerator:

\[ \frac{5}{3} \cdot \frac{25}{3} \cdot 270 = \frac{5 \cdot 25 \cdot 270}{9} = \frac{33750}{9} = 3750 \]

Now, substituting back into the velocity equation:

\[ v = \sqrt{\frac{3750}{20 \times 10^{-3}}} = \sqrt{\frac{3750}{0.02}} = \sqrt{187500} = 433.01 \, \text{m/s} \]

### Step 4: Calculate the length \( L \)
Now substituting \( v \) back into the fundamental frequency equation:

\[ f = \frac{1}{2L} v \implies L = \frac{v}{2f} \]

Substituting the values:

\[ L = \frac{433.01}{2 \cdot 300} = \frac{433.01}{600} \approx 0.72168 \, \text{m} \]

### Step 5: Final expression for length
To express the length in a simplified form, we can calculate:

\[ L \approx 5 \cdot \sqrt{\frac{3}{12}} \, \text{m} \]

### Conclusion
Thus, the length of the tube is approximately:

\[ L \approx 0.72168 \, \text{m} \]