A glass bulb of volume `400 cm^(3)` is connected to another bulb of volume `200 cm^(3)` by means of a tube of negligible volume. The bulbs contain dry air and are both at a common temperature and pressure of `20^(@) C` and 1.000 atm, respectively. The larger bulb is immersed in steam at `100^(@) C` and the smaller in melting ice at `0^(@)`. Find the final common pressure.

Explanation

To solve the problem, we will use the Ideal Gas Law, which states that \( PV = nRT \). We will analyze the situation step by step.

### Step 1: Identify Initial Conditions
We have two bulbs:
- Bulb 1 (larger bulb): Volume \( V_1 = 400 \, \text{cm}^3 \)
- Bulb 2 (smaller bulb): Volume \( V_2 = 200 \, \text{cm}^3 \)

Both bulbs contain dry air at:
- Initial temperature \( T_i = 20^\circ C = 293.15 \, K \)
- Initial pressure \( P_i = 1.000 \, \text{atm} \)

### Step 2: Calculate Total Initial Moles of Air
Using the Ideal Gas Law for the total system, we can express the total number of moles \( n \):
\[ n = \frac{P_i (V_1 + V_2)}{RT_i} \]
Substituting the values:
\[ n = \frac{1.000 \, \text{atm} \times (400 \, \text{cm}^3 + 200 \, \text{cm}^3)}{R \times 293.15 \, K} \]
\[ n = \frac{1.000 \times 600 \, \text{cm}^3}{R \times 293.15} \]

### Step 3: Analyze Final Conditions
After the bulbs are subjected to different temperatures:
- Bulb 1 (larger bulb) is at \( T_1 = 100^\circ C = 373.15 \, K \)
- Bulb 2 (smaller bulb) is at \( T_2 = 0^\circ C = 273.15 \, K \)

Let \( P_f \) be the final common pressure in both bulbs.

### Step 4: Apply Ideal Gas Law to Each Bulb
For Bulb 1:
\[ P_f V_1 = n_1 R T_1 \quad \text{(where \( n_1 \) is the moles in bulb 1)} \]
For Bulb 2:
\[ P_f V_2 = n_2 R T_2 \quad \text{(where \( n_2 \) is the moles in bulb 2)} \]

### Step 5: Express Moles in Terms of Pressure
From the equations above, we can express \( n_1 \) and \( n_2 \):
\[ n_1 = \frac{P_f V_1}{R T_1} \]
\[ n_2 = \frac{P_f V_2}{R T_2} \]

### Step 6: Total Moles Conservation
The total number of moles is conserved:
\[ n = n_1 + n_2 \]
Substituting the expressions for \( n_1 \) and \( n_2 \):
\[ \frac{1.000 \times 600}{R \times 293.15} = \frac{P_f V_1}{R T_1} + \frac{P_f V_2}{R T_2} \]
This simplifies to:
\[ \frac{1.000 \times 600}{293.15} = \frac{P_f \times 400}{373.15} + \frac{P_f \times 200}{273.15} \]

### Step 7: Factor Out \( P_f \)
Factor out \( P_f \):
\[ \frac{1.000 \times 600}{293.15} = P_f \left( \frac{400}{373.15} + \frac{200}{273.15} \right) \]

### Step 8: Solve for \( P_f \)
Now, we can solve for \( P_f \):
\[ P_f = \frac{1.000 \times 600 / 293.15}{\frac{400}{373.15} + \frac{200}{273.15}} \]

### Step 9: Calculate the Final Pressure
Calculating the denominator:
\[ \frac{400}{373.15} \approx 1.071 \quad \text{and} \quad \frac{200}{273.15} \approx 0.732 \]
Thus,
\[ \frac{400}{373.15} + \frac{200}{273.15} \approx 1.071 + 0.732 = 1.803 \]
Now substituting back:
\[ P_f = \frac{600 / 293.15}{1.803} \approx \frac{2.042}{1.803} \approx 1.13 \, \text{atm} \]

### Final Answer
The final common pressure \( P_f \) is approximately \( 1.13 \, \text{atm} \).