Animated Solution for Chemistry - Chemical Thermodynamics: One mole of an ideal gas at $300\text{ K}$ in thermal contact with surroundings expands isothermally from $1.0\text{ L}$ to $2.0\text{ L}$ against a constant pressure of $3.0\text{ atm}$. In this process, the change in entropy of surroundings ($\Delta S_{\text{surr}}$) in ${\text{J K}}^{-1}$ is - (Given: $1\text{ L atm}=101.3\text{ J}$)

Welcome to a fascinating journey into the heart of chemical thermodynamics! Imagine you are standing in a laboratory, observing a perfectly sealed cylinder containing exactly one mole of an ideal gas. The gas is maintained at a cozy, constant temperature of $300\text{ K}$. Suddenly, the piston is released, and the gas expands from a volume of $1.0\text{ L}$ to $2.0\text{ L}$. But it doesn't expand freely; it pushes against a stubborn, constant external pressure of $3.0\text{ atm}$.

Our mission? To uncover the hidden story of the surroundings. Specifically, we need to calculate the change in entropy of the surroundings ($\Delta S_{\text{surr}}$) during this process. I know thermodynamics can sometimes feel like a maze of abstract equations, but let's take a breath and break this down logically, step by step.

First, let's extract the vital clues from our problem statement. We are dealing with an ideal gas undergoing an isothermal (constant temperature) expansion. The expansion happens against a constant external pressure ($P_{\text{ext}}$). This last detail is a massive clue! Whenever a gas expands against a constant external pressure, the process is inherently irreversible.

Why does this matter? Because the path the gas takes determines how much work it does and, consequently, how much heat it exchanges with the surroundings.

First Law of Thermodynamics

To understand the heat exchange, we must invoke the First Law of Thermodynamics, the ultimate law of energy conservation:

Here, $\Delta U$ is the change in internal energy of the gas, $q_{\text{sys}}$ is the heat absorbed by the gas, and $w$ is the work done on the gas.

Now, let's apply our specific conditions. For an ideal gas, the internal energy depends only on its temperature. Since our process is isothermal ($\Delta T=0$), the internal energy remains perfectly constant. Therefore, $\Delta U=0$.

Substituting this into our master equation, we get a beautiful relationship:

This tells us that any work the gas does on the surroundings is exactly compensated by the heat it absorbs from the surroundings.

Let's calculate the work done. For an irreversible expansion against a constant external pressure, the work done is given by:

Substituting this back into our heat equation:

Now, let's plug in the raw numbers. The external pressure $P_{\text{ext}}$ is $3.0\text{ atm}$, the final volume $V_2$ is $2.0\text{ L}$, and the initial volume $V_1$ is $1.0\text{ L}$.

We have the heat in liter-atmospheres, but standard entropy is measured in Joules per Kelvin. The problem kindly provides the conversion factor: $1\text{ L atm}=101.3\text{ J}$. Let's convert our heat:

So, the system absorbs $303.9\text{ J}$ of heat from the surroundings.

Now we arrive at the core of the problem: the entropy change of the surroundings. The surroundings are massive, acting as a giant thermal reservoir. When they exchange heat, their temperature doesn't change. Therefore, the entropy change of the surroundings is always calculated simply as the heat absorbed by the surroundings divided by their temperature:

Here is where we must be careful with our signs. By the law of conservation of energy, the heat absorbed by the system must be exactly equal to the heat lost by the surroundings.

We are in the home stretch! Let's substitute our values into the entropy equation. The temperature of the surroundings is in thermal equilibrium with the system, so $T=300\text{ K}$.

And there we have it! The entropy of the surroundings decreases by $1.013{\text{ J K}}^{-1}$. The negative sign makes perfect physical sense because the surroundings lost heat to the system.

This is a classic JEE Advanced problem that beautifully weaves together the First Law of Thermodynamics, work calculations, and the concept of entropy. Always remember to carefully track your signs and units, and you'll master these concepts in no time!