Animated Solution for Physics - Thermodynamics: An ideal gas expands isothermally from a volume $V_1$ to $V_2$ and then compressed to original volume $V_1$ adiabatically. Initial pressure is $p_1$ and final pressure is $p_3$. The total work done is $W$. Then,

Imagine you are observing a gas trapped inside a cylinder with a movable piston. The problem describes a two-step thermodynamic journey, and the best way to understand it is by drawing a $p-V$ (pressure-volume) diagram.

We start at an initial state, let's call it $A$, where the gas has a volume $V_1$ and a pressure $p_1$.

In the first step, the gas expands isothermally to a new, larger volume $V_2$. During an isothermal process, the temperature remains constant. As the gas expands, its pressure drops. On our $p-V$ diagram, this is represented by a smooth, downward-sloping curve from state $A$ to a new state $B$.

Isothermal vs. Adiabatic

Now comes the crucial second step. The gas is compressed back to its original volume $V_1$, but this time, the process is adiabatic. This means no heat is allowed to enter or leave the gas ($Q=0$).

When you compress a gas adiabatically, you are doing work on it, and all that work goes directly into increasing its internal energy. As a result, the gas heats up. Because the temperature is rising during the compression, the pressure increases much faster than it would if the temperature were kept constant.

Mathematically, the slope of an adiabatic curve is $\gamma$ times steeper than the slope of an isothermal curve at any given point (where $\gamma >1$). Therefore, as we trace the path from state $B$ back to volume $V_1$, the adiabatic curve rises steeply, landing at a final state $C$.

Because the adiabatic curve is strictly steeper than the isothermal curve, state $C$ will always lie vertically above state $A$. This immediately tells us our first key result: the final pressure $p_3$ is strictly greater than the initial pressure $p_1$.

Next, we need to determine the sign of the total work done, $W$. In a $p-V$ diagram, the work done during a process is simply the area under the curve.

During the isothermal expansion from $A$ to $B$, the volume increases. The gas does positive work on the surroundings. This is the area under the curve $AB$.

During the adiabatic compression from $B$ to $C$, the volume decreases. Work is done on the gas by the surroundings, making this work negative. This is the area under the curve $BC$.

Now, look at the two areas. Because the adiabatic curve $BC$ lies above the isothermal curve $AB$, the area under $BC$ is significantly larger than the area under $AB$. This means the magnitude of the negative work is greater than the magnitude of the positive work.

The total work done in the entire process is the algebraic sum of the work done in each step:

Since the negative term dominates, the net total work $W$ must be less than zero.

Combining our two findings, we conclude that the final pressure is greater than the initial pressure, and the total work done is negative. This perfectly matches option (c).