Animated Solution for Chemistry - Chemical Thermodynamics: A piston filled with $0.04$ mole of an ideal gas expands reversibly from $50.0\text{ mL}$ to $375\text{ mL}$ at a constant temperature of $37.0^{\circ }\text{C}$. As it does so, it absorbs $208\text{ J}$ of heat. The values of $q$ and $W$ for the process will be ($R=8.314\text{ J / mol K}$, $\ln 7.5=2.01$)

## The Isothermal Illusion: Navigating the First Law of Thermodynamics

Have you ever looked at a physics or chemistry problem and felt completely overwhelmed by the sheer amount of data provided? You see moles, initial volumes, final volumes, temperatures, and even logarithmic values. Your brain immediately starts scrambling to find a formula that fits all these puzzle pieces together. But what if I told you that sometimes, the examiner is just playing a psychological game with you?

Welcome to one of the most classic traps in JEE and NEET thermodynamics. In this problem, we are going to learn how to see through the noise, identify the core physical principle, and solve a seemingly complex calculation in mere seconds without touching a single decimal point.

The Expanding Gas

Imagine you are standing in a laboratory, looking at a classic piston-cylinder apparatus. Inside this cylinder, we have $0.04$ moles of an ideal gas. The gas is currently occupying a volume of $50.0\text{ mL}$ and is sitting at a comfortable temperature of $37.0^{\circ }\text{C}$.

Suddenly, the system starts to absorb heat from its surroundings. We are told it absorbs exactly $208\text{ J}$ of heat. As the gas absorbs this thermal energy, it gets "excited" and begins to expand, pushing the piston upwards until it reaches a new volume of $375\text{ mL}$.

The question asks us to find the values of the heat $q$ and the work done $W$ for this process.

At first glance, your instinct might be to grab the formula for reversible isothermal work:

You have $n=0.04\text{ mol}$, $R=8.314\text{ J / mol K}$, $T=310\text{ K}$ (which is $37.0^{\circ }\text{C}+273$), $V_1=50.0\text{ mL}$, and $V_2=375\text{ mL}$. The problem even generously provides the value of $\ln (7.5)=2.01$.

You could absolutely plug all these numbers in and start multiplying. But let's take a step back. Is there a smarter, more elegant way to approach this?

The First Law of Thermodynamics

Whenever a thermodynamic system undergoes a process involving heat and work, our ultimate guiding principle is the First Law of Thermodynamics. This law is essentially the principle of conservation of energy applied to thermal systems.

It states that the change in the internal energy of a system ($\Delta U$) is equal to the sum of the heat exchanged with the surroundings ($q$) and the work done on or by the system ($W$).

Mathematically, it is expressed as:

Think of the internal energy $U$ as the system's bank account balance. Heat $q$ and work $W$ are the transactions—deposits and withdrawals. If you add heat to the system, you are depositing energy. If the system does work on the surroundings (like expanding and pushing a piston), it is spending energy, which is a withdrawal.

The Isothermal Condition

Now, let's look closely at the problem statement again. There is a phrase hidden in plain sight that completely breaks this problem wide open: "at a constant temperature".

This phrase tells us that the process is isothermal.

Why is this so crucial? Because for an ideal gas, the internal energy is purely a function of its kinetic energy, which in turn is directly proportional to its absolute temperature. The molecules of an ideal gas don't have intermolecular forces (potential energy), so their entire internal energy is just the energy of their random motion.

If the temperature of an ideal gas does not change ($\Delta T=0$), then its internal energy cannot change either.

Therefore, for any isothermal process involving an ideal gas:

This is the golden key! We don't need to know the initial volume, the final volume, or the number of moles to know that the internal energy change is zero. The word "isothermal" guarantees it.

Final Calculation

Let's take our golden key ($\Delta U=0$) and insert it into the First Law of Thermodynamics.

By simply rearranging this equation, we get a beautiful, direct relationship between heat and work for this specific process:

This equation tells us a profound physical story: in an isothermal expansion, every single joule of heat that the gas absorbs is immediately spent doing work on the surroundings. The gas doesn't keep any of the energy for itself (which is why its internal energy doesn't change).

Now, let's apply the standard IUPAC sign conventions. The problem states that the system absorbs $208\text{ J}$ of heat. Heat added to the system is positive.

Substituting this into our derived relationship:

And just like that, we have our answer! The heat $q$ is $+208\text{ J}$, and the work $W$ is $-208\text{ J}$. The negative sign for work perfectly aligns with the physical reality that the gas is expanding and doing work by the system, which means energy is leaving the system as work.

Let's take a moment to appreciate what just happened. We solved the problem using pure conceptual logic and one simple addition equation.

But what about all that other data? The $0.04$ moles, the $50.0\text{ mL}$, the $375\text{ mL}$, the $R$ value, and the $\ln (7.5)$?

They were a trap. A brilliantly designed distractor meant to test your conceptual clarity.

If you had taken the bait and used the work formula, your calculation would have looked like this:

Due to slight rounding in the provided $\ln (7.5)$ value and the temperature, you get approximately $-207.2\text{ J}$, which rounds to $-208\text{ J}$.

Yes, the long method works. The data provided is physically consistent. But look at the time and effort it takes! In a high-pressure exam like JEE or NEET, time is your most valuable resource.

This problem is a masterclass in why you should always read the question carefully and look for conceptual shortcuts before diving into heavy calculations. The First Law of Thermodynamics isn't just a formula; it's a powerful lens through which you can view and simplify the physical world.

Always trust the concepts, watch out for the distractors, and remember: sometimes the smartest way to calculate is to not calculate at all.