Ideal Gas Law and Piston Equilibrium

Hello! Welcome back to our course on thermodynamics.

In the last lesson, we tackled problems involving gases in connected vessels, establishing a solid foundation in using mole conservation and equilibrium principles. We saw how a movable partition settles when forces on it balance. Today, we'll zero in on that concept and master a cornerstone of JEE Advanced thermodynamics: problems combining the ideal gas law with the mechanical equilibrium of pistons.

These problems require a seamless blend of mechanics (force analysis) and thermodynamics (state equations). By the end of this lesson, you will be able to construct and solve the equations that describe the equilibrium of piston-cylinder systems, even when they involve complex elements like springs, gravity, or multiple gas compartments.

1. The Core Principle: Force Balance on the Piston

The starting point for every piston problem is drawing a Free-Body Diagram (FBD) for the piston and applying Newton's First Law. At equilibrium, the piston is stationary, so the net force acting on it is zero.

Let's identify the forces typically involved:

Gas Pressure Force: The gas inside the cylinder exerts an outward force $F_{g a s} = P_{g a s} \cdot A$ , where $P_{g a s}$ is the absolute pressure of the gas and $A$ is the piston's area.
Atmospheric Pressure Force: The atmosphere exerts an inward force $F_{a t m} = P_{a t m} \cdot A$ .
Gravitational Force: If the cylinder is vertical, the piston's own weight, $m g$ , acts downwards.
Spring Force: If a spring is attached, it exerts a force $F_{s p r in g} = k x$ , where $x$ is the displacement from the spring's natural length.

The core of your task is to write the force balance equation:
$\sum F = 0$

Ideal Gas in Cylinder with Movable Piston and Spring Problem — This diagram illustrates the key forces acting on a piston connected to a spring. In equilibrium, the outward force from the gas pressure must balance the inward forces from the atmosphere and the compressed spring.

2. Linking Mechanics and Thermodynamics

The force balance equation is mechanical. To bring in thermodynamics, we use the ideal gas law to connect the gas pressure $P_{g a s}$ to its state variables (n, V, T).

$P_{g a s} = \frac{n R T}{V}$

The volume $V$ is almost always a function of the piston's position. For a cylindrical container, if $x$ is the length of the gas column, then $V = A \cdot x$ .

By substituting the ideal gas law into the force balance equation, you create a powerful single equation that links the mechanical equilibrium position of the piston to the thermodynamic state of the gas.

Example: A Vertical Piston with Mass

Consider a vertical cylinder with a piston of mass $m$ . The gas is at the bottom.

Force Balance: The upward force from the gas must balance the downward forces from the atmosphere and the piston's weight.
$P_{g a s} \cdot A = P_{a t m} \cdot A + m g$
Isolate Gas Pressure:
$P_{g a s} = P_{a t m} + \frac{m g}{A}$
Link to Thermodynamics: If the gas has $n$ moles and occupies a volume $V$ at temperature $T$ , we can substitute $P_{g a s}$ :
$\frac{n R T}{V} = P_{a t m} + \frac{m g}{A}$
This final equation is your tool to solve for any unknown variable ( $n, R, T, V$ , or $m$ ) if the others are given.

3. Analyzing a JEE Advanced Problem Setup

Let's apply this to a more complex, exam-level problem. This problem involves two gas compartments separated by a movable piston attached to a spring.

JEE Advanced Problem: Piston with Spring Between Two Compartments

This problem from a past JEE Advanced paper is an excellent example of applying equilibrium principles to a multi-component system. Please read the problem statement carefully.

Locate the second numerical question in the resource. It begins with 'The left and right compartments of a thermally isolated container...'. Study the diagram and the problem description to understand the initial setup.

Let's break down how to formulate the equations for this problem.

1. Identify the Equilibrium Conditions:

Mechanical Equilibrium: The piston is stationary. This means the net force on it is zero.
- Force from left gas: $P_{L} \cdot A$ (acting to the right).
- Force from right gas: $P_{R} \cdot A$ (acting to the left).
- Force from spring: The spring's natural length is $\frac{2 L}{5}$ . The piston is at $x = \frac{L}{2}$ from the left edge. The spring is stretched by $x - L_{na t u r a l} = \frac{L}{2} - \frac{2 L}{5} = \frac{L}{10}$ . The spring force is $k \cdot \frac{L}{10}$ (acting to the left, as it's stretched).
- Force balance equation: $P_{L} \cdot A = P_{R} \cdot A + k \frac{L}{10}$
Thermal Equilibrium: The piston is thermally conducting. This means the temperatures of the gases in the left ( $T_{L}$ ) and right ( $T_{R}$ ) compartments must be equal. Let's call this common temperature $T$ .
$T_{L} = T_{R} = T$

2. Apply the Ideal Gas Law:

Left Compartment: $n_{L} = 3/2$ moles. Volume $V_{L} = A \cdot \frac{L}{2}$ .
$P_{L} = \frac{n _{L} R T}{V _{L}} = \frac{( 3/2 ) R T}{A ( L /2 )} = \frac{3 R T}{A L}$
Right Compartment: $n_{R} = 1$ mole. Volume $V_{R} = A \cdot \frac{L}{2}$ .
$P_{R} = \frac{n _{R} R T}{V _{R}} = \frac{( 1 ) R T}{A ( L /2 )} = \frac{2 R T}{A L}$

3. Combine and Solve:
Now, substitute the expressions for $P_{L}$ and $P_{R}$ into the force balance equation:
$(\frac{3 R T}{A L}) A = (\frac{2 R T}{A L}) A + \frac{k L}{10}$
$\frac{3 R T}{L} = \frac{2 R T}{L} + \frac{k L}{10}$
$\frac{R T}{L} = \frac{k L}{10} ⟹ R T = \frac{k L ^{2}}{10}$

The problem asks for an expression for the pressure in the right compartment, $P_{R}$ . We found $P_{R} = \frac{2 R T}{A L}$ . Substituting the expression for $R T$ :
$P_{R} = \frac{2}{A L} (\frac{k L ^{2}}{10}) = \frac{2 k L}{10 A} = \frac{k L}{5 A}$
The problem states $P_{R} = \frac{k L}{A} α$ , so by comparison, $α = 1/5 = 0.2$ .

This systematic approach—Force Balance, Thermal Equilibrium, Ideal Gas Law, Combine & Solve—is the blueprint for cracking these problems.

Test your understanding!

In the problem we just solved, suppose the spring's natural length was $L /2$ instead of $2 L /5$ . What would be the relationship between $P_{L}$ and $P_{R}$ at equilibrium?

Show answer

If the natural length of the spring is $L /2$ and the piston is at $x = L /2$ , the spring is at its natural length. Therefore, the spring exerts zero force ( $F_{s p r in g} = k \cdot 0 = 0$ ). The force balance equation simplifies to:
$P_{L} \cdot A = P_{R} \cdot A ⟹ P_{L} = P_{R}$
This means the pressures in the two compartments would be equal.

4. Advanced Application: Processes and the First Law

The most complex problems involve a process where the system moves from one equilibrium state to another, often triggered by heating or external work. These problems require you to use the First Law of Thermodynamics in conjunction with the equilibrium equations.

Let's watch a video that solves a fantastic JEE Advanced problem integrating thermodynamics, mechanics, and even a bit of electromagnetism.

[JEE ADVANCED] GAS PISTON CYLINDER WITH SPRINGS AND CAPACITOR [ ADVANCE PROBLEMS IN SCHOOL PHYSICS]

This video from 'INDIAN SCHOOL OF PHYSICS' solves a brilliant problem where the energy from a discharging capacitor heats a gas, causing a piston-spring system to expand to a new equilibrium. This demonstrates the full synthesis of concepts required at the highest level.

Watch the video from 0:38 to 8:06. As you watch, focus on the three distinct parts of the solution strategy: (01:28 - 02:49) Energy Input: How the initial energy stored in the capacitor (Q_0^2/2C) is identified as the heat (Q) supplied to the gas. (02:49 - 04:17) Final Mechanical Equilibrium: How the force balance equation is set up for the piston's final position to determine the final gas pressure, P_{gas}. (04:17 - 07:32) The First Law of Thermodynamics: How the equation Q = \Delta U + W_{by gas} is used to link everything together. Pay special attention to the careful calculation of work done by the gas (W_{by gas}), which includes work against the atmosphere and the energy stored in the springs. This is a common place to make mistakes.

Deconstructing the Video's Logic

Let's summarize the elegant problem-solving architecture from the video:

Energy Input (The "Why"): The process starts because the capacitor discharges, releasing its stored energy $E = \frac{Q _{0}^{2}}{2 C}$ as heat. This heat is absorbed by the gas. So, for the thermodynamic process, we have $Q = \frac{Q _{0}^{2}}{2 C}$ .
Mechanical Analysis (The "Where"): The system moves to a final equilibrium state where the piston has moved by $x_{0}$ . At this final state, the forces on the piston must be balanced.
- $F_{g a s, f ina l} = F_{a t m} + F_{s p r in g}$
- $P_{f ina l} \cdot A = P_{a t m} \cdot A + k x_{0} ⟹ P_{f ina l} = P_{a t m} + \frac{k x _{0}}{A}$
- The final volume is also determined: $V_{f ina l} = V_{ini t ia l} + 2 A x_{0}$ (since both piston and cylinder move).
Thermodynamic Analysis (The "How"): The First Law connects the energy input to the change in state. The convention used here is $Δ U = Q - W_{b y_g a s}$ , or $Q = Δ U + W_{b y_g a s}$ .
- $Q$ : We already know this is $\frac{Q _{0}^{2}}{2 C}$ .
- $Δ U$ : For an ideal gas, $Δ U = n C_{v} Δ T$ . A more direct way, as shown in the video, is to use $U = \frac{f}{2} n R T = \frac{f}{2} P V$ . For a monatomic gas, $f = 3$ , so $Δ U = U_{f} - U_{i} = \frac{3}{2} (P_{f} V_{f} - P_{i} V_{i})$ . We know all these terms from the mechanical analysis.
- $W_{b y_g a s}$ : This is the crucial part. The work done by the expanding gas has two components:
  1. Work done against the constant atmospheric pressure on both sides: $W_{a t m} = P_{a t m} Δ V = P_{a t m} (2 A x_{0})$ .
  2. Work done to compress the two springs, which is stored as potential energy: $W_{s p r in g s} = 2 \times \frac{1}{2} k x_{0}^{2} = k x_{0}^{2}$ .
    So, $W_{b y_g a s} = W_{a t m} + W_{s p r in g s} = 2 P_{a t m} A x_{0} + k x_{0}^{2}$ .

By calculating $Δ U$ and $W_{b y_g a s}$ and plugging them into $Q = Δ U + W$ , you arrive at the final equation to solve for the unknown $Q_{0}$ .

Conclusion

Mastering piston-cylinder problems is about having a clear, structured approach. You are no longer just solving for 'x' in a formula; you are building a mathematical model of a physical system from first principles.

Key Takeaways:

Always start with a Free-Body Diagram and the force balance equation for the piston. This is the mechanical heart of the problem.
Use the Ideal Gas Law to translate the mechanical quantity of force/pressure into thermodynamic state variables ( $V, T$ ).
For problems involving a process (heating, cooling, work input), apply the First Law of Thermodynamics to the gas as your system to connect the initial and final equilibrium states.
Be meticulous when calculating work done ( $W$ ). It's the total work done by the gas against all external forces (atmosphere, springs, gravity, etc.).

Preview of the Next Lesson:
Our next lesson will introduce a fascinating and counter-intuitive process: the free expansion of a gas. We will analyze why the work done is zero and what happens to the temperature of an ideal gas during this process. This will lead us directly to the first law of thermodynamics and the critical topic of sign conventions.