Work Done on P-V Diagrams

Hello! Welcome to the third lesson in our module on the First Law of Thermodynamics.

In our last session, we demystified the different sign conventions for work and established a robust method to translate between them. We concluded that no matter the convention, the physical reality of a process remains the same. Our key rule was $W_{on the system} = - W_{by the system}$ .

Today, we move from interpreting the sign of work to calculating its magnitude. You'll learn how the abstract concept of thermodynamic work is visually and mathematically represented using Pressure-Volume (P-V) diagrams. This is a fundamental skill for solving a vast range of thermodynamics problems in JEE Advanced. By the end of this lesson, you will be able to calculate the work done for various thermodynamic processes directly from their P-V graphs.

1. From Force and Displacement to Pressure and Volume

Let's begin from first principles. You know from mechanics that the work done by a force $F$ over a small displacement $d x$ is $d W = F \cdot d x$ . How does this apply to a gas in a cylinder expanding against a piston?

To understand this fundamental connection, let's watch a clear explanation from Prof. H.C. Verma. He breaks down how the mechanical definition of work translates directly to the context of a gas.

First Law of Thermodynamics : sign convention for WORK

This video explains how the universal definition of work (force times displacement) is applied to a gas in a piston-cylinder system. This helps to ground the thermodynamic work formula in fundamental physics.

Watch the segment from 00:46 to 08:29. Focus on how the sign of work is determined by the relative direction of the force exerted by the gas and the displacement of the piston. Note that this definition is independent of any convention.

As Prof. Verma explained, the analysis is straightforward:

The force exerted by the gas on the piston of area $A$ is $F = P_{gas} \times A$ .
If the piston moves outwards by a small distance $d x$ , the displacement is in the same direction as the force. The work done by the gas is $d W = F d x = (P_{gas} A) d x$ .
Since $A \cdot d x$ is the small change in volume $d V$ , we arrive at the cornerstone equation for mechanical work in thermodynamics:
$d W = P d V$
For a finite process from an initial volume $V_{i}$ to a final volume $V_{f}$ , the total work done by the gas is the integral:
$W_{by} = \int_{V_{i}}^{V_{f}} P d V$

Given your background in calculus, you will immediately recognize this integral as the area under the curve on a Pressure-Volume (P-V) diagram. This graphical interpretation is incredibly powerful.

P-V Diagrams and Work Done — These diagrams illustrate how the work done (the shaded area) is calculated for various processes. For expansion (a, b, c), where volume increases, the work done by the system is positive. For compression (e), where volume decreases, the work done by the system is negative. For a cyclic process (f), the net work is the area enclosed by the loop.

The sign of the work done by the gas is naturally determined by the integration limits:

Expansion: $V_{f} > V_{i}$ , so $d V > 0$ . The area is swept from left to right, and $W_{by}$ is positive.
Compression: $V_{f} < V_{i}$ , so $d V < 0$ . The area is swept from right to left, and $W_{by}$ is negative. This makes perfect sense: negative work done by the gas is equivalent to positive work done on the gas.

2. Calculating Work for Standard Thermodynamic Processes

Now, let's apply the formula $W = \int P d V$ to the standard processes you'll encounter in JEE problems. The following video provides a clear, step-by-step guide for each case. We will use the physics convention, where $W$ denotes work done by the system.

PV Diagrams, How To Calculate The Work Done By a Gas, Thermodynamics & Physics

This video from The Organic Chemistry Tutor demonstrates how to calculate work for isobaric, isochoric, isothermal, and adiabatic processes. It provides the formulas and works through numerical examples for each.

Watch the first three examples from 00:00 to 04:47. Pay attention to how the area under the P-V curve corresponds to the work calculation for each process.

Let's summarize and formalize the results from the video.

a) Isochoric Process (Constant Volume)

This is the simplest case.

On a P-V diagram, this is a vertical line.
Since $V$ is constant, $d V = 0$ .
Therefore, $W = \int P d V = 0$ .
There is no area under a vertical line, confirming that no work is done.

b) Isobaric Process (Constant Pressure)

On a P-V diagram, this is a horizontal line. The area underneath is a simple rectangle.
Since $P$ is constant, we can take it out of the integral:
$W = P \int_{V_{i}}^{V_{f}} d V = P (V_{f} - V_{i}) = PΔV$

c) Isothermal Process (Constant Temperature)

On a P-V diagram, this is a hyperbolic curve (for an ideal gas).
To evaluate the integral, we must express $P$ as a function of $V$ . For an ideal gas, $P V = n R T$ , so $P = \frac{n R T}{V}$ .
The work done is the area under this hyperbola:
$W = \int_{V_{i}}^{V_{f}} \frac{n R T}{V} d V$
Since $n, R, T$ are constant, they come out of the integral:
$W = n R T \int_{V_{i}}^{V_{f}} \frac{d V}{V} = nRT ln (\frac{V _{f}}{V _{i}})$

Test your understanding!

A cylinder contains 2 moles of an ideal gas at 300 K. It undergoes two different expansions from 10 L to 20 L.

Process 1: An isobaric expansion at a constant pressure of 2 atm.
Process 2: An isothermal expansion.

Calculate the work done by the gas in each case. Use $R = 8.314 J/(mol\cdotK)$ and $1 L\cdotatm = 101.3 J$ . Which process does more work?

Show answer

Process 1 (Isobaric):

$W = P Δ V$
$P = 2$ atm
$Δ V = V_{f} - V_{i} = 20 L - 10 L = 10 L$
$W = (2 atm) \times (10 L) = 20 L\cdotatm$
Converting to Joules: $W = 20 \times 101.3 J = 2026 J$ .

Process 2 (Isothermal):

$W = n R T ln (V_{f} / V_{i})$
$n = 2$ mol, $R = 8.314$ J/(mol·K), $T = 300$ K
$V_{f} / V_{i} = 20/10 = 2$
$W = (2) \times (8.314) \times (300) \times ln (2)$
$W \approx 4988.4 \times 0.693 \approx 3457 J$ .

Conclusion: The isothermal expansion does significantly more work than the isobaric expansion in this scenario. This makes sense graphically: the isothermal curve $P = n R T / V$ would lie above the horizontal line of the isobaric process for the entire expansion (since the initial pressure for the isothermal process would be $P_{i} = n R T / V_{i} = 2 \times 0.0821 \times 300/10 \approx 4.9$ atm, which is greater than 2 atm).

d) Adiabatic Process (No Heat Exchange, Q=0)

The relationship for a reversible adiabatic process is $P V^{γ} = K$ , where $K$ is a constant and $γ = C_{P} / C_{V}$ .
The work integral is $W = \int_{V_{i}}^{V_{f}} K V^{- γ} d V$ . The integration gives:
$W = \frac{K}{1 - γ} [V^{1 - γ}]_{V_{i}}^{V_{f}} = \frac{K ( V _{f}^{1 - γ} - V _{i}^{1 - γ} )}{1 - γ}$
Using $K = P_{f} V_{f}^{γ} = P_{i} V_{i}^{γ}$ , we can substitute and simplify to a more useful form:
$W = \frac{P _{f} V _{f} - P _{i} V _{i}}{1 - γ}$
Finally, using the ideal gas law $P V = n R T$ , this becomes the most common formula for work in an adiabatic process:
$W = \frac{n R ( T _{f} - T _{i} )}{1 - γ} = \frac{nR ( T _{i} - T _{f} )}{γ - 1}$

3. Work in Cyclic Processes

Many JEE problems involve thermodynamic cycles, where a system undergoes several processes and returns to its initial state.

PV Diagrams, How To Calculate The Work Done By a Gas, Thermodynamics & Physics

Let's now look at cyclic processes. This part of the video explains the key rule for finding the net work done in a cycle and the sign convention associated with it.

Watch the segments from 07:09 to 09:03 and from 15:57 to 20:18. The first part introduces the concept for a clockwise cycle. The second part confirms the rule by calculating work for a counter-clockwise cycle, showing that the net work is negative.

For any cyclic process, the net work done is the sum of the work from each step. Graphically, this corresponds to the area enclosed by the cycle on the P-V diagram.

The sign of the net work follows a simple rule:

Clockwise Cycle: The work of expansion (top part of the loop) is greater than the work of compression (bottom part). The net work is positive ( $W_{n e t} > 0$ ). The system does net work on the surroundings. This is characteristic of a heat engine.
Counter-clockwise Cycle: The work of compression (top part) is greater in magnitude than the work of expansion (bottom part). The net work is negative ( $W_{n e t} < 0$ ). Net work is done on the system. This is characteristic of a refrigerator or heat pump.

4. Application to a JEE Advanced Problem

Let's see how these concepts are pieced together in a multi-step problem. The following resource from IIT Kanpur's SATHEE platform presents a typical advanced problem. We will not solve it entirely, but we'll focus on the work calculation part.

Advanced Thermodynamics Problem - SATHEE - IIT Kanpur

This document outlines an advanced thermodynamics problem involving a three-stage cycle. We will examine how the work is calculated for each stage using the formulas we've just learned.

Read the sections 'Stage 1: Isothermal Expansion', 'Stage 2: Adiabatic Expansion', and 'Stage 3: Isobaric Compression'. Note the formula used in each case to calculate work (W_1, W_2, and W_3). Also, look at the sign of W_3 (compression work). Finally, see how they are summed up in the 'Net work' calculation under 'PV Diagram Analysis'.

This problem perfectly illustrates our discussion:

Stage 1 (Isothermal Expansion): Work is positive and calculated using $W_{1} = n R T_{1} ln (V_{B} / V_{A})$ .
Stage 2 (Adiabatic Expansion): Work is positive and calculated using $W_{2} = \frac{n R ( T _{B} - T _{C} )}{γ - 1}$ .
Stage 3 (Isobaric Compression): Work is negative, as expected for compression. It's calculated as $W_{3} = P_{C} (V_{A} - V_{C})$ . Since $V_{A} < V_{C}$ , the result is negative.
Net Work: The total work $W_{n e t} = W_{1} + W_{2} + W_{3}$ is the algebraic sum of the work done in each stage.

Note that the problem resource also flags "Incorrect signs for work and heat" as a common mistake. By mastering the concepts from this and the previous lesson, you are directly addressing this frequent pitfall.

Conclusion

Today, we've connected the fundamental physics of work to its practical calculation in thermodynamics using P-V diagrams. This graphical method is not just a tool; it's a way of thinking that helps you visualize and solve complex problems.

Key Takeaways:

The work done by a gas during a volume change is $W_{by} = \int P d V$ , which is the area under the P-V curve.
Work is positive for expansion ( $V_{f} > V_{i}$ ) and negative for compression ( $V_{f} < V_{i}$ ) when using the physics convention (work done by the system).
You can now calculate work for isochoric ( $0$ ), isobaric ( $P Δ V$ ), isothermal ( $n R T ln (V_{f} / V_{i})$ ), and adiabatic ( $\frac{P _{f} V _{f} - P _{i} V _{i}}{1 - γ}$ ) processes.
For cyclic processes, the net work is the area enclosed by the loop: positive for clockwise cycles (engines) and negative for counter-clockwise cycles (refrigerators).

Preview of the Next Lesson:
We've covered $Q$ (heat signs), $W$ (work signs and calculation), and the relationship $Δ U = Q - W$ . The final missing piece is calculating the change in internal energy, $Δ U$ . In the next lesson, we will focus on the crucial formula $Δ U = n C_{v} Δ T$ for an ideal gas and explore why this simple equation is valid for any process, not just constant volume ones. This will equip us to solve for any of the three quantities in the First Law of Thermodynamics.

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