A balloon containing 1 mole air at 1 atm initially is filled further with air till pressure increases to 4 atm. The intial diameter of the ballon is 1 m and the pressure at each stage is proportional to diameter of the balloon. How many moles of air added to change the pressure from 1 atm to 4 atm.

Options

A. 80
B. 257
C. 255
D. 256

Explanation

To solve the problem step by step, we will follow the logic presented in the video transcript and apply the principles of the gas laws.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- We have a balloon containing 1 mole of air at an initial pressure \( P_1 = 1 \, \text{atm} \).
- The initial diameter of the balloon is \( d_1 = 1 \, \text{m} \).

2. **Final Conditions**:
- The pressure is increased to \( P_2 = 4 \, \text{atm} \).
- We need to find the new diameter \( d_2 \) when the pressure is 4 atm.

3. **Using the Relationship Between Pressure and Diameter**:
- Since pressure is proportional to the diameter, we can write:
\[ \frac{P_2}{P_1} = \frac{d_2}{d_1} \]
- Substituting the known values:
\[ \frac{4 \, \text{atm}}{1 \, \text{atm}} = \frac{d_2}{1 \, \text{m}} \implies d_2 = 4 \, \text{m} \]

4. **Calculating the Volume of the Balloon**:
- The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
- The radius \( r \) is half of the diameter, so:
- Initial radius \( r_1 = \frac{d_1}{2} = \frac{1}{2} \, \text{m} \)
- Final radius \( r_2 = \frac{d_2}{2} = \frac{4}{2} = 2 \, \text{m} \)
- Now, calculate the initial and final volumes:
- Initial volume \( V_1 \):
\[ V_1 = \frac{4}{3} \pi \left(\frac{1}{2}\right)^3 = \frac{4}{3} \pi \left(\frac{1}{8}\right) = \frac{\pi}{6} \, \text{m}^3 \]
- Final volume \( V_2 \):
\[ V_2 = \frac{4}{3} \pi (2)^3 = \frac{4}{3} \pi (8) = \frac{32}{3} \pi \, \text{m}^3 \]

5. **Using the Ideal Gas Law**:
- The ideal gas law states:
\[ PV = nRT \]
- We can express the number of moles at both stages:
- Initial moles \( n_1 = 1 \) mole at \( P_1 = 1 \, \text{atm} \) and \( V_1 = \frac{\pi}{6} \, \text{m}^3 \):
\[ n_1 = \frac{P_1 V_1}{RT} = \frac{1 \cdot \frac{\pi}{6}}{RT} \]
- Final moles \( n_2 \) at \( P_2 = 4 \, \text{atm} \) and \( V_2 = \frac{32}{3} \pi \, \text{m}^3 \):
\[ n_2 = \frac{P_2 V_2}{RT} = \frac{4 \cdot \frac{32}{3} \pi}{RT} \]

6. **Finding the Number of Moles Added**:
- From the relationship of volumes and pressures, we can derive:
\[ n_2 = 4 n_1 \]
- Since \( n_1 = 1 \):
\[ n_2 = 4 \cdot 1 = 4 \, \text{moles} \]
- The number of moles added:
\[ \text{Moles added} = n_2 - n_1 = 4 - 1 = 3 \, \text{moles} \]

### Final Answer:
The number of moles of air added to change the pressure from 1 atm to 4 atm is **3 moles**.

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