Isothermal Free Expansion & the First Law

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

In our previous session, we dissected cyclic processes, establishing the crucial rule that for any complete cycle, the net change in internal energy is zero ( $Δ U_{cy c l e} = 0$ ), leading to the conclusion that net heat absorbed equals net work done ( $Q_{n e t} = W_{n e t}$ ).

Today, we will pivot from cycles to a very specific and conceptually important irreversible process: free expansion. This is a classic thought experiment in thermodynamics that is essential for a deep understanding of work, internal energy, and the First Law. Our goal is to understand what constitutes a free expansion and, most importantly, to justify why the work done during this process is always zero.

1. Defining Free Expansion

At its core, free expansion is the process of a gas expanding into a vacuum. Imagine a rigid, insulated container divided into two compartments by a partition. One compartment contains a gas, and the other is a complete vacuum. If we suddenly remove the partition, the gas will spontaneously expand to fill the entire volume.

This process has two key characteristics:

Expansion into a vacuum: There is no external pressure opposing the expansion.
Irreversibility: It's a spontaneous, chaotic process. The system is not in equilibrium during the expansion, so intermediate states (P, V, T) are not well-defined. You can't simply reverse the process by infinitesimally changing conditions.

Let's watch a short video that introduces this concept and its immediate consequences.

Class 11 Physics | Laws of Thermodynamics | #28 Free Expansion of a Gas | For JEE & NEET

This video from Physics Galaxy clearly defines free expansion and explains the fundamental reason why work done is zero. It also connects this to the First Law for a thermally insulated container.

Please watch from the beginning to 02:40. Focus on: The definition of free expansion as expansion "without any opposition." The conclusion that if there is no opposition, no work is done. How the First Law of Thermodynamics is applied when the container is also thermally insulated (adiabatic).

2. Justifying Why Work Done is Zero

As the video explained, the work done is zero. Let's formalize this with the definition of work you're familiar with. The work done by a gas on its surroundings is given by the integral:

$W = \int_{V_{i}}^{V_{f}} P_{e x t} d V$

where $P_{e x t}$ is the external pressure the gas is expanding against.

In the case of free expansion, the gas expands into a vacuum. By definition, a vacuum exerts no pressure. Therefore:

$P_{e x t} = 0$

Substituting this into the work equation, we get:

$W = \int_{V_{i}}^{V_{f}} (0) d V = 0$

This is the rigorous justification. No matter how much the volume changes, if there is no external pressure to push against, no work is done.

Another way to think about this, as described in the MIT resource 4.2 Difference between Free and Isothermal Expansions, is that work is done when the boundaries of a system move against an external force. In free expansion into a rigid container, the overall boundary of the system (the container walls) does not move. The gas simply redistributes itself within the fixed volume. Since there is no displacement of the external boundary, no work is exchanged with the surroundings.

Test your understanding!

Two moles of an ideal gas, initially confined to a 1-liter flask at 300 K, are allowed to expand spontaneously into an evacuated 2-liter flask connected to it. The entire system is rigid. What is the work done by the gas during this expansion?

Show answer

The work done is zero.

This is a classic free expansion scenario. The gas expands into an evacuated space, meaning the external pressure ( $P_{e x t}$ ) it is expanding against is zero. Since $W = \int P_{e x t} d V$ , the work done must be zero. The amount of gas, its temperature, and the volumes are all irrelevant to the calculation of work in a free expansion.

3. The First Law and Adiabatic Free Expansion

Now, let's apply the First Law of Thermodynamics, $Δ U = Q - W$ , to this process. A common scenario in problems, particularly in JEE Advanced, is adiabatic free expansion. This means the expansion is not only free ( $W = 0$ ) but also occurs in a perfectly insulated container ( $Q = 0$ ).

Let's plug these values into the First Law:

$W = 0$ (because it's a free expansion)
$Q = 0$ (because it's adiabatic)

Therefore:
$Δ U = Q - W = 0 - 0 = 0$

This is a powerful result: In an adiabatic free expansion, the change in the internal energy of the gas is zero.

This principle holds true regardless of whether the gas is ideal, real, or even a photon gas. Let's see how this is applied in a JEE Advanced problem involving a non-standard system.

ORIGINALS VIDEO | PHOTON GAS THERMODYNAMICS | JEE ADVANCED SCHOOL PHYSICS PROBLEMS |

This video by Janardhan Sir tackles a JEE Advanced problem on the adiabatic free expansion of a photon gas. This is a great example of applying the core principles to an unfamiliar system.

Please watch from 04:10 to 05:41. Pay close attention to the reasoning: The process is irreversible, so intermediate steps are not defined. Why the total heat transfer, Q, is zero. The justification for why the work done, W, is zero (P_{external} is zero). The final conclusion that \Delta U must be zero.

As you saw, the logic is identical. The instructor correctly identifies that $Q = 0$ (adiabatic) and $W = 0$ (free expansion), which forces the conclusion that $Δ U = 0$ . He then uses this fact ( $U_{ini t ia l} = U_{f ina l}$ ) to solve the rest of the problem.

4. Conceptual Check

Let's consolidate these ideas.

Test your understanding!

For an adiabatic free expansion of an ideal gas, which of the following quantities are zero?

Heat exchanged ( $Q$ )
Work done ( $W$ )
Change in Internal Energy ( $Δ U$ )
Change in Temperature ( $Δ T$ )

Choose all that apply.

Show answer

All four quantities are zero.

$Q = 0$ : The process is explicitly stated as adiabatic.
$W = 0$ : The process is a free expansion.
$Δ U = 0$ : From the First Law, $Δ U = Q - W = 0 - 0 = 0$ .
$Δ T = 0$ : For an ideal gas, internal energy is a function of temperature only ( $U = f (T)$ ). Since $Δ U = n C_{v} Δ T$ , if $Δ U = 0$ (and $n, C_{v}$ are non-zero), then $Δ T$ must also be zero. The temperature of an ideal gas does not change during an adiabatic free expansion.

Conclusion

In this lesson, we've isolated and analyzed the important case of free expansion. By understanding its definition and consequences, you've added a key tool to your problem-solving arsenal.

Key Takeaways:

Free expansion is the expansion of a gas into a vacuum, where the external pressure $P_{e x t}$ is zero.
The work done during any free expansion is always zero because $W = \int P_{e x t} d V = 0$ .
For an adiabatic free expansion, both heat transfer ( $Q$ ) and work done ( $W$ ) are zero.
Consequently, from the First Law ( $Δ U = Q - W$ ), the change in internal energy ( $Δ U$ ) during an adiabatic free expansion is also zero.

Preview of the Next Lesson:
We've established that for an adiabatic free expansion, $Δ U = 0$ . In our final mini-exercise, we saw that for an ideal gas, this implies $Δ T = 0$ . But what about a real gas? In the next lesson, we will explore this question in detail, analyzing the temperature change (or lack thereof) during a free expansion for both ideal and real gases. This will reveal some interesting effects that arise from intermolecular forces, a key distinction between real and ideal gases.

Recommended resources

1. Defining Free Expansion

2. Justifying Why Work Done is Zero

3. The First Law and Adiabatic Free Expansion

4. Conceptual Check

Conclusion