Animated Solution for Physics - Thermodynamics: The $p$-$T$ diagram for an ideal gas is shown in the figure, where $AC$ is an adiabatic process, find the corresponding $p$-$V$ diagram

Let's embark on a thrilling journey to decode the given $p$-$T$ diagram. At first glance, it looks like a standard thermodynamic cycle, but as we dig deeper, we will uncover a fascinating physical contradiction.

First, let's analyze the process from $A$ to $B$. On the $p$-$T$ diagram, this is a vertical line pointing downwards. This indicates that the temperature $T$ remains constant while the pressure $p$ decreases. By the ideal gas law, $pV=nRT$, if $T$ is constant and $p$ drops, the volume $V$ must increase. This is an isothermal expansion, meaning $V_B>V_A$.

Next, we look at the process from $B$ to $C$. This is a horizontal line pointing to the right, meaning the pressure $p$ is constant while the temperature $T$ increases. According to Charles's Law, at constant pressure, volume is directly proportional to temperature ($V\propto T$). Therefore, as the gas heats up, it expands. This is an isobaric expansion, which tells us that $V_C>V_B$.

Finally, the problem states that the process from $C$ back to $A$ is an adiabatic process. In an adiabatic process, no heat is exchanged, and the relationship between pressure and volume is governed by $p{V}^{\gamma }=\text{constant}$. Since the pressure increases from $C$ to $A$, the volume must decrease.

Now, let's visualize this on a $p$-$V$ diagram. At point $A$, two curves intersect: the isothermal curve $AB$ and the adiabatic curve $CA$. We know a fundamental rule of thermodynamics: an adiabatic curve is always steeper than an isothermal curve passing through the same point. Mathematically, the slope of an adiabatic curve is $\gamma$ times the slope of an isothermal curve, and since $\gamma >1$, it drops faster.

Because the adiabatic curve $CA$ is steeper than the isothermal curve $AB$, the adiabatic curve must lie to the left of the isothermal curve for lower pressures. This implies that for any given pressure below $p_A$, the volume on the adiabatic curve must be strictly less than the volume on the isothermal curve.

Since points $B$ and $C$ are at the same pressure ($p_B=p_C$), the volume at $C$ (on the adiabatic curve) must be less than the volume at $B$ (on the isothermal curve). So, the slope condition demands that $V_C<V_B$.

But wait! Our initial analysis of the $p$-$T$ diagram clearly proved that $V_C>V_B$ due to the isobaric expansion.

These two conditions are mutually exclusive. It is physically impossible for process $CA$ to be adiabatic while simultaneously satisfying the given $p$-$T$ diagram. The question itself contains a fundamental flaw, which is why none of the provided options can be correct!