Animated Solution for Physics - Thermodynamics: The $p-V$ diagram of a diatomic ideal gas system going under cyclic process as shown in figure. The work done during an adiabatic process $CD$ is (use, $\gamma =1.4$)

Welcome to this fascinating journey into the world of thermodynamics! Today, we are going to dissect a classic problem involving a cyclic process on a $p-V$ (pressure-volume) diagram. This problem is a beautiful blend of graphical interpretation and fundamental thermodynamic principles.

Imagine you are looking at the heartbeat of a heat engine. The $p-V$ diagram provided in the question represents exactly that—a cyclic process where a diatomic ideal gas undergoes a series of transformations. The cycle consists of four distinct paths: $A\to B$, $B\to C$, $C\to D$, and $D\to A$.

However, the question is highly specific. It doesn't ask for the net work done in the entire cycle; instead, it directs our focus to just one segment: the adiabatic process $CD$.

This is a crucial skill in physics—filtering out the noise and zeroing in on exactly what is asked. Let's look closely at the path from $C$ to $D$. The arrow clearly indicates the direction of the process.

At the starting point $C$, the gas is at a lower pressure but a higher volume. By reading the axes, we can extract the exact coordinates: Initial pressure, $p_i=100{\text{ N/m}}^2$ Initial volume, $V_i=4{\text{ m}}^3$

As the gas moves to point $D$, it is compressed. The volume decreases, and the pressure shoots up. The coordinates for the final state are: Final pressure, $p_f=200{\text{ N/m}}^2$ Final volume, $V_f=3{\text{ m}}^3$

Now that we have our initial and final states, we need the right tool for the job. The process is explicitly stated to be adiabatic. In an adiabatic process, the gas is perfectly insulated from its surroundings—no heat enters, and no heat leaves ($Q=0$).

The work done $W$ by an ideal gas during an adiabatic process is given by the elegant formula:

Here, $\gamma$ (gamma) is the ratio of specific heats ($C_p/C_v$). The problem generously provides us with $\gamma =1.4$, which is the standard value for a diatomic gas like oxygen or nitrogen at room temperature.

Let's bring our extracted values into the master equation. This is where we must be careful with our arithmetic.

Substituting the values, we get:

Let's break this down step-by-step. First, we evaluate the numerator, which represents the difference in the product of pressure and volume:

Subtracting the initial from the final gives us:

Now, let's look at the denominator. This is where many students make a silly mistake.

Notice the negative sign! It is absolutely critical. Now, we divide the numerator by the denominator:

We arrived at a final answer of $-500\text{ J}$. But what does that negative sign actually mean in the physical world?

In thermodynamics, the sign convention is paramount. Positive work means the gas is expanding and doing work on its surroundings (like pushing a piston outward). Negative work means the surroundings are doing work on the gas, compressing it into a smaller volume.

Looking back at our $p-V$ diagram, the volume decreased from $4{\text{ m}}^3$ to $3{\text{ m}}^3$. It was a compression! Therefore, the work done by the gas must be negative, which perfectly aligns with our calculated result.

Furthermore, since this is an adiabatic compression ($Q=0$), the First Law of Thermodynamics ($\Delta U=Q-W$) tells us that $\Delta U=-(-500)=+500\text{ J}$. The work done on the gas goes entirely into increasing its internal energy, which means the gas heats up significantly during this process.

And there you have it! By carefully reading the graph, selecting the correct formula, and respecting the sign conventions, we've not only solved the problem but also understood the beautiful physics behind it.