Gases in Connected Vessels: Constrained Systems

Hello! Welcome to the next lesson in our journey through thermodynamics.

In our last session, we zoomed in on the microscopic world, deriving the mean free path and understanding how molecular collisions govern gas properties. Today, we'll zoom back out to the macroscopic level to tackle a classic category of JEE Advanced problems: gases in connected vessels with constraints.

This topic requires you to be a master integrator, combining the ideal gas law with principles of conservation and equilibrium. By the end of this lesson, you will be able to systematically analyze and solve problems involving gases in multiple compartments, whether they are connected by a valve or separated by a movable partition.

1. The Core Principles

Problems involving connected vessels might seem varied, but they are all governed by a few fundamental principles. Your first step in any such problem should be to identify which principles apply.

Conservation of Moles: If you have a closed system (no gas can leak out), the total number of moles of gas remains constant, no matter how the gas redistributes itself between compartments.
$n_{total} = n_{1} + n_{2} + \dots = constant$
Using the ideal gas law, $P V = n R T$ , you can express this as:
$\frac{P _{1} V _{1}}{T _{1}} + \frac{P _{2} V _{2}}{T _{2}} + \dots = constant$
Mechanical Equilibrium: When two gas compartments are separated by a freely movable, frictionless piston, the piston will settle at a position where the forces on it are balanced. In the simplest case, this means the pressures on both sides are equal:
$P_{1} = P_{2}$
If other forces are present (like the piston's weight, or a spring), the force balance equation must include them. For a vertical piston of mass $m$ and area $A$ , equilibrium means:
$P_{bottom} \cdot A = P_{top} \cdot A + m g$
Thermal Equilibrium: If the barrier separating the compartments is a diathermic wall (a good thermal conductor), heat will flow until the temperatures on both sides are equal.
$T_{1} = T_{2}$
If the vessels themselves are not insulated and are in a common thermal environment, they will eventually reach the temperature of that environment.

Let's see how to apply these principles in different scenarios.

2. Scenario 1: Interconnected Rigid Vessels

This is the most straightforward case. Two or more vessels with fixed volumes are connected by a valve. Initially, they may have different pressures and temperatures. When the valve is opened, gas flows until a final equilibrium state is reached.

The key here is applying the conservation of moles. Let's walk through a solved example.

Two Connected Vessels - Solved Problem

The following resource provides a clear, step-by-step solution to a classic problem of two vessels connected by a tap. It perfectly demonstrates the application of mole conservation.

Please read through the 'Task', 'Analysis', and 'Solution' sections of this problem. Focus on how the total number of moles before opening the tap (n_A + n_B) is equated to the total number of moles after (n_1 + n_2). Notice how the ideal gas law is used to substitute for each 'n'.

As you saw, the strategy is:

Write expressions for the initial number of moles in each vessel: $n_{A} = \frac{P _{A} V _{A}}{R T _{A}}$ and $n_{B} = \frac{P _{B} V _{B}}{R T _{B}}$ .
Write expressions for the final number of moles in each vessel. The final pressure $P_{f}$ is the same for both, but the temperatures might remain different. $n_{A f} = \frac{P _{f} V _{A}}{R T _{A}}$ and $n_{B f} = \frac{P _{f} V _{B}}{R T _{B}}$ .
Set the total initial moles equal to the total final moles: $n_{A} + n_{B} = n_{A f} + n_{B f}$ .
Solve for the unknown, which is usually the final pressure $P_{f}$ .

Test your understanding!

Two identical vessels are connected by a narrow tube. Initially, they both contain an ideal gas at STP (Standard Temperature and Pressure: 0°C and 1 atm). One vessel is then heated to 100°C while the other is kept at 0°C. Find the final pressure in the vessels.

Show answer

Let the volume of each vessel be $V$ , initial temperature be $T_{0} = 273 K$ , and initial pressure be $P_{0} = 1 atm$ .
The initial total number of moles is $n_{i} = n_{1} + n_{2} = \frac{P _{0} V}{R T _{0}} + \frac{P _{0} V}{R T _{0}} = \frac{2 P _{0} V}{R T _{0}}$ .

Let the final pressure be $P_{f}$ . The temperatures are $T_{1} = 273 K$ and $T_{2} = 273 + 100 = 373 K$ .
The final total number of moles is $n_{f} = n_{f 1} + n_{f 2} = \frac{P _{f} V}{R T _{1}} + \frac{P _{f} V}{R T _{2}}$ .

By mole conservation, $n_{i} = n_{f}$ :
$\frac{2 P _{0} V}{R T _{0}} = \frac{P _{f} V}{R T _{1}} + \frac{P _{f} V}{R T _{2}}$
Since $T_{0} = T_{1}$ , we can simplify:
$\frac{2 P _{0}}{T _{1}} = P_{f} (\frac{1}{T _{1}} + \frac{1}{T _{2}})$
$\frac{2 P _{0}}{T _{1}} = P_{f} (\frac{T _{2} + T _{1}}{T _{1} T _{2}})$
Solving for $P_{f}$ :
$P_{f} = P_{0} \frac{2 T _{2}}{T _{1} + T _{2}}$
Substituting the values:
$P_{f} = (1 atm) \frac{2 \times 373}{273 + 373} = \frac{746}{646} \approx 1.155 atm$

3. Scenario 2: Movable Partitions

This scenario introduces mechanical equilibrium. The partition moves until the forces on it balance.

Example A: Partition with Mass

Consider the setup from a JEE Advanced 2020 problem described in resource LINK. A thermally isolated vertical cylinder is divided into two equal parts by a diathermic, frictionless partition of mass $m = 8.3 kg$ . Each part contains 0.1 moles of an ideal gas at 300 K. The partition is released and settles at a new equilibrium position.

Here's how to structure the analysis:

Identify Equilibrium Conditions:
- Thermal: The partition is diathermic, so the final temperature $T_{f}$ will be the same in both compartments. Since the entire cylinder is isolated, no energy is lost, but the distribution of potential energy changes. However, in many such problems (unless significant gravitational potential energy changes for the gas itself are considered), the final gas temperature remains the same, $T_{f} = T_{i}$ . Let's assume $T_{f} = 300 K$ .
- Mechanical: The partition has mass. At equilibrium, the pressure in the lower compartment ( $P_{2}$ ) must support both the pressure from the upper compartment ( $P_{1}$ ) and the weight of the piston.
  $P_{2} \cdot A = P_{1} \cdot A + m g ⟹ P_{2} = P_{1} + \frac{m g}{A}$
Apply the Ideal Gas Law:
Let the final position of the partition be a distance $h$ from the top. The total height is $H = 8 m$ .
- Upper compartment: $V_{1} = A \cdot h$ , $n_{1} = 0.1 mol$ . So, $P_{1} = \frac{n _{1} R T _{f}}{A h}$ .
- Lower compartment: $V_{2} = A \cdot (H - h)$ , $n_{2} = 0.1 mol$ . So, $P_{2} = \frac{n _{2} R T _{f}}{A ( H - h )}$ .
Combine and Solve:
Substitute the pressure expressions into the mechanical equilibrium equation:
$\frac{n _{2} R T _{f}}{A ( H - h )} = \frac{n _{1} R T _{f}}{A h} + \frac{m g}{A}$
Since $n_{1} = n_{2} = n$ , we can simplify and solve for $h$ .
$\frac{n R T _{f}}{H - h} - \frac{n R T _{f}}{h} = m g$
$n R T_{f} (\frac{h - ( H - h )}{h ( H - h )}) = m g$
$n R T_{f} \frac{2 h - H}{h ( H - h )} = m g$
All values are known except for $h$ . Plugging them in (n=0.1, R=8.3, T=300, H=8, m=8.3, g=10) leads to a quadratic equation for $h$ , which you can then solve.

Example B: Partition with a Spring

Now let's add another common constraint: a spring.

Ideal Gas in Insulated Cylinder with Movable Piston and Spring — In this setup, a movable piston separates a gas from the atmosphere. A spring connects the piston to the cylinder wall. The equilibrium position is determined by the balance of forces from the gas pressure, atmospheric pressure, and the spring.

The key is again the force balance equation. If the piston is at a position $x$ from its equilibrium spring position:
$F_{gas} = F_{atm} + F_{spring}$
$P_{gas} \cdot A = P_{atm} \cdot A + k x$
You can then use $P_{gas} = \frac{n R T}{V}$ where $V = A \cdot L$ (with L being the length of the gas column) to relate the thermodynamic state to the mechanical equilibrium.

4. Advanced Synthesis: Combining Mole Conservation and the First Law

The most challenging JEE Advanced problems require you to synthesize all these principles. They might involve a connection opening and a piston moving, with external work being done. The strategy is to form as many equations as you have unknowns by systematically applying the principles.

Let's analyze a top-tier problem that does just this.

Pathfinder Solutions | Thermodynamics | Advanced Problem | Two Connected Vessels

This video from 'Physics with Akash Goyal' solves an advanced problem from the Pathfinder textbook. It involves two vessels connected by a valve, where one vessel has a movable piston acted on by an external force. This problem is an excellent test of your ability to combine mole conservation with the first law of thermodynamics.

Watch the explanation from the beginning up to 12:10. You don't need to follow the final algebraic solution, but focus intensely on how the two main equations are formulated: (05:30 - 07:43) The first equation from the conservation of moles. Understand how the initial total moles are equated to the final total moles. (07:43 - 12:10) The second equation from the First Law of Thermodynamics (\Delta U = Q + W) applied to the entire system of gas. Pay close attention to how the work done by the external agent (W_{ext}) is calculated and related to the change in internal energy (\Delta U) of the whole system.

Deconstruction of the Problem Strategy

This problem is complex, so let's break down the logic shown in the video.

The Setup:

Vessel A: Movable piston, initial state $(P_{0}, V_{0}, T_{0})$ .
Vessel B: Rigid, initial state $(P_{0}, 3 V_{0}, T_{0})$ .
An external agent keeps the pressure in A at $2 P_{0}$ after the valve is opened.
The vessels are in thermal contact, so they reach a common final temperature $T_{f}$ .
We need to find the final volume of A ( $V_{A}$ ) and the final temperature $T_{f}$ . We have two unknowns, so we need two equations.

Equation 1: Conservation of Moles

Initial moles: $n_{i} = n_{A, i} + n_{B, i} = \frac{P _{0} V _{0}}{R T _{0}} + \frac{P _{0} ( 3 V _{0} )}{R T _{0}} = \frac{4 P _{0} V _{0}}{R T _{0}}$
Final moles: $n_{f} = n_{A, f} + n_{B, f}$ .
- In A: Pressure is maintained at $2 P_{0}$ , final volume is $V_{A}$ , final temp is $T_{f}$ . So, $n_{A, f} = \frac{( 2 P _{0} ) V _{A}}{R T _{f}}$ .
- In B: Pressure equalizes to $2 P_{0}$ , volume is fixed at $3 V_{0}$ , final temp is $T_{f}$ . So, $n_{B, f} = \frac{( 2 P _{0} ) ( 3 V _{0} )}{R T _{f}}$ .
Equating $n_{i} = n_{f}$ :
$\frac{4 P _{0} V _{0}}{R T _{0}} = \frac{2 P _{0} V _{A}}{R T _{f}} + \frac{6 P _{0} V _{0}}{R T _{f}}$
This gives us our first equation relating the two unknowns, $V_{A}$ and $T_{f}$ .

Equation 2: First Law of Thermodynamics

The system is the total gas in both vessels. The cylinder walls are insulating, so $Q = 0$ .
The first law is $Δ U = Q + W_{o n_sy s t e m}$ . Here, the only work is done on the gas by the external agent pushing the piston.
Work done by the gas in vessel A is $W_{b y_g a s} = \int P d V = 2 P_{0} (V_{0} - V_{A})$ . The work done on the gas by the external agent is the negative of this: $W_{o n_g a s} = - W_{b y_g a s} = 2 P_{0} (V_{A} - V_{0})$ .
Change in Internal Energy $Δ U$ : The entire system of gas ( $n_{i}$ moles) changes temperature from $T_{0}$ to $T_{f}$ . For a monatomic gas, $C_{V} = \frac{3}{2} R$ .
$Δ U = n_{i} C_{V} (T_{f} - T_{0}) = (\frac{4 P _{0} V _{0}}{R T _{0}}) (\frac{3}{2} R) (T_{f} - T_{0})$
Equating $Δ U = W_{o n_g a s}$ :
$(\frac{4 P _{0} V _{0}}{R T _{0}}) (\frac{3}{2} R) (T_{f} - T_{0}) = 2 P_{0} (V_{A} - V_{0})$
This gives our second equation relating $V_{A}$ and $T_{f}$ . With two equations and two unknowns, the system can be solved.

The key takeaway is the methodical application of fundamental laws to generate the necessary equations.

Conclusion

Today we've developed a systematic framework for one of the most common and challenging types of problems in thermodynamics for JEE Advanced.

Key Takeaways:

Identify the System: Are the vessels rigid? Is there a movable piston? Is it diathermic or adiabatic? Are there springs or other external forces?
Apply Conservation of Moles: This is your go-to principle for any closed system where gas is redistributed. $n_{initial} = n_{final}$ .
Apply Equilibrium Conditions: For movable partitions, use mechanical equilibrium ( $P_{1} = P_{2}$ or force balance). For diathermic partitions, use thermal equilibrium ( $T_{1} = T_{2}$ ).
Use the First Law When Needed: If work is done or heat is transferred, and you need to relate it to temperature change, apply $Δ U = Q + W$ . Remember to apply it to the correct system (e.g., the total amount of gas).

Preview of the Next Lesson:
We've touched upon mechanical equilibrium with pistons today. In the next lesson, we will dive deeper into this specific topic: "Solve problems combining the ideal gas law with the mechanical equilibrium of pistons." We will explore more intricate scenarios involving multiple pistons, complex linkages, and how to analyze the states of the gas in equilibrium under various mechanical loads.

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