In a tyre of "Ferrari" car of Mr. Obama, a tube having a volume of 12.3 litres is filled with air at a pressure of 4 atm at 300 K.Due to travelling, the temperature of the tube and air inside it raised to 360 K and pressure reduced to 3.6 atm in 20 minutes. If the porosity (number of pores per unit area) of the tube material is `5xx10^3` pores/`cm^2` and each pore can transfer air from inside to outside of the tube with the rate of `6.023xx10^8` molecules per minute.Calculate the total surface area `(m^2)` of the tube.(R=0.082 Lt-atm/mole-K)
Give your answer divide by 100.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the Initial Moles of Air in the Tube
Using the Ideal Gas Law, we can find the initial moles of air in the tube:
\[ PV = nRT \]
Where:
- \( P = 4 \, \text{atm} \)
- \( V = 12.3 \, \text{L} \)
- \( R = 0.082 \, \text{L atm/(mol K)} \)
- \( T = 300 \, \text{K} \)

Rearranging for \( n \):
\[ n = \frac{PV}{RT} \]
Substituting the values:
\[ n = \frac{4 \, \text{atm} \times 12.3 \, \text{L}}{0.082 \, \text{L atm/(mol K)} \times 300 \, \text{K}} \]
Calculating:
\[ n = \frac{49.2}{24.6} \approx 2 \, \text{moles} \]

### Step 2: Calculate the Moles of Air After 20 Minutes
Now, we calculate the moles of air after the temperature and pressure change:
- New Pressure \( P = 3.6 \, \text{atm} \)
- New Temperature \( T = 360 \, \text{K} \)

Using the Ideal Gas Law again:
\[ n' = \frac{PV}{RT} \]
Substituting the new values:
\[ n' = \frac{3.6 \, \text{atm} \times 12.3 \, \text{L}}{0.082 \, \text{L atm/(mol K)} \times 360 \, \text{K}} \]
Calculating:
\[ n' = \frac{44.28}{29.592} \approx 1.5 \, \text{moles} \]

### Step 3: Calculate the Moles of Air Leaked in 20 Minutes
The moles of air that leaked in 20 minutes:
\[ \text{Moles leaked} = n - n' = 2 - 1.5 = 0.5 \, \text{moles} \]

### Step 4: Calculate the Total Molecules of Air Leaked
Using Avogadro's number \( N_A = 6.022 \times 10^{23} \, \text{molecules/mole} \):
\[ \text{Molecules leaked} = 0.5 \times N_A = 0.5 \times 6.022 \times 10^{23} \approx 3.011 \times 10^{23} \, \text{molecules} \]

### Step 5: Calculate the Total Surface Area of the Tube
Given:
- Porosity = \( 5 \times 10^3 \, \text{pores/cm}^2 \)
- Rate of air transfer per pore = \( 6.023 \times 10^8 \, \text{molecules/min} \)

Let \( A \) be the total surface area in cm². The total molecules escaping in 20 minutes is:
\[ \text{Total molecules} = \text{Pores} \times \text{Rate} \times \text{Time} \]
\[ 3.011 \times 10^{23} = A \times (5 \times 10^3) \times (6.023 \times 10^8) \times 20 \]

Now, rearranging for \( A \):
\[ A = \frac{3.011 \times 10^{23}}{(5 \times 10^3) \times (6.023 \times 10^8) \times 20} \]

Calculating the denominator:
\[ = 5 \times 10^3 \times 6.023 \times 10^8 \times 20 \]
\[ = 6.023 \times 10^{12} \]

Now calculating \( A \):
\[ A = \frac{3.011 \times 10^{23}}{6.023 \times 10^{12}} \approx 5 \times 10^{10} \, \text{cm}^2 \]

### Step 6: Convert to m² and Divide by 100
Converting cm² to m²:
\[ A = 5 \times 10^{10} \, \text{cm}^2 = 5 \times 10^{6} \, \text{m}^2 \]

Finally, dividing by 100:
\[ \frac{5 \times 10^{6}}{100} = 5 \times 10^{4} \, \text{m}^2 \]

### Final Answer
The total surface area of the tube is \( 50000 \, \text{m}^2 \).