Gases Separated by Movable Partitions

1. The Core Principles of Equilibrium

When two or more gases are separated by a movable partition, the final equilibrium state is governed by a few key principles. The properties of the partition itself are critical.

Mechanical Equilibrium

This is the principle we learned last time, applied to the partition. At equilibrium, the partition is stationary, so the net force on it is zero. For a frictionless, massless piston in a horizontal cylinder, this means the pressure on both sides must be equal.

$P_1=P_2$

If the partition has mass $m$ and the cylinder is vertical, you must include its weight in the force balance. If the lower compartment is '2' and the upper is '1', the equation becomes:

$P_2\cdot A=P_1\cdot A+mg$

Thermal Equilibrium

This depends on whether the partition allows heat to pass through it.

• Diathermic Partition (Conducting): Heat can flow freely across the partition. At equilibrium, the temperatures of the gases in both compartments must be equal.
  $T_1=T_2$
• Adiabatic Partition (Insulating): No heat can flow across the partition. The temperatures on either side can be, and often are, different at equilibrium.
  $T_1\ne T_2(\text{in general})$

By combining these conditions with the ideal gas law for each compartment ($PV=nRT$) and any geometric constraints (e.g., $V_1+V_2=V_{total}$), we can solve for the final state of the system.

2. Strategy for Diathermic (Conducting) Partitions

Let's consider the most common case: a rigid, insulated outer container with two gas compartments separated by a movable, diathermic partition. Since the outer container is insulated, the total system (gas 1 + gas 2) does not exchange energy with the surroundings.

A very effective strategy to solve this involves two main steps. A great tutorial from the University of Waterloo demonstrates this approach.

As you saw in the document, the logic is as follows:

Step 1: Find the Final Equilibrium Temperature ($T_f$)
The total system is isolated, so its total internal energy is conserved. The work done by one gas is done on the other, so the net work for the combined system is zero.
$\Delta U_{total}=Q-W=0-0=0$
This means the change in internal energy of the first gas plus the change in internal energy of the second gas is zero.
$\Delta U_1+\Delta U_2=0$
$n_1{C}_{v1}(T_f-T_{1,i})+n_2{C}_{v2}(T_f-T_{2,i})=0$
This equation allows you to solve for the final common temperature, $T_f$, which is essentially a weighted average of the initial temperatures, with the weighting factors being $n{C}_v$.

Step 2: Find the Final Equilibrium Pressure ($P_f$)
At equilibrium, we know $P_1=P_2=P_f$. We can apply the ideal gas law to the entire system. The total volume $V_{total}$ is the sum of the initial volumes and remains constant.
$V_{total}=V_{1,i}+V_{2,i}$
From the initial states, we have $V_{1,i}=\frac{n_{1}R{T}_{1,i}}{P_{1,i}}$ and $V_{2,i}=\frac{n_{2}R{T}_{2,i}}{P_{2,i}}$.

Now, apply the ideal gas law to the final state of the whole system:
$P_f{V}_{total}=(n_1+n_2)R{T}_f$
Since you found $T_f$ in Step 1 and can calculate $V_{total}$ from initial conditions, you can now easily solve for $P_f$.

This two-step method is a powerful and efficient way to solve problems with diathermic partitions.

3. Strategy for Adiabatic (Insulating) Partitions

When the partition is insulating, the problem changes significantly. The temperatures on each side will likely be different. We can no longer use $\Delta U_{total}=0$ to find a common final temperature.

Instead, these problems usually involve a slow process where one side is actively changed (e.g., heated), forcing the other side to react. The key is to identify the nature of the process occurring on the "passive" side.

Let's watch a video that solves a JEE Advanced problem where this exact scenario plays out.

The crucial insight from the video is that because the partition is insulating ($Q_R=0$) and the process is slow (quasi-static and reversible), the compression of the gas on the right side is an adiabatic process.

This allows the use of the powerful adiabatic equations for the right-side gas:
$T_i{V}_i^{\gamma -1}=T_f{V}_f^{\gamma -1}$
Since the initial state ($T_0,V_0$) and the final volume ($V_0/2$) were known, the final temperature $T_R$ could be directly calculated.

This highlights the main difference in strategy:

• Diathermic partition: Think about the equilibrium of the whole system ($\Delta U_{total}=0$).
• Adiabatic partition (in a process): Think about the process on each side separately. The link between them is the mechanical equilibrium ($P_L=P_R$ at all times for a slow process).

4. Practice with a JEE Advanced Problem

Let's apply these principles to set up another classic JEE Advanced problem. This one involves a massive, diathermic partition in a vertical cylinder.

Your task is to find the final equilibrium position of the partition. Let's build the system of equations together. Let $h$ be the final distance of the partition from the top. The total height is 8 m.

1. Mechanical Equilibrium: The partition has mass. The pressure from the lower gas ($P_{lower}$) must support both the upper gas pressure ($P_{upper}$) and the partition's weight.
   $P_{lower}\cdot A=P_{upper}\cdot A+mg⟹P_{lower}=P_{upper}+\frac{mg}{A}$

2. Thermal Equilibrium: The partition is diathermic. The final temperatures are equal. Let's call the final temperature $T_f$.
   $T_{lower}=T_{upper}=T_f$
   The initial temperatures were both 300 K. Since the entire vessel is isolated and the only work is internal, $\Delta U_{total}=0$. Because the initial temperatures are the same, the final equilibrium temperature will also be the same. $T_f=300K$. (If initial temperatures were different, we'd use the method from Section 2).

3. Ideal Gas Law & Volume Constraints:

   • Upper compartment: $n_{upper}=0.1$ mol, $V_{upper}=A\cdot h$.
     $P_{upper}=\frac{n_{upper}R{T}_f}{V_{upper}}=\frac{0.1\cdot R\cdot 300}{A\cdot h}$
   • Lower compartment: $n_{lower}=0.1$ mol, $V_{lower}=A\cdot (8-h)$.
     $P_{lower}=\frac{n_{lower}R{T}_f}{V_{lower}}=\frac{0.1\cdot R\cdot 300}{A\cdot (8-h)}$

4. Combine and Solve: Now substitute the expressions for $P_{upper}$ and $P_{lower}$ into the force balance equation:
   $\frac{0.1\cdot R\cdot 300}{A(8-h)}=\frac{0.1\cdot R\cdot 300}{Ah}+\frac{mg}{A}$
   The area $A$ cancels out nicely:
   $\frac{30R}{8-h}=\frac{30R}{h}+mg$

Given $R=8.3$ J/mol·K, $m=8.3$ kg, and $g=10$ m/s², you can plug in the values and solve this equation for $h$. This problem elegantly combines all the principles we've discussed. (The full solution is on page 17 of the document if you wish to check your final calculation).

Conclusion

You have now learned the systematic approach to deconstructing problems with movable partitions. The core strategy is to translate the physical description of the system into a set of algebraic equations and solve them.

Key Takeaways:

• Start with Equilibrium Conditions: First, establish the mechanical equilibrium ($P_1=P_2$ + other forces) and thermal equilibrium ($T_1=T_2$ for diathermic, $T_1\ne T_2$ for adiabatic).
• Apply the Ideal Gas Law: Write $PV=nRT$ for each compartment.
• Use Conservation Laws: Total volume and total moles are usually conserved. For isolated systems with diathermic partitions, total internal energy is conserved ($\Delta U_{total}=0$).
• Identify the Process: For problems involving change (like heating), determine the thermodynamic process on each side (e.g., adiabatic compression).

Preview of the Next Lesson:
We will shift our focus back to calorimetry, but with a significant increase in complexity. Our next lesson will be on solving problems involving simultaneous phase changes in multiple substances. This will challenge you to track energy balance meticulously when, for example, hot steam is mixed with ice, and the final state could be a mix of ice and water, just water, or a mix of water and steam.