Animated Solution for Physics - Thermodynamics: An engine operates by taking a monatomic ideal gas through the cycle shown in the figure. The percentage efficiency of the engine is close to ......... .

The quest for the perfect engine has driven physicists and engineers for centuries. While we can't build a perfectly efficient engine due to the relentless laws of thermodynamics, we can certainly calculate the efficiency of idealized cycles. In this problem, we are presented with a classic thermodynamic puzzle: a monatomic ideal gas undergoing a rectangular cycle on a $p-V$ diagram. Our mission is to uncover its percentage efficiency.

Let's embark on this journey step-by-step, breaking down the physics behind the graph.

We are given a $p-V$ (pressure-volume) diagram depicting a cyclic process $ABCDA$. The gas inside our theoretical engine is a monatomic ideal gas. This is a crucial piece of information because it dictates the internal energy and heat capacities of the gas. For a monatomic gas, the translational kinetic energy is the only form of internal energy, giving it 3 degrees of freedom.

Consequently, the molar heat capacity at constant volume is $C_V=\frac{3}{2}R$, and the molar heat capacity at constant pressure is $C_p=\frac{5}{2}R$.

The cycle consists of four distinct processes: 1. $A\to B$: Isochoric (constant volume) heating. 2. $B\to C$: Isobaric (constant pressure) expansion. 3. $C\to D$: Isochoric cooling. 4. $D\to A$: Isobaric compression.

The efficiency $\eta$ of any heat engine is defined as the ratio of the net work output to the total heat input. Mathematically, it is expressed as:

To express this as a percentage, we simply multiply by 100. Our roadmap is clear: we need to calculate the net work done $W$ and the total heat absorbed $Q_{\text{in}}$.

In thermodynamics, the work done by a gas during a small expansion $dV$ is $dW=pdV$. For a complete cycle on a $p-V$ diagram, the net work done is beautifully represented by the area enclosed by the loop.

Looking at our rectangular cycle $ABCDA$, the width of the rectangle represents the change in volume, and the height represents the change in pressure.

The area of this rectangle gives us the net work done:

So, our engine performs $2p_0{V}_0$ amount of work in one complete cycle.

Next, we must determine $Q_{\text{in}}$, the total heat absorbed by the gas. Heat is absorbed when the temperature of the gas increases. According to the ideal gas law, $pV=nRT$, the temperature is directly proportional to the product of pressure and volume ($T\propto pV$).

Let's evaluate the $pV$ product at each state: State A: $p_A{V}_A=p_0{V}_0$ State B: $p_B{V}_B=(3p_0)(V_0)=3p_0{V}_0$ State C: $p_C{V}_C=(3p_0)(2V_0)=6p_0{V}_0$ State D: $p_D{V}_D=(p_0)(2V_0)=2p_0{V}_0$

Notice how the $pV$ product (and therefore the temperature) strictly increases from A to B, and from B to C. This means heat is absorbed during processes $A\to B$ and $B\to C$.

Conversely, the temperature drops from C to D and D to A, meaning heat is rejected to the surroundings. For efficiency, we only care about the heat we have to "pay for", which is the heat absorbed.

Let's calculate the exact amount of heat absorbed in these two processes.

1. Process $A\to B$ (Isochoric Heating) Since the volume is constant, the work done is zero. All the heat added goes into increasing the internal energy. We use the formula for heat at constant volume:

Using the ideal gas law ($nRT=pV$), we can elegantly rewrite this in terms of pressure and volume:

Substituting our values:

2. Process $B\to C$ (Isobaric Expansion) Here, the pressure is constant. The gas does work while its internal energy increases. We use the formula for heat at constant pressure:

Again, substituting $nRT$ with $pV$:

Plugging in the values:

Total Heat Absorbed ($Q_{\text{in}}$) We simply add the heat absorbed in both processes:

We now have all the pieces of the puzzle. Let's substitute $W$ and $Q_{\text{in}}$ back into our master efficiency equation.

The $p_0{V}_0$ terms cancel out perfectly, demonstrating that the efficiency is independent of the specific initial pressure and volume, depending only on the geometry of the cycle and the nature of the gas.

To find the percentage efficiency, we multiply by 100:

Rounding to the nearest integer, as requested by the problem format, we get 19%.

This problem is a beautiful synthesis of graphical analysis and thermodynamic principles. By understanding how to extract work from the area of a $p-V$ loop and how to track heat flow using specific heat capacities, we successfully determined the engine's efficiency.

As a thought experiment, consider how this result would change if the working fluid were a diatomic gas like Nitrogen or Oxygen. The degrees of freedom would shift, altering $C_V$ and $C_p$, and ultimately leading to a different efficiency. Physics is all about these interconnected relationships!