Assume end correction approximately equals to `(0.3) xx` (diameter of tube), estimate the approximate number of moles of ir present inside the tube (Assume tube is at `NTP`, and at `NTP, 22.4` litre contains `1` mole)

Options

A. `(100pi)/(36 xx 22.4)`
B. `(10pi)/(18 xx 22.4)`
C. `(10pi)/(72 xx 22.4)`
D. `(10pi)/(60 xx 22.4)`

Explanation

To solve the problem step by step, we will follow the given information and perform the necessary calculations.

### Step 1: Determine the End Correction
We are given that the end correction \( E \) is approximately equal to \( 0.3 \times \text{diameter of the tube} \). We also know that \( E = 1 \) cm.

Using the formula for end correction:
\[ E = 0.3 \times d \]
Substituting the value of \( E \):
\[ 1 = 0.3 \times d \]
To find the diameter \( d \):
\[ d = \frac{1}{0.3} = \frac{10}{3} \text{ cm} \approx 3.33 \text{ cm} \]

### Step 2: Calculate the Radius of the Tube
The radius \( r \) is half of the diameter:
\[ r = \frac{d}{2} = \frac{10/3}{2} = \frac{10}{6} = \frac{5}{3} \text{ cm} \approx 1.67 \text{ cm} \]

### Step 3: Calculate the Volume of the Tube
The volume \( V \) of a cylinder (tube) is given by:
\[ V = \pi r^2 h \]
Where \( h \) is the length of the tube. Given that the length \( h = 99 \) cm (approximately 100 cm), we can substitute the values:
\[ V = \pi \left(\frac{5}{3}\right)^2 \times 100 \]
Calculating \( r^2 \):
\[ r^2 = \left(\frac{5}{3}\right)^2 = \frac{25}{9} \]
Now substituting back into the volume formula:
\[ V = \pi \times \frac{25}{9} \times 100 = \frac{2500\pi}{9} \text{ cm}^3 \]

### Step 4: Convert Volume from cm³ to Liters
Since \( 1 \text{ liter} = 1000 \text{ cm}^3 \), we convert the volume:
\[ V = \frac{2500\pi}{9} \text{ cm}^3 \times \frac{1 \text{ liter}}{1000 \text{ cm}^3} = \frac{2500\pi}{9000} \text{ liters} \]
Simplifying:
\[ V = \frac{250\pi}{900} = \frac{25\pi}{90} \text{ liters} \]

### Step 5: Calculate the Number of Moles
At NTP, \( 22.4 \) liters contains \( 1 \) mole. Therefore, the number of moles \( n \) in the volume \( V \) is given by:
\[ n = \frac{V}{22.4} \]
Substituting the volume we found:
\[ n = \frac{\frac{25\pi}{90}}{22.4} \]
Calculating:
\[ n = \frac{25\pi}{90 \times 22.4} \]
Calculating \( 90 \times 22.4 = 2016 \):
\[ n = \frac{25\pi}{2016} \]

### Final Answer
The approximate number of moles of gas present inside the tube is:
\[ n \approx \frac{25\pi}{2016} \text{ moles} \]

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