Animated Solution for Chemistry - Chemical Thermodynamics: Five moles of an ideal gas at $1\text{ bar}$ and $298\text{ K}$ is expanded into vacuum to double the volume. The work done is

Imagine you are trapped in a crowded room, pushing against a heavy door to get out. You sweat, you strain, and you expend a massive amount of energy. That is what a gas feels like when it expands against an external atmospheric pressure. It has to do work to push the atmosphere out of the way.

But what if you open the door, and there is absolutely nothing on the other side? No air, no pressure, just an infinite, empty void. Would you need to push? Not at all! You would just walk right in. This is the exact physical reality of free expansion.

In thermodynamics, the work done by a gas during expansion or compression is governed by a very simple, yet profound equation:

Here, $W$ is the work done, $p_{\text{ext}}$ is the external pressure opposing the expansion, and $\Delta V$ is the change in volume. The negative sign is a convention: if the gas expands ($\Delta V>0$), it does work on the surroundings, losing energy, so $W$ is negative.

The problem states that five moles of an ideal gas at $1\text{ bar}$ and $298\text{ K}$ expand into a vacuum to double its volume.

The word "vacuum" is the ultimate trap for the unwary student. A vacuum, by definition, is a space entirely devoid of matter. If there is no matter, there are no particles colliding, and therefore, the pressure is exactly zero.

The problem also throws in extra numbers: $5\text{ moles}$, $1\text{ bar}$, $298\text{ K}$, and "double the volume". These are classic distractors! They are there to tempt you into using complex formulas like $-nRT\ln (V_2/V_1)$. But remember, that formula is for a reversible isothermal expansion against a gradually changing external pressure, not a sudden expansion into nothingness.

Let's substitute our raw values into the master equation. We know there is a change in volume ($\Delta V=2V_1-V_1=V_1$), but look at what happens when we plug in the external pressure:

The math perfectly mirrors the physical reality. Because there is no opposing force, the gas doesn't have to expend any energy to expand. The work done is exactly zero.

This simple result has profound implications when we look at the First Law of Thermodynamics:

Since $W=0$, the equation simplifies to $\Delta U=q$. For an ideal gas, the internal energy $U$ depends solely on its temperature. If the expansion happens in an insulated container (adiabatic, $q=0$), then $\Delta U=0$, which means the temperature remains perfectly constant ($\Delta T=0$). This is why ideal gases do not cool down or heat up during free expansion!