A closed tank has two compartments A and B, both filled with oxygen (assumed to be an ideal gas). The partition separating the two compartments is fixed and is a perfect heat insulator (Figure 1). If the old partition is replaced by a new partition, which can slide and conduct heat does not allow the gas to leak across (Figure 2), the volume (in {{m}^{3}}) of the compartment A after the system attains equilibrium is ______________

To solve this question, let's break it down step by step, considering the properties of the system and the laws of thermodynamics.

### Initial Setup:
- **Compartment A and B** are filled with oxygen, which is assumed to be an ideal gas.
- The partition initially is **fixed** and is a **perfect heat insulator** (Figure 1), meaning no heat exchange occurs between the two compartments.

### Final Setup:
- The old partition is replaced by a **new partition**, which can **slide** and **conduct heat**, but it **does not allow gas to leak** between the compartments (Figure 2).

### Key Assumptions and Principles:
- The gases in compartments A and B are ideal, so we will use the **ideal gas law** $PV=nRT$, where $P$ is pressure, $V$ is volume, $n$ is the number of moles of gas, $R$ is the universal gas constant, and $T$ is temperature.
- **Thermal equilibrium** is reached, meaning the final temperatures in compartments A and B will be the same.
- The **total volume** of the tank (compartments A and B together) is constant, but the volume distribution between A and B may change due to the sliding partition.
- **No gas leaks**, so the number of moles of gas in each compartment remains the same.

### Step-by-Step Solution:

1. **Initial State**:
In the initial state, the partition is fixed and insulated. The gases in compartments A and B can have different temperatures and pressures.

2. **Final State (Equilibrium)**:
When the partition is replaced by a sliding, heat-conducting partition, the system will evolve to a state of **mechanical equilibrium** and **thermal equilibrium**:
- **Thermal Equilibrium**: The temperature of the gas in compartments A and B must be equal. Let this final temperature be $T_f$.
- **Mechanical Equilibrium**: The pressure in both compartments must be equal. Let this final pressure be $P_f$.

3. **Ideal Gas Law for Each Compartment**:
For compartment A, the ideal gas law is:
$$P_f{V}_A=n_{A}R{T}_f$$
For compartment B, the ideal gas law is:
$$P_f{V}_B=n_{B}R{T}_f$$
Here, $n_A$ and $n_B$ are the moles of oxygen in compartments A and B, respectively, and $V_A$ and $V_B$ are the final volumes of compartments A and B.

4. **Volume Conservation**:
The total volume of the system remains constant. If the initial total volume is $V_{\text{total}}$, then:
$$V_A+V_B=V_{\text{total}}$$

5. **Mole Conservation**:
The number of moles of gas in each compartment remains the same because no gas leaks across the partition:
- The moles of gas in compartment A is $n_A$.
- The moles of gas in compartment B is $n_B$.

6. **Pressure Equality**:
Since the final pressures are equal, we can equate the ideal gas law equations for compartments A and B:
$$\frac{n_A}{V_A}=\frac{n_B}{V_B}$$
This relationship shows that the ratio of moles to volume in each compartment is the same at equilibrium.

7. **Solution**:
Given the conservation of volume and the relationship between moles and volumes, we can solve for the final volume of compartment A $V_A$. However, since specific values for the moles $n_A$ and $n_B$ or the total volume $V_{\text{total}}$ are not provided in the question, the exact numerical value of $V_A$ cannot be determined without additional information.

If the total volume and the ratio of moles $n_A/n_B$ were given, we could use the above equations to find the final volume of compartment A after equilibrium is reached.

### Conclusion:
To find the exact volume of compartment A after equilibrium, more specific numerical information (such as the total volume or the ratio of moles in compartments A and B) is needed. The approach relies on applying the ideal gas law, conservation of volume, and equilibrium conditions for temperature and pressure.