In the reported figure, heat energy absorbed by a system in going through a cyclic process is ......

Q: Why do we use the formula for the area of an ellipse ($\pi r_1 r_2$) instead of a circle ($\pi r^2$)?

Even though the shape looks like a perfect circle on the graph, the x and y axes represent different physical quantities (Volume and Pressure) with different units. Therefore, the "radii" in the horizontal and vertical directions have different physical meanings and values, making it an ellipse in physical space.

Q: How do I know if the work done is positive or negative?

In a $p-V$ diagram, a clockwise cycle means the gas expands at higher pressures and compresses at lower pressures, resulting in net positive work done by the gas. A counter-clockwise cycle would mean net negative work.

Q: Why is the change in internal energy zero?

Internal energy is a state function, meaning it only depends on the current state of the system (its temperature, pressure, volume), not how it got there. In a cyclic process, the initial and final states are the same, so the net change is zero.

Q: How do we convert Liters to cubic meters?

One Liter is equal to one cubic decimeter ($1 \text{ dm}^3$). Since $1 \text{ m} = 10 \text{ dm}$, $1 \text{ m}^3 = 1000 \text{ dm}^3 = 1000 \text{ L}$. Therefore, $1 \text{ L} = 10^{-3} \text{ m}^3$.

Q: What if the cycle was not a perfect circle or ellipse, but a weird shape?

The principle remains exactly the same: the work done is the area enclosed. For irregular shapes, you would need to use integration ($W = \int p \, dV$) to find the exact area.

Question

In the reported figure, heat energy absorbed by a system in going through a cyclic process is ...... $\pi$J.

Sigma Percentile · Accepted Answer

The Geometry of Thermodynamics: Decoding the Cyclic Process

Thermodynamics is not just about abstract equations; it is deeply visual. When we plot the state of a gas on a $p-V$ (pressure-volume) diagram, the shapes we draw directly translate to physical energy.

In this problem, we are presented with a cyclic process that looks like a perfect circle. Let's embark on a journey to decode what this shape tells us about the heat absorbed by the system.

### The First Law in a Loop

Imagine a gas expanding and compressing, eventually returning to the exact same state it started from. This is the essence of a cyclic process.

Because the initial and final states are identical, the temperature, pressure, and volume are unchanged at the end of the cycle. Consequently, the change in internal energy, $\Delta U$, must be zero.

**$\Delta U = 0$**

Now, we bring in the heavy artillery: the First Law of Thermodynamics. It states that the heat supplied to a system equals the change in its internal energy plus the work done by it.

**$\Delta Q = \Delta U + W$**

Since $\Delta U$ vanishes, the equation simplifies beautifully. The heat absorbed by the system is entirely converted into work.

**$\Delta Q = W$**

### The Illusion of the Circle

To find the heat absorbed, we just need to find the work done. On a $p-V$ diagram, the work done is simply the area enclosed by the cycle.

Looking at the graph, the cycle traces a circle. However, we must be incredibly careful here. While it looks like a circle geometrically, the x-axis and y-axis represent entirely different physical quantities with different units.

In the physical world, this shape is an ellipse. The area of an ellipse is given by:

**$	ext{Area} = \pi r_1 r_2$**

Here, $r_1$ and $r_2$ are the semi-major and semi-minor axes, representing the "radius" in the pressure dimension and the volume dimension, respectively.

### The Trap of Units

This is where many students make a fatal error. We cannot simply multiply the numbers on the axes; we must convert them into standard SI units (Pascals and cubic meters) to get the energy in Joules.

Let's extract the vertical radius, $r_1$, which represents the change in pressure. The pressure goes from $20 	ext{ kPa}$ to $40 	ext{ kPa}$, so the diameter is $20 	ext{ kPa}$. The radius is half of that:

**$r_1 = 10 	ext{ kPa} = 10 	imes 10^3 	ext{ Pa}$**

Next, we extract the horizontal radius, $r_2$, which represents the change in volume. The volume goes from $20 	ext{ L}$ to $40 	ext{ L}$, so the radius is $10 	ext{ L}$. We must convert Liters to cubic meters:

**$r_2 = 10 	ext{ L} = 10 	imes 10^{-3} 	ext{ m}^3$**

### Final Calculation

With our units perfectly aligned, we are ready for the final execution. We substitute $r_1$ and $r_2$ into our area formula:

**$W = \pi 	imes (10 	imes 10^3) 	imes (10 	imes 10^{-3})$**

Notice the sheer elegance of the numbers. The $10^3$ and $10^{-3}$ perfectly cancel each other out!

**$W = \pi 	imes 10 	imes 10 = 100\pi 	ext{ J}$**

Since the cycle is clockwise, the work done is positive. And as we established earlier, the heat absorbed equals the work done.

**$\Delta Q = 100\pi 	ext{ J}$**

The question asks for the integer value that precedes $\pi 	ext{ J}$. Therefore, our final answer is simply 100.

Animated Solution for Physics - Thermodynamics: In the reported figure, heat energy absorbed by a system in going through a cyclic process is ...... $π$ J.

Enter Numerical Value:

Visualized Solution

$C y c l i c P r ocess & I n t er na l E n er g y$

$F i r s t L a w o f T h er m o d y nami cs$

$W or k D o n e a s A r e a$

$A r e a o f t h e E l l i p se$

$E x t r a c t in g R a d ii$

$C a l c u l a t in g W or k D o n e$

$F ina l A n s w er$

The Sigma Insight: First Law of Thermodynamics

The First Law in a Loop

The Illusion of the Circle

The Trap of Units

Final Calculation

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Animated Solution for Physics - Thermodynamics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In the reported figure, heat energy absorbed by a system in going through a cyclic process is ...... πJ.