Askiitians Tutor Team
Last Activity: 9 Months ago
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.
