Animated Solution for Chemistry - Chemical Thermodynamics: The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of $10{\text{ dm}}^3$ to a volume of $100{\text{ dm}}^3$ at $27^{\circ }\text{C}$ is

The concept of entropy often feels abstract, like a ghost in the machine of thermodynamics. But at its core, entropy is simply a measure of how spread out or dispersed the energy of a system is.

Imagine you are standing in a small, crowded room. You can barely move. This is our gas at $10{\text{ dm}}^3$. Now, the walls suddenly expand, and the room becomes ten times larger! You and everyone else can spread out, run around, and occupy more space. This spreading out is exactly what happens during an isothermal expansion, and it corresponds to an increase in entropy.

When an ideal gas expands isothermally (meaning the temperature remains perfectly constant), its internal energy doesn't change. All the heat it absorbs from the surroundings is converted entirely into work to push the piston outward.

To quantify this increase in randomness, we use the beautiful relation derived from the Second Law of Thermodynamics:

Here, $n$ is the number of moles, $R$ is the universal gas constant, and $V_1$ and $V_2$ are the initial and final volumes, respectively.

Let's bring our specific problem into focus. We have $n=2\text{ moles}$ of an ideal gas. It expands from $V_1=10{\text{ dm}}^3$ to $V_2=100{\text{ dm}}^3$. The temperature is $27^{\circ }\text{C}$, but notice how temperature doesn't even appear in our final equation! That's the magic of isothermal processes for ideal gases.

Before we plug in the numbers, it's often easier to work with base-10 logarithms instead of natural logarithms. We can convert it using the factor $2.303$:

Now, let's substitute our values:

The volume ratio is incredibly clean: $\frac{100}{10}=10$. And we all know that ${\log }_{10}(10)=1$.

So, our equation simplifies to a straightforward multiplication:

Rounding to one decimal place, we get $38.3{\text{ J K}}^{-1}$.

Here is a fascinating thought experiment: What if the gas didn't expand reversibly? What if it expanded suddenly into a vacuum (a free expansion)?

Because entropy is a state function, the entropy change of the gas would be exactly the same! It only cares about the initial and final states, not the journey taken to get there. The universe's total entropy would change differently, but the gas itself would still experience an entropy increase of $38.3{\text{ J K}}^{-1}$.