At a pressure of `3` atm air (treated as an ideal gas) is pumped into the tubes of a cycle rickshaw. The volume of each tube at given pressure is `0.004 m^(3)`. One of the tubes gets punctured and the volume of the tube reduces to `0.0008 m^(3)`. Find the number of moles of air that have leaked out? Assume that the temperature remains constant at `300K. (R = 25//3J mol^(-1)K^(-1))`

Answer

To solve the problem, we will use the Ideal Gas Law, which states that:

\[ PV = nRT \]

where:
- \( P \) = pressure,
- \( V \) = volume,
- \( n \) = number of moles,
- \( R \) = ideal gas constant,
- \( T \) = temperature.

### Step 1: Identify the initial conditions before the puncture

Given:
- Initial pressure \( P_1 = 3 \) atm
- Initial volume \( V_1 = 0.004 \, m^3 \)
- Temperature \( T = 300 \, K \)
- Ideal gas constant \( R = \frac{25}{3} \, J \, mol^{-1} \, K^{-1} \)

### Step 2: Calculate the number of moles before the puncture

Using the ideal gas equation:

\[ n_1 = \frac{P_1 V_1}{RT} \]

Substituting the values:

\[ n_1 = \frac{(3 \, \text{atm}) (0.004 \, m^3)}{\left(\frac{25}{3} \, J \, mol^{-1} \, K^{-1}\right)(300 \, K)} \]

Convert pressure from atm to Pa (1 atm = 101325 Pa):

\[ P_1 = 3 \times 101325 \, Pa = 303975 \, Pa \]

Now substituting:

\[ n_1 = \frac{(303975 \, Pa)(0.004 \, m^3)}{\left(\frac{25}{3} \, J \, mol^{-1} \, K^{-1}\right)(300 \, K)} \]

Calculating \( R \times T \):

\[ R \times T = \frac{25}{3} \times 300 = 2500 \, J \, mol^{-1} \]

Now substituting back:

\[ n_1 = \frac{1215.9}{2500} \approx 0.48636 \, mol \]

### Step 3: Identify the conditions after the puncture

After the puncture:
- New volume \( V_2 = 0.0008 \, m^3 \)
- Pressure \( P_2 = 1 \, atm \) (atmospheric pressure)

### Step 4: Calculate the number of moles after the puncture

Using the ideal gas equation again:

\[ n_2 = \frac{P_2 V_2}{RT} \]

Convert \( P_2 \) to Pa:

\[ P_2 = 1 \times 101325 \, Pa = 101325 \, Pa \]

Now substituting:

\[ n_2 = \frac{(101325 \, Pa)(0.0008 \, m^3)}{\left(\frac{25}{3} \, J \, mol^{-1} \, K^{-1}\right)(300 \, K)} \]

Using \( R \times T \) calculated earlier:

\[ n_2 = \frac{81.06}{2500} \approx 0.032424 \, mol \]

### Step 5: Calculate the number of moles that leaked out

The number of moles that leaked out is given by:

\[ n_{\text{leaked}} = n_1 - n_2 \]

Substituting the values:

\[ n_{\text{leaked}} = 0.48636 - 0.032424 \approx 0.453936 \, mol \]

### Final Answer

The number of moles of air that have leaked out is approximately:

\[ n_{\text{leaked}} \approx 0.454 \, mol \]

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