Animated Solution for Physics - Thermodynamics: Helium gas goes through a cycle ABCDA (consisting of two isochoric and isobaric lines) as shown in figure. Efficiency of this cycle is nearly (Assume the gas to be close to ideal gas)

Imagine you are looking at the heartbeat of an engine. The $p-V$ diagram in front of us represents a thermodynamic cycle, specifically the cycle $ABCDA$.

The working substance is Helium, which is a monoatomic gas. This is a critical piece of information because it dictates the specific heat capacities we will use later.

For a monoatomic gas, the molar specific heat at constant volume is $C_V=\frac{3}{2}R$, and at constant pressure, it is $C_p=\frac{5}{2}R$.

Our goal is to find the efficiency of this cycle. But what exactly is efficiency?

In thermodynamics, efficiency $\eta$ is the ratio of the net work done by the engine to the total heat supplied to it.

It is a measure of how well the engine converts the fuel (heat) into useful output (work).

Let's start with the numerator: the net work done $W$.

In a $p-V$ diagram, the net work done in a cyclic process is beautifully simple. It is exactly equal to the area enclosed by the cycle.

Here, our cycle forms a perfect rectangle. The width of this rectangle is the change in volume, and the height is the change in pressure.

This is the total useful energy we extract from one complete cycle.

Now, we need to find the denominator: the total heat supplied $Q_{\text{in}}$.

Heat is supplied to the gas only when its temperature increases. We need to identify which processes involve heating.

In process $A\to B$, the volume is constant but the pressure increases. According to the ideal gas law ($pV=nRT$), an increase in pressure at constant volume means the temperature must rise. Thus, heat is absorbed here.

Similarly, in process $B\to C$, the pressure is constant but the volume increases. Again, the temperature must rise, meaning heat is absorbed.

In the remaining processes ($C\to D$ and $D\to A$), the temperature drops, meaning heat is rejected. We do not include rejected heat in $Q_{\text{in}}$.

Let's calculate the exact amount of heat supplied during the isochoric process $A\to B$.

Since the volume is constant, we use the formula for heat at constant volume:

Using the ideal gas law, we can rewrite $nR\Delta T$ as $\Delta (pV)$.

Substituting the coordinates from our diagram:

Next, we calculate the heat supplied during the isobaric process $B\to C$.

Since the pressure is constant, we use the formula for heat at constant pressure:

Again, rewriting this in terms of pressure and volume:

Substituting the coordinates:

We now have all the pieces of the puzzle. Let's find the total heat supplied by adding the heat from both processes.

Finally, we plug the work done and the total heat supplied back into our master equation for efficiency.

The $p_0{V}_0$ terms cancel out beautifully, leaving us with a pure number.

Converting this to a percentage, we get our final answer: $15.4\%$.

Before we wrap up, let's do a quick thought experiment. What if the gas inside the engine was diatomic, like Oxygen or Nitrogen?

A diatomic gas has more degrees of freedom, meaning its specific heat capacities are higher ($C_V=\frac{5}{2}R$ and $C_p=\frac{7}{2}R$).

This means the gas would require more heat to achieve the same changes in pressure and volume. Since the work done (the area) remains the same, a larger heat input would result in a lower overall efficiency. Always think about how the physical properties of the system govern its behavior!