Animated Solution for Physics - Thermodynamics: Three different processes that can occur in an ideal monoatomic gas are shown in the $p$ versus $V$ diagram. The paths are labelled as $A\to B$, $A\to C$ and $A\to D$. The change in internal energies during these process are taken as $E_{AB}$, $E_{AC}$ and $E_{AD}$ and the work done as $W_{AB}$, $W_{AC}$ and $W_{AD}$. The correct relation between these parameters are

Thermodynamics is a beautiful interplay of state variables and path-dependent functions. In this problem, we are presented with a classic $p-V$ diagram showing three distinct thermodynamic processes originating from a single state and terminating on a common isotherm. Let's break down the physics step-by-step to uncover the absolute truth, even if the given options try to trick us!

Take a close look at the $p-V$ diagram. We have an initial state $A$ lying on a lower isotherm, $T_2$. From this point, the gas undergoes three different processes: $A\to B$, $A\to C$, and $A\to D$.

The crucial observation here is that all three final states—$B$, $C$, and $D$—lie on the exact same upper isotherm, $T_1$. We are explicitly given that $T_1>T_2$. This geometric setup immediately tells us that regardless of the path taken, the initial and final temperatures for all three processes are identical.

Now, let's talk about the change in internal energy, denoted as $E$ (or $\Delta U$). For an ideal gas, the internal energy is a state function. This means it depends exclusively on the absolute temperature of the gas and is completely blind to the path taken to reach that state.

The mathematical expression for this is:

Since all three processes start at temperature $T_2$ and end at temperature $T_1$, the change in temperature $\Delta T=T_1-T_2$ is exactly the same for paths $A\to B$, $A\to C$, and $A\to D$. Consequently, the change in internal energy must be identical across the board:

Next, we evaluate the work done, $W$. Unlike internal energy, work is a path function. Geometrically, the work done by a gas is the area under the $p-V$ curve. Mathematically, it is given by the integral:

The sign of the work done is dictated by the change in volume. Let's analyze each path:

1. Path $A\to B$: The arrow points to the right, indicating an expansion. The volume increases ($\Delta V>0$), so the work done by the gas is positive. Thus, $W_{AB}>0$. 2. Path $A\to C$: This is a perfectly vertical line. The volume remains constant ($\Delta V=0$), making it an isochoric process. With no area under the curve, the work done is zero. Thus, $W_{AC}=0$. 3. Path $A\to D$: The arrow points to the left, indicating a compression. The volume decreases ($\Delta V<0$), so the work done by the gas is negative. Thus, $W_{AD}<0$.

Bringing our findings together, we have established the absolute truth for this system:

If we carefully inspect the given options, we realize that none of the options perfectly match this correct set of relations. Option (b) comes close but incorrectly states $W_{AD}>0$. In competitive exams like JEE, such anomalies can occur, and they are often treated as bonus questions. The key takeaway is to trust your fundamental concepts!

As a thought experiment, consider the heat supplied ($\Delta Q$) for these processes. According to the First Law of Thermodynamics ($\Delta Q=\Delta U+W$), since $\Delta U$ is constant, the path with the most positive work will absorb the most heat. Therefore, path $A\to B$ absorbs the maximum heat, while path $A\to D$ absorbs the least. Always keep pushing your conceptual boundaries!