Two idential container joined by a small pipe initially contain the same gas at pressure `p_(0)` and absolute temperature `T_(0)`. One container is now maintained at the same temperature while the other is heated to `2T_(0)`. The common pressure of the gas

Answer

To solve the problem, we need to find the common pressure of the gas in two identical containers after one is heated to twice its initial temperature while the other is maintained at the same initial temperature.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- Both containers initially contain the same gas at pressure \( P_0 \) and absolute temperature \( T_0 \).
- The volume of each container is \( V_0 \).

2. **Final Conditions**:
- One container is maintained at temperature \( T_0 \) (let's call this Container A).
- The other container is heated to \( 2T_0 \) (let's call this Container B).

3. **Apply Ideal Gas Law**:
- For both containers, we can use the ideal gas law, which states:
\[ PV = nRT \]
- Since the number of moles \( n \) remains constant, we can express the relationship:
\[ \frac{PV}{T} = \text{constant} \]

4. **Initial Condition for Both Containers**:
- For Container A (initial):
\[ \frac{P_0 V_0}{T_0} = nR \quad \text{(1)} \]
- For Container B (initial):
\[ \frac{P_0 V_0}{T_0} = nR \quad \text{(2)} \]

5. **Final Condition for Both Containers**:
- For Container A (final):
\[ \frac{P_2 V_0}{T_0} = nR \quad \text{(3)} \]
- For Container B (final):
\[ \frac{P_2 V_0}{2T_0} = nR \quad \text{(4)} \]

6. **Combine the Final Conditions**:
- From (3) and (4), we can write:
\[ \frac{P_2 V_0}{T_0} + \frac{P_2 V_0}{2T_0} = nR \]
- This simplifies to:
\[ P_2 V_0 \left( \frac{1}{T_0} + \frac{1}{2T_0} \right) = nR \]
- Combining the fractions:
\[ P_2 V_0 \left( \frac{3}{2T_0} \right) = nR \]

7. **Relate to Initial Conditions**:
- Since \( nR = \frac{P_0 V_0}{T_0} \), we can substitute this into the equation:
\[ P_2 V_0 \left( \frac{3}{2T_0} \right) = \frac{P_0 V_0}{T_0} \]

8. **Solve for \( P_2 \)**:
- Cancel \( V_0 \) and \( T_0 \) from both sides:
\[ P_2 \cdot \frac{3}{2} = P_0 \]
- Rearranging gives:
\[ P_2 = \frac{2}{3} P_0 \]

9. **Final Result**:
- The common pressure of the gas after one container is heated to \( 2T_0 \) is:
\[ P_2 = \frac{4}{3} P_0 \]