Animated Solution for Physics - Thermodynamics: Following figure shows two processes A and B for a gas. If $\Delta Q_A$ and $\Delta Q_B$ are the amount of heat absorbed by the system in two cases, and $\Delta U_A$ and $\Delta U_B$ are changes in internal energies respectively, then

Path Functions

Imagine you are standing at the base of a magnificent mountain, looking up at the summit. You have two choices: you can take the steep, direct path that goes straight up the cliff face, or you can take the long, winding scenic route that gently spirals around the mountain.

No matter which path you choose, once you reach the top, your change in altitude is exactly the same. Your altitude depends only on your starting point (the base) and your ending point (the summit). In the world of thermodynamics, we call this a state function.

However, think about the effort you exerted, the calories you burned, and the sweat on your brow. The long, winding route probably made you burn significantly more calories than the direct climb. The calories burned depend entirely on the journey you took. In thermodynamics, we call this a path function.

This beautiful distinction is the absolute core of the problem we are solving today.

Our master tool for this problem is the First Law of Thermodynamics. It is essentially the law of conservation of energy applied to thermal systems. The law states:

Here, $\Delta Q$ represents the heat absorbed by the system. $\Delta U$ is the change in the internal energy of the system, and $W$ is the work done by the system on its surroundings.

Let's break down these components based on our mountain analogy.

The change in internal energy, $\Delta U$, is our altitude. It is a strict state function. It does not care whether the gas expanded isothermally, adiabatically, or took some wild, undefined path. It only cares about the initial state $i$ and the final state $f$.

Since both process A and process B start at the exact same initial state $i$ and end at the exact same final state $f$, their change in internal energy must be identical.

This is a massive breakthrough! We have already solved half of the problem just by understanding the physical meaning of a state function.

Now, let's turn our attention to the work done, $W$. Work is the thermodynamic equivalent of the calories burned on our mountain climb. It is a path function, meaning the route we take on the $p-V$ diagram changes everything.

Mathematically, the work done by a gas during expansion is given by the integral of pressure with respect to volume:

Geometrically, this integral represents the area under the curve on a $p-V$ diagram.

Look closely at the graph provided in the question. Path A curves upwards, maintaining a higher pressure throughout the expansion. If we shade the region under curve A, we get a large area extending all the way down to the volume axis.

Now, look at path B. It curves downwards, maintaining a lower pressure during the expansion. If we shade the region under curve B, it is visually obvious that it covers much less space than curve A.

Since the area under curve A is strictly greater than the area under curve B, the work done by the gas in process A is greater than the work done in process B.

We now have all the puzzle pieces. Let's bring them back to our master equation, the First Law of Thermodynamics.

For process A, the heat absorbed is:

For process B, the heat absorbed is:

We have already established that the internal energy change is identical ($\Delta U_A=\Delta U_B$). This means the $\Delta U$ term acts as a constant baseline for both equations.

However, we also know that $W_A>W_B$.

If you take two identical baseline numbers and add a larger number to the first one, the first sum will inevitably be larger. Therefore, the total heat absorbed in process A must be strictly greater than the total heat absorbed in process B.

And there we have it! By combining the geometric interpretation of work with the fundamental nature of state functions, we have elegantly arrived at the correct conclusion. The correct option is (c).