The magnitude of work done by a gas that undergoes a reversible expansion along the path ABC shown i

Q: Why do we calculate the area under the curve to find work done?

Work is defined as the integral of pressure with respect to volume ($W = \int P dV$). Geometrically, a definite integral represents the area under the curve on a P-V diagram.

Q: What if the path was a curve instead of a straight line?

If the path was a curve, we couldn't use the simple trapezium formula. We would need the exact mathematical function of the curve to integrate it, or we would have to estimate the area by counting grid squares.

Q: Why is the work done positive here?

The volume of the gas is increasing from $4 \text{ m}^3$ to $10 \text{ m}^3$. This is an expansion process. When a gas expands, it does work on the surroundings, which is conventionally taken as positive work done *by* the gas.

Q: Does the point B affect the calculation?

In this specific problem, the path from A to C is a single straight line, and B is just a point on that line. The area under the entire line segment AC is what matters. If the path was two distinct line segments (A to B, then B to C) with different slopes, we would calculate the area of two separate shapes and add them.

Q: What does the origin (2, 2) mean in this graph?

It simply means the axes intersect at $P=2$ and $V=2$ instead of $(0,0)$. This is a common way to save space on a graph when the relevant data points are far from zero. You must read the coordinates relative to the axis ticks, not just by counting grid boxes from the corner.

Question

The magnitude of work done by a gas that undergoes a reversible expansion along the path ABC shown in the figure is ……… .

Sigma Percentile · Accepted Answer

### The Visual Anchor

When dealing with thermodynamic processes, a Pressure-Volume (P-V) graph is your best friend. It translates abstract gas behaviors into tangible geometry. In this problem, we are asked to find the magnitude of work done by a gas as it undergoes a reversible expansion along a specific path, denoted as ABC.

The first step is to carefully observe the graph. The path ABC is a straight line connecting point A to point C. The fundamental principle we rely on here is that the **work done by a gas during a reversible process is equal to the area under the P-V curve**, projected down to the volume axis.

### Decoding the Graph

Before we calculate anything, we must extract the exact coordinates from the graph. This requires careful attention to the axes. Notice that the origin of this graph is not $(0,0)$, but rather $(2,2)$.

Looking at point A, we trace a horizontal line to the y-axis to find the initial pressure, $P_A = 10 	ext{ Pa}$. Tracing a vertical line down to the x-axis gives us the initial volume, $V_A = 4 	ext{ m}^3$.

Similarly, for point C, the final state of the gas, we find the pressure $P_C = 6 	ext{ Pa}$ and the volume $V_C = 10 	ext{ m}^3$. The gas is expanding because the volume is increasing.

### The Mathematics of the Trapezium

Now, let's look at the shape formed by the area under the line segment AC. It is bounded by the vertical lines at $V=4$ and $V=10$, the horizontal axis, and the slanted line AC. This shape is a **trapezium**.

The formula for the area of a trapezium is:

$$ 	ext{Area} = \frac{1}{2} 	imes (	ext{Sum of parallel sides}) 	imes 	ext{Height} $$

In the context of our P-V graph, the parallel sides are the vertical pressure lines at A and C, and the height is the horizontal change in volume.

$$ |W| = \frac{1}{2} 	imes (P_A + P_C) 	imes (V_C - V_A) $$

### The Final Calculation

Substituting the values we extracted from the graph into our formula:

$$ |W| = \frac{1}{2} 	imes (10 + 6) 	imes (10 - 4) $$

First, we evaluate the terms inside the parentheses. The sum of the pressures is $16 	ext{ Pa}$, and the change in volume is $6 	ext{ m}^3$.

$$ |W| = \frac{1}{2} 	imes 16 	imes 6 $$

Multiplying these together, half of $16$ is $8$, and $8 	imes 6$ gives us $48$.

$$ |W| = 48 	ext{ J} $$

Therefore, the magnitude of the work done by the gas during this expansion is exactly **48 Joules**. Because it is an expansion, the actual work done by the gas is positive, but since the question only asked for the magnitude, $48$ is our final answer.

Animated Solution for Chemistry - Chemical Thermodynamics: The magnitude of work done by a gas that undergoes a reversible expansion along the path ABC shown in the figure is ……… .

Enter Numerical Value:

Visualized Solution

$∣ W ∣ = ?$ \text{ for path ABC}

$∣ W ∣ = Area under P-V curve$

\text{Shape: Trapezium}

\text{Extracting Values}

\text{Substitution}

\text{Calculation}

\text{Final Answer}

\text{The Way Forward}

The Sigma Insight: First Law of Thermodynamics

The Visual Anchor

Decoding the Graph

The Mathematics of the Trapezium

The Final Calculation

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