An ideal gas is allowed to expand against a constant pressure of 2 bar from 10 L to 50 L in one step. Calculat

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Q. An ideal gas is allowed to expand against a constant pressure of 2 bar from 10 L to 50 L in one step. Calculate the amount of work done by the gas. If the same expansion were carried out reversibly, will the work done be higher or lower than the earlier case? (Given that 1 L bar = 100 J)

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Concept understanding — Irreversible Expansion Work

Irreversible Expansion Work – From Intuition to Precision

Imagine you have a gas trapped inside a cylinder with a piston. If you suddenly pull the piston outward, the gas expands rapidly into the newly available space. That's an irreversible expansion — the gas doesn't pass through a series of equilibrium states; it rushes, swirls, and settles only at the end.

Now, think about the work done by the gas during this process. Work, in physics, is force times displacement. For a piston, force is pressure times area, so work becomes $P \Delta V$ . But here's the catch: during an irreversible expansion, the pressure of the gas is not uniform throughout the cylinder. There are pressure gradients, turbulence, and the gas near the piston face may be at a different pressure than the gas deeper inside.

So how do we calculate the work done?

The Key Insight

The work done by the gas is determined by the external pressure it pushes against — not its own internal pressure. Why? Because the piston moves only in response to the net force acting on it. That net force comes from the external pressure on the other side of the piston.

Important

For any expansion (reversible or irreversible), the work done by the gas is:

$W = \int P_{\text{ext}} \, dV$

where $P_{\text{ext}}$ is the pressure exerted on the gas by the surroundings (the piston face).

During a reversible expansion, the gas is always in equilibrium with the surroundings, so $P_{\text{gas}} = P_{\text{ext}}$ at every instant. That's why you can replace $P_{\text{ext}}$ with $P_{\text{gas}}$ and integrate using the gas's equation of state.

During an irreversible expansion, $P_{\text{gas}}$ is not equal to $P_{\text{ext}}$ — and often, $P_{\text{ext}}$ is held constant (like when you suddenly release the piston against atmospheric pressure). In that case, the work simplifies dramatically:

W = P_{\text{ext}} \Delta V

The Intuitive Picture

Think of pushing a heavy box across a rough floor. The work you do depends on the force you apply (your "external" force), not on the internal stresses inside the box. Similarly, the gas does work against the external resistance it meets — the piston's opposing force.

If the external pressure is constant (say, 1 atm), the gas does work equal to $P_{\text{ext}} \times$ (change in volume), regardless of how chaotically it expands. The gas might have been at 10 atm initially, but it only does work against the 1 atm it actually pushes.

Watch out

A common mistake: using the gas's own pressure to calculate irreversible work. Unless the process is reversible, $P_{\text{gas}} \neq P_{\text{ext}}$ , and using $P_{\text{gas}}$ gives the wrong answer.

The Precise Statement

Irreversible expansion work is the work done by a gas when it expands through a series of non-equilibrium states. It is calculated using the external pressure that opposes the expansion:

W_{\text{irr}} = \int_{V_1}^{V_2} P_{\text{ext}} \, dV

For the most common case — expansion against a constant external pressure (like the atmosphere or a fixed weight on the piston): …

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The work done by the gas during irreversible expansion against a constant external pressure is calculated using $W = -P_{ext} \Delta V$ . For the given conditions, the work done is $\mathbf{-8000 \text{ J}}$ . If the same expansion were carried out reversibly, the work done would be lower (more negative) than in the irreversible case.

When a gas expands, it pushes against its surroundings and performs work. In thermodynamics, work done by the system (the gas) is conventionally assigned a negative sign. The amount of work done depends significantly on the path taken during the expansion. We will consider two distinct paths: irreversible expansion against a constant external pressure and reversible expansion.

Understanding Work Done During Expansion

Irreversible Expansion: This is a common process where the gas expands rapidly against a constant external pressure that is significantly less than the internal pressure of the gas. The system is not in equilibrium with its surroundings throughout the process. Since the external pressure is constant, the calculation of work is straightforward.
Reversible Expansion: This is an idealized process where the expansion occurs in an infinite number of infinitesimal steps. At each step, the external pressure is infinitesimally smaller than the internal pressure of the gas, ensuring that the system is always in equilibrium with its surroundings. This theoretical process yields the maximum possible work done by the system for a given change in volume.

1. Calculating Work Done in Irreversible Expansion

For an irreversible expansion against a constant external pressure, the work done ( $W$ ) is given by the formula:

W = -P_{ext} \Delta V

Let's identify the given values:

Constant external pressure, $P_{ext} = 2 \text{ bar}$
Initial volume, $V_1 = 10 \text{ L}$
Final volume, $V_2 = 50 \text{ L}$

Now, we can calculate the change in volume:

$\Delta V = V_2 - V_1 = 50 \text{ L} - 10 \text{ L} = 40 \text{ L}$

Substitute these values into the work formula:

$W = -(2 \text{ bar}) \times (40 \text{ L})$

$W = -80 \text{ L bar}$

The problem provides a conversion factor: $1 \text{ L bar} = 100 \text{ J}$ . We use this to convert the work done from L bar to Joules:

$W = -80 \text{ L bar} \times \frac{100 \text{ J}}{1 \text{ L bar}}$

$W = -8000 \text{ J}$

The negative sign indicates that work is done by the gas (the system) on the surroundings.

2. Comparing with Reversible Expansion

For any expansion process between the same initial and final states, the work done by the gas is always maximized when the process is carried out reversibly. This means that the magnitude of work done in a reversible expansion is greater than that in an irreversible expansion. …

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Q.An ideal gas is allowed to expand against a constant pressure of 2 bar from 10 L to 50 L in one step. Calculate the amount of work done by the gas. If the same expansion were carried out reversibly, will the work done be higher or lower than the earlier case? (Given that 1 L bar = 100 J)

Understanding Work Done During Expansion

1. Calculating Work Done in Irreversible Expansion

2. Comparing with Reversible Expansion

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