Two containers `A` and `B` of equal volume `V_(0)//2` each are connected by a narrow tube which can be closed by a value. The containers are fitted with pistons which can be moved to change the volumes. Initially the value is open and the containers contain an ideal gas `(C_(p)//C_(v)=gamma)` at atmospheric pressure `P_(0)` and atmospheric temperature `2T_(0)`. The walls of the containers `A` are highly conducting and of `B` are non-conducting. The value is now closed and the pistons are slowly pulled out to increase the volumes of the containers to double the original value. (a) Calculate the temperature and pressure in the two containers. (b) The value is now opened for sufficient time so that the gases acquire a common temperature and pressure. Find the new values of the temperature and the pressure.

Explanation

To solve the problem step by step, we will break it down into parts as described in the question.

### Part (a): Calculate the temperature and pressure in the two containers after the pistons are pulled out.

1. **Initial Conditions**:
- Both containers A and B have the same initial pressure \( P_0 \), volume \( V_0/2 \), and temperature \( 2T_0 \).
- Container A is diathermic (can exchange heat) and container B is adiabatic (cannot exchange heat).

2. **Container A (Diathermic)**:
- Since the container is diathermic, the temperature remains constant during the expansion.
- Therefore, the final temperature \( T_A = 2T_0 \).
- Using the ideal gas law \( PV = nRT \), we can write:
\[ P_1 V_1 = P_2 V_2 \]
where \( P_1 = P_0 \), \( V_1 = V_0/2 \), and \( V_2 = V_0 \) (double the original volume).
- Substituting the values:
\[ P_0 \cdot \frac{V_0}{2} = P_2 \cdot V_0 \]
- Solving for \( P_2 \):
\[ P_2 = \frac{P_0}{2} \]

3. **Container B (Adiabatic)**:
- For an adiabatic process, the relationship between temperature and volume is given by:
\[ T_1 V^{\gamma - 1} = T_2 V^{\gamma - 1} \]
- Here, \( T_1 = 2T_0 \), \( V_1 = V_0/2 \), and \( V_2 = V_0 \).
- Substituting the values:
\[ 2T_0 \left(\frac{V_0}{2}\right)^{\gamma - 1} = T_2 \left(V_0\right)^{\gamma - 1} \]
- Rearranging gives:
\[ T_2 = 2T_0 \cdot \frac{(V_0/2)^{\gamma - 1}}{(V_0)^{\gamma - 1}} = 2T_0 \cdot \left(\frac{1}{2^{\gamma - 1}}\right) \]
- Thus,
\[ T_2 = \frac{2T_0}{2^{\gamma - 1}} = \frac{2T_0}{2^{\gamma - 1}} = \frac{2T_0}{2^{\gamma - 1}} = \frac{2T_0}{2^{\gamma - 1}} = 2^{2 - \gamma} T_0 \]

4. **Pressure in Container B**:
- The relationship between pressure and volume for an adiabatic process is given by:
\[ P_1 V^\gamma = P_2 V^\gamma \]
- Substituting the known values:
\[ P_0 \left(\frac{V_0}{2}\right)^\gamma = P_2 (V_0)^\gamma \]
- Rearranging gives:
\[ P_2 = P_0 \cdot \frac{(V_0/2)^\gamma}{(V_0)^\gamma} = P_0 \cdot \frac{1}{2^\gamma} = \frac{P_0}{2^\gamma} \]

### Summary of Part (a):
- For Container A:
- Temperature \( T_A = 2T_0 \)
- Pressure \( P_A = \frac{P_0}{2} \)

- For Container B:
- Temperature \( T_B = 2^{2 - \gamma} T_0 \)
- Pressure \( P_B = \frac{P_0}{2^\gamma} \)

### Part (b): Find the new values of temperature and pressure when the valve is opened.

1. **Common Temperature**:
- When the valve is opened, the gases will reach thermal equilibrium.
- Since container A is diathermic, its temperature remains \( 2T_0 \).
- Container B will adjust to the same temperature, so the final common temperature \( T_f = 2T_0 \).

2. **Common Pressure**:
- The pressure in container A remains constant at \( P_0 \).
- For container B, we can use Dalton's law of partial pressures:
\[ P_{total} = P_A + P_B \]
- Since \( P_A = P_0 \) and \( P_B = \frac{P_0}{2^\gamma} \):
\[ P_{total} = P_0 + \frac{P_0}{2^\gamma} \]
- The total pressure can be expressed as:
\[ P_f = \frac{P_0 (1 + \frac{1}{2^{\gamma - 1}})}{2} \]

### Summary of Part (b):
- Final common temperature \( T_f = 2T_0 \)
- Final common pressure \( P_f = P_0 \left(1 + \frac{1}{2^{\gamma - 1}}\right) \)