Two identical containers joned by a small pipe initially contain the same gas at pressue `p_(o)` and abosolute temperature `T_(o^.)` One container is now mantained at the same temperature while the other is heated to `2T_(0^.)` The commmon pressure of the gases will be

Options

A. `3/2p_(o)`
B. `4/3p_(o)`
C. `3/5p_(o)`
D. `2p_(o)`

Explanation

To solve the problem, we will use the ideal gas law and the concept of conservation of moles. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- Both containers have the same gas with initial pressure \( P_0 \) and absolute temperature \( T_0 \).
- Let the volume of each container be \( V_0 \).

2. **Determine Moles in Each Container Initially**:
- For Container 1 (maintained at \( T_0 \)):
\[ n_1 = \frac{P_0 V_0}{R T_0} \]
- For Container 2 (heated to \( 2T_0 \)):
\[ n_2 = \frac{P_0 V_0}{R (2T_0)} = \frac{P_0 V_0}{2R T_0} \]

3. **Total Moles in the System**:
- The total number of moles \( n_{total} \) in the system is:
\[ n_{total} = n_1 + n_2 = \frac{P_0 V_0}{R T_0} + \frac{P_0 V_0}{2R T_0} \]
- Simplifying this gives:
\[ n_{total} = \frac{P_0 V_0}{R T_0} \left(1 + \frac{1}{2}\right) = \frac{P_0 V_0}{R T_0} \cdot \frac{3}{2} = \frac{3 P_0 V_0}{2 R T_0} \]

4. **Final Condition After Equilibrium**:
- After reaching equilibrium, let the common pressure be \( P \) and the temperature of Container 1 remains \( T_0 \) while Container 2 is at \( 2T_0 \).
- The moles in each container at equilibrium can be expressed as:
- For Container 1:
\[ n_1' = \frac{P V_0}{R T_0} \]
- For Container 2:
\[ n_2' = \frac{P V_0}{R (2T_0)} = \frac{P V_0}{2R T_0} \]

5. **Equating Total Moles**:
- At equilibrium, the total moles must remain the same:
\[ n_1' + n_2' = n_{total} \]
- Substituting the expressions:
\[ \frac{P V_0}{R T_0} + \frac{P V_0}{2R T_0} = \frac{3 P_0 V_0}{2 R T_0} \]

6. **Simplifying the Equation**:
- Factor out \( \frac{V_0}{R T_0} \):
\[ \left(P + \frac{P}{2}\right) = \frac{3 P_0}{2} \]
- This simplifies to:
\[ \frac{3P}{2} = \frac{3P_0}{2} \]

7. **Solving for Common Pressure \( P \)**:
- Multiply both sides by \( \frac{2}{3} \):
\[ P = P_0 \cdot \frac{4}{3} \]

### Final Answer:
The common pressure of the gases after equilibrium is:
\[ P = \frac{4P_0}{3} \]