Animated Solution for Chemistry - Chemical Thermodynamics: The combination of plots which does not represent isothermal expansion of an ideal gas is

Imagine you are holding a perfectly sealed syringe filled with an ideal gas. If you slowly pull the plunger out while keeping the syringe in a massive water bath, the temperature of the gas remains perfectly constant. This is the essence of an isothermal expansion.

In the world of thermodynamics, the word "isothermal" is a massive cheat code. It immediately tells us that the change in temperature, $\Delta T$, is zero.

For an ideal gas, the relationship between pressure, volume, and temperature is governed by the legendary equation $p{V}_m=RT$. Since $T$ is constant, the entire right side of the equation becomes a constant. Let's call it $K$.

Therefore, we have our master equation for this problem: $p{V}_m=K$. Now, let's put on our detective hats and interrogate each graph to see if it obeys this law.

Let's start with Graph (a). Here, we are plotting pressure $p$ on the y-axis against $1/V_m$ on the x-axis.

If we rearrange our master equation, we get $p=K\left(\frac{1}{V_m}\right)$. Notice the structure? It perfectly mirrors the equation of a straight line, $y=mx$, where our slope $m$ is $K$.

Since there is no y-intercept, this line must pass straight through the origin. Graph (a) shows exactly this, making it a correct representation.

Now, shift your focus to Graph (b). This time, $p$ is plotted directly against $V_m$.

Our equation is $p=\frac{K}{V_m}$. This tells us that pressure and volume are inversely proportional. As volume increases, pressure must decrease, tracing out a beautiful curve known as a rectangular hyperbola.

However, Graph (b) audaciously displays a straight line passing through the origin, implying $p\propto V_m$. This is a blatant violation of Boyle's Law! Thus, Graph (b) is incorrect.

Moving on to Graph (c), we see $p{V}_m$ plotted against $p$.

We already established that $p{V}_m=K$. This means that no matter how much you crank up the pressure $p$, the product $p{V}_m$ will stubbornly remain at the value $K$.

Geometrically, this is represented by a perfectly horizontal line parallel to the pressure axis. Graph (c) depicts this flawlessly, so it is correct.

Finally, let's examine Graph (d), which plots internal energy $U$ against volume $V_m$. This is where many students fall into a trap.

You might think, "If the gas is expanding, surely its internal energy is changing?" But remember, we are dealing with an ideal gas.

In an ideal gas, there are no intermolecular forces of attraction or repulsion. The internal energy is purely kinetic, and kinetic energy is a direct measure of temperature. Mathematically, $U=f(T)$.

Since our process is isothermal, the temperature is locked in place. If $T$ is constant, then $U$ must also be absolutely constant, regardless of how much the volume changes.

The graph should be a horizontal line. Instead, Graph (d) shows internal energy increasing with volume. This is physically impossible for an ideal gas in an isothermal process. Therefore, Graph (d) is incorrect.

Our investigation reveals that Graphs (b) and (d) fail to represent the isothermal expansion of an ideal gas.

When tackling such graphical questions in JEE, never rely on just the visual shape. Always write down the mathematical relationship between the y-axis and x-axis variables, and let the math dictate the geometry!