Real-Life Hook: The Heart is a Heat Engine!
Your heart works like a thermodynamic engine, completing about 100,000 cycles per day! Each heartbeat follows a cycle on a pressure-volume diagram:
- Filling (diastole): Volume increases at low pressure
- Contraction (systole): Pressure rises at nearly constant volume
- Ejection: Blood pushed out at high pressure
- Relaxation: Return to initial state
Just like the Carnot cycle or Otto cycle (in car engines), your heart’s PV diagram forms a closed loop. The area inside? That’s the work done per beat—literally the mechanical energy pumping blood through your body!
Understanding PV diagrams helps us analyze:
- Car engines (Otto, Diesel cycles)
- Refrigerators and ACs
- Steam turbines
- Any cyclic thermodynamic process
PV Diagrams: The Basics
What is a PV Diagram?
A PV diagram (Pressure-Volume diagram) plots:
- X-axis: Volume (V)
- Y-axis: Pressure (P)
- Curve: Shows how gas state changes during a process
Why PV Diagrams?
- Visual representation of thermodynamic processes
- Work done = Area under the curve
- Easy comparison of different processes
- Cyclic processes shown as closed loops
- State changes clearly visible
Work from PV Diagrams
Fundamental Principle
Work done BY gas:
$$W = \int_{V_1}^{V_2} P \, dV$$Graphically:
$$W = \text{Area under the P-V curve}$$Sign Convention
- Expansion ($V$ increases, move right): W > 0 (gas does work)
- Compression ($V$ decreases, move left): W < 0 (work done on gas)
Visual Rules
- Area above the curve (moving right) = Positive work
- Area below the curve (moving left) = Negative work (in magnitude)
- Net work in cycle = Area enclosed by loop
- Clockwise cycle = Net positive work (heat engine)
- Counter-clockwise cycle = Net negative work (refrigerator/heat pump)
The Four Basic Processes on PV Diagram
1. Isothermal Process
Curve: Rectangular hyperbola
$$PV = \text{constant}$$Shape: Smooth curve, decreasing P as V increases
Work:
$$W = nRT\ln\frac{V_2}{V_1}$$On diagram: Area under hyperbola
2. Adiabatic Process
Curve: Steeper hyperbola
$$PV^\gamma = \text{constant}$$Shape: Steeper than isothermal (γ > 1)
Work:
$$W = \frac{P_1V_1 - P_2V_2}{\gamma - 1}$$Key: Adiabatic slope = γ × Isothermal slope
3. Isobaric Process
Curve: Horizontal line
$$P = \text{constant}$$Work:
$$W = P(V_2 - V_1) = P\Delta V$$On diagram: Area of rectangle (height P, width ΔV)
4. Isochoric Process
Curve: Vertical line
$$V = \text{constant}$$Work:
$$W = 0$$(no area under vertical line!)
On diagram: No work (can’t integrate when V doesn’t change)
Comparing Processes on Same PV Diagram
All Starting from Same Point
If all four processes start at point A:
- Isochoric: Vertical line (up or down)
- Adiabatic: Steepest curve
- Isothermal: Moderate curve
- Isobaric: Horizontal line
Slope magnitude order:
$$\text{Isochoric (∞)} > \text{Adiabatic} > \text{Isothermal} > \text{Isobaric (0)}$$Expansion Work Comparison
For same final volume:
$$W_{\text{isobaric}} > W_{\text{isothermal}} > W_{\text{adiabatic}}$$Why? Higher pressure throughout → More area under curve → More work
Cyclic Processes
Definition
Cyclic process: System returns to initial state after completing a sequence of processes.
Key Properties
- ΔU = 0 (internal energy is state function)
- Q = W (from First Law: 0 = Q - W)
- Net heat absorbed = Net work done
- Closed loop on PV diagram
Work in Cyclic Process
Net work = Area enclosed by cycle
- Clockwise: W > 0 (heat engine)
- Counter-clockwise: W < 0 (refrigerator)
Clockwise vs Counter-Clockwise
Clockwise (Heat Engine):
- Absorbs heat at high temperature
- Rejects heat at low temperature
- Produces net positive work
- Example: Car engine, steam turbine
Counter-Clockwise (Refrigerator/Heat Pump):
- Absorbs heat at low temperature
- Rejects heat at high temperature
- Requires net work input
- Example: Refrigerator, AC
Calculating Work: Different Methods
Method 1: Area Counting (for Simple Shapes)
For cycles made of straight lines:
- Divide into rectangles, triangles, trapezoids
- Calculate each area
- Sum with proper signs
Example: Rectangle
$$W = \text{base} \times \text{height} = \Delta V \times \Delta P$$Method 2: Integration
For curved processes:
$$W = \int_{V_1}^{V_2} P \, dV$$Need to express P as function of V.
Method 3: Process-by-Process
For cycles:
- Calculate work for each leg
- Sum: $W_{\text{net}} = W_1 + W_2 + W_3 + ...$
- Use appropriate formula for each process
Method 4: Coordinates
For polygon on PV diagram:
$$W = \oint P \, dV$$Can use shoelace formula for polygon area.
Important PV Diagram Configurations
Type 1: Rectangular Cycle
Process:
- A→B: Isobaric expansion (high P)
- B→C: Isochoric cooling
- C→D: Isobaric compression (low P)
- D→A: Isochoric heating
Work:
$$W = P_{\text{high}}(V_B - V_A) - P_{\text{low}}(V_B - V_A)$$ $$W = (P_{\text{high}} - P_{\text{low}})(V_B - V_A)$$ $$W = \Delta P \times \Delta V$$Area of rectangle!
Type 2: Triangular Cycle
Three processes, forms triangle.
Work = Area of triangle:
$$W = \frac{1}{2} \times \text{base} \times \text{height}$$Type 3: Isothermal-Adiabatic Cycle (Carnot)
- Two isothermals at $T_H$ and $T_C$
- Two adiabatics connecting them
- Most efficient cycle
- Work = Area of curved quadrilateral
Multi-Process Cycles: Step-by-Step
Example: Cycle ABCA
Given:
- A (1 L, 10 atm)
- B (4 L, 10 atm)
- C (4 L, 2.5 atm)
- A→B: Isobaric
- B→C: Isochoric
- C→A: Straight line
Find net work.
Solution:
Process AB (Isobaric):
$$W_{AB} = P(V_B - V_A) = 10 \times 10^5 \times (4-1) \times 10^{-3}$$ $$W_{AB} = 3000 \text{ J}$$Process BC (Isochoric):
$$W_{BC} = 0$$Process CA (Straight line): Need to integrate or use geometry.
Straight line from C(4, 2.5) to A(1, 10):
Equation of line:
$$P - P_C = \frac{P_A - P_C}{V_A - V_C}(V - V_C)$$ $$P - 2.5 = \frac{10 - 2.5}{1 - 4}(V - 4)$$ $$P - 2.5 = \frac{7.5}{-3}(V - 4)$$ $$P = 2.5 - 2.5(V - 4) = 2.5 - 2.5V + 10$$ $$P = 12.5 - 2.5V$$Work CA:
$$W_{CA} = \int_4^1 P \, dV = \int_4^1 (12.5 - 2.5V) \, dV$$ $$W_{CA} = \left[12.5V - 1.25V^2\right]_4^1$$ $$W_{CA} = (12.5 - 1.25) - (50 - 20)$$ $$W_{CA} = 11.25 - 30 = -18.75 \text{ (atm·L)}$$Convert: $-18.75 \times 10^5 \times 10^{-3} = -1875$ J
Or geometrically: Area of trapezoid
$$W_{CA} = -\frac{1}{2}(P_A + P_C)(V_A - V_C)$$ $$W_{CA} = -\frac{1}{2}(10 + 2.5)(1 - 4) \times 10^5 \times 10^{-3}$$ $$W_{CA} = -\frac{1}{2}(12.5)(-3) \times 100$$ $$W_{CA} = -(-1875) = -1875 \text{ J}$$Wait, sign confusion. For compression (V decreasing), work is negative on the gas.
Correct approach: Going from C to A, V decreases (4 → 1), so moving left = compression = negative work.
Area of trapezoid (base1 = 10, base2 = 2.5, height = 3):
$$\text{Area} = \frac{1}{2}(10 + 2.5) \times 3 = \frac{1}{2} \times 12.5 \times 3 = 18.75 \text{ (atm·L)}$$Since compression: $W_{CA} = -1875$ J
Net work:
$$W_{\text{net}} = 3000 + 0 + (-1875) = 1125 \text{ J}$$Clockwise → Positive work ✓
Common PV Diagram Scenarios in JEE
Scenario 1: Free Expansion
Gas expands into vacuum.
- No external pressure: $P_{\text{ext}} = 0$
- $W = \int P_{\text{ext}} dV = 0$
- On PV diagram: Jump from one point to another (not quasi-static)
- Not a valid PV curve (process is irreversible)
Scenario 2: Two Gases Separated by Piston
When piston is free to move:
- Both gases do equal and opposite work
- $W_1 = -W_2$
- Net work on system (both gases) = 0 (if isolated)
Scenario 3: Gas in Cylinder with Weight on Piston
External pressure = atmospheric + weight/area
$$P_{\text{ext}} = P_0 + \frac{mg}{A}$$Process is isobaric at this pressure.
Practice Problems
Level 1: JEE Main Basics
Q1. A gas expands isobarically at 2 atm from 2 L to 5 L. Find work done.
Solution
$$W = P\Delta V = P(V_2 - V_1)$$ $$W = 2 \times 10^5 \times (5 - 2) \times 10^{-3}$$ $$W = 2 \times 10^5 \times 3 \times 10^{-3}$$ $$W = 600 \text{ J}$$On PV diagram: Rectangle with height 2 atm, width 3 L.
Q2. In a cyclic process ABCA, $W_{AB} = 100$ J, $W_{BC} = -50$ J. Find $W_{CA}$.
Solution
For cyclic process: $W_{\text{net}} = Q_{\text{net}}$
But we can directly use: If cycle is closed,
$$W_{\text{net}} = W_{AB} + W_{BC} + W_{CA}$$Actually, we need more information. Let me reconsider.
If the problem asks only for $W_{CA}$ and the cycle is given, we need heat information or to know it’s a specific type.
However, if question states “find $W_{CA}$ if net work is zero”:
$$W_{AB} + W_{BC} + W_{CA} = 0$$ $$100 + (-50) + W_{CA} = 0$$ $$W_{CA} = -50 \text{ J}$$If net work is 200 J:
$$100 + (-50) + W_{CA} = 200$$ $$W_{CA} = 150 \text{ J}$$Need additional constraint. Typically JEE provides cycle closure or heat values.
Q3. On a PV diagram, an isothermal and adiabatic curve intersect at point (2 L, 5 atm). Which curve is steeper at this point if γ = 1.4?
Solution
Isothermal slope:
$$\left|\frac{dP}{dV}\right|_{iso} = \frac{P}{V} = \frac{5}{2} = 2.5 \text{ atm/L}$$Adiabatic slope:
$$\left|\frac{dP}{dV}\right|_{ad} = \gamma\frac{P}{V} = 1.4 \times 2.5 = 3.5 \text{ atm/L}$$Adiabatic is steeper (as always, since γ > 1).
Level 2: JEE Main/Advanced
Q4. A gas undergoes cycle shown: A(1,3) → B(4,3) → C(4,1) → A. (Coordinates are (V in L, P in atm))
Find: (a) Work in each process (b) Net work (c) Direction (clockwise or counter-clockwise) (d) Is it heat engine or refrigerator?
Solution
Process AB: (1,3) → (4,3)
- Isobaric expansion at P = 3 atm $$W_{AB} = P\Delta V = 3 \times (4-1) = 9 \text{ atm·L}$$ $$W_{AB} = 9 \times 10^5 \times 10^{-3} = 900 \text{ J}$$
Process BC: (4,3) → (4,1)
- Isochoric (V = 4 L constant) $$W_{BC} = 0$$
Process CA: (4,1) → (1,3)
- Linear path, need to calculate
Area under line from V = 4 to V = 1 (going left, so negative):
Equation of line: From C(4,1) to A(1,3)
$$P - 1 = \frac{3-1}{1-4}(V - 4)$$ $$P - 1 = \frac{2}{-3}(V - 4)$$ $$P = 1 - \frac{2}{3}(V - 4) = 1 - \frac{2V}{3} + \frac{8}{3}$$ $$P = \frac{11}{3} - \frac{2V}{3}$$Work (compression, V from 4 to 1):
$$W_{CA} = \int_4^1 P \, dV = \int_4^1 \left(\frac{11}{3} - \frac{2V}{3}\right) dV$$Or use geometry: Trapezoid area
Bases: $P_C = 1$, $P_A = 3$; Height: $|V_C - V_A| = 3$
$$\text{Area} = \frac{1}{2}(1 + 3) \times 3 = 6 \text{ atm·L}$$Since going left (compression): $W_{CA} = -6 \times 100 = -600$ J
(b) Net work:
$$W_{\text{net}} = 900 + 0 + (-600) = 300 \text{ J}$$(c) Direction: Plot points: A(1,3) → B(4,3) → C(4,1) → A(1,3)
- Right (expansion)
- Down
- Left-up (back to start)
This is clockwise.
(d) Type: Clockwise + Positive work → Heat Engine
Q5. A gas expands from state A to state B along two different paths:
- Path 1: Directly along $P = 5 - V$ (P in atm, V in L)
- Path 2: A→C (isochoric) then C→B (isobaric)
A is at (1, 4) and B is at (4, 1). Find work done in each path.
Solution
Path 1: Direct, $P = 5 - V$
$$W_1 = \int_1^4 P \, dV = \int_1^4 (5 - V) \, dV$$ $$W_1 = \left[5V - \frac{V^2}{2}\right]_1^4$$ $$W_1 = \left(20 - 8\right) - \left(5 - 0.5\right)$$ $$W_1 = 12 - 4.5 = 7.5 \text{ atm·L}$$ $$W_1 = 750 \text{ J}$$Path 2: A→C→B
A→C: Isochoric at V = 1, from P = 4 to P = 1
$$W_{AC} = 0$$C→B: Isobaric at P = 1, from V = 1 to V = 4
$$W_{CB} = P\Delta V = 1 \times (4 - 1) = 3 \text{ atm·L}$$ $$W_{CB} = 300 \text{ J}$$Total Path 2:
$$W_2 = 0 + 300 = 300 \text{ J}$$Comparison:
$$W_1 = 750 \text{ J} > W_2 = 300 \text{ J}$$Work depends on path! (Not a state function)
Higher path on PV diagram → More area → More work.
Q6. A Carnot engine operates between 400 K and 300 K. The engine absorbs 1000 J at high temperature. Find: (a) Work done by engine (b) Heat rejected at low temperature (c) If the cycle is represented by a loop on PV diagram, what is the area?
Solution
(a) Carnot efficiency:
$$\eta = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{400} = 1 - 0.75 = 0.25$$ $$W = \eta \times Q_H = 0.25 \times 1000 = 250 \text{ J}$$(b) Heat rejected:
$$Q_C = Q_H - W = 1000 - 250 = 750 \text{ J}$$(c) Area on PV diagram:
Area enclosed = Work done = 250 J
Convert to atm·L: $250 \text{ J} = 250/100 = 2.5$ atm·L
The area of the Carnot cycle loop equals 2.5 atm·L.
Note: We don’t know the specific P and V values, but area still equals work!
Level 3: JEE Advanced
Q7. A gas follows path ABC where:
- A: (V₀, P₀)
- B: (2V₀, P₀/2)
- C: (2V₀, P₀)
- AB is isothermal
- BC is isochoric
Then returns to A via CA (straight line). Find: (a) Work in AB (b) Work in BC (c) Work in CA (d) Net work in cycle ABCA
Solution
Given:
- A: $(V_0, P_0)$
- B: $(2V_0, P_0/2)$
- C: $(2V_0, P_0)$
(a) Work AB (Isothermal):
Check if AB is isothermal: $P_AV_A = P_BV_B$?
$$P_0 \cdot V_0 = \frac{P_0}{2} \cdot 2V_0 = P_0V_0$$✓
$$W_{AB} = P_AV_A \ln\frac{V_B}{V_A} = P_0V_0 \ln(2)$$ $$W_{AB} = 0.693 P_0V_0$$(b) Work BC (Isochoric): Volume constant at $2V_0$:
$$W_{BC} = 0$$(c) Work CA (Straight line): From C$(2V_0, P_0)$ to A$(V_0, P_0)$:
Wait, both C and A are at pressure $P_0$!
So CA is isobaric (not just straight line):
$$W_{CA} = P_0(V_A - V_C) = P_0(V_0 - 2V_0)$$ $$W_{CA} = -P_0V_0$$(d) Net work:
$$W_{net} = W_{AB} + W_{BC} + W_{CA}$$ $$W_{net} = 0.693P_0V_0 + 0 + (-P_0V_0)$$ $$W_{net} = (0.693 - 1)P_0V_0$$ $$W_{net} = -0.307P_0V_0$$Negative net work → Counter-clockwise → Refrigerator/Heat Pump
Check direction:
- A→B: Right-down (expansion)
- B→C: Up (pressure increases)
- C→A: Left (compression)
This forms counter-clockwise loop ✓
Q8. On a PV diagram, a cycle consists of:
- Isothermal expansion from (1, 8) to (4, 2) (V in L, P in atm)
- Adiabatic compression from (4, 2) to (1, P₂)
- Isochoric process back to (1, 8)
Find: (a) P₂ (pressure after adiabatic compression) (b) γ for the gas (c) Work done in each process (d) Net work
Solution
State A: (1, 8) State B: (4, 2) State C: (1, P₂)
(a) Find P₂:
Process BC is adiabatic: $P_BV_B^\gamma = P_CV_C^\gamma$
Process CA is isochoric from C(1, P₂) to A(1, 8), so both at V = 1 L.
This means C and A are at same volume! So:
- C: (1, P₂)
- A: (1, 8)
For isochoric from C to A, pressure increases from P₂ to 8.
So P₂ < 8 (will find exact value using adiabatic equation).
(b) Find γ:
Process AB is isothermal: $P_AV_A = P_BV_B$
$$8 \times 1 = 2 \times 4 = 8$$✓
Temperature: $T_A = T_B$ (isothermal AB)
For adiabatic BC:
$$P_BV_B^\gamma = P_CV_C^\gamma$$ $$2 \times 4^\gamma = P_2 \times 1^\gamma$$ $$P_2 = 2 \times 4^\gamma = 2 \times 2^{2\gamma} = 2^{1+2\gamma}$$Also, temperature relation for BC (adiabatic):
$$T_BV_B^{\gamma-1} = T_CV_C^{\gamma-1}$$ $$T_B \times 4^{\gamma-1} = T_C \times 1$$ $$T_C = T_B \times 4^{\gamma-1}$$For isochoric CA at V = 1:
$$\frac{P_C}{T_C} = \frac{P_A}{T_A}$$ $$\frac{P_2}{T_C} = \frac{8}{T_A}$$Since $T_A = T_B$:
$$\frac{P_2}{T_C} = \frac{8}{T_B}$$ $$P_2 = 8 \times \frac{T_C}{T_B}$$Substitute $T_C = T_B \times 4^{\gamma-1}$:
$$P_2 = 8 \times 4^{\gamma-1} = 8 \times 2^{2(\gamma-1)} = 2^3 \times 2^{2\gamma-2} = 2^{2\gamma+1}$$From earlier: $P_2 = 2^{1+2\gamma}$
Both expressions match! ✓
Now we need another equation. The cycle must close, so we need consistency.
Actually, let me use the fact that this forms a valid cycle.
Alternative approach: Use given information directly.
We know:
- AB: Isothermal at temperature T (from PV = 8)
- BC: Adiabatic
- CA: Isochoric
From $P_AV_A = nRT$: $8 \times 1 = nRT$ → $T = 8/nR$ (in atm·L units)
Actually, without more constraints, we can’t uniquely determine γ.
Let me assume γ = 1.5 (common value) and solve:
$$P_2 = 2 \times 4^{1.5} = 2 \times 8 = 16 \text{ atm}$$But this gives P₂ = 16 > P_A = 8, which means pressure increases during adiabatic compression, which makes sense!
(c) Work in each process:
AB (Isothermal expansion):
$$W_{AB} = P_AV_A\ln\frac{V_B}{V_A} = 8 \times 1 \times \ln(4)$$ $$W_{AB} = 8 \times 1.386 = 11.09 \text{ atm·L} = 1109 \text{ J}$$BC (Adiabatic compression with γ = 1.5):
$$W_{BC} = \frac{P_BV_B - P_CV_C}{\gamma - 1}$$ $$W_{BC} = \frac{2 \times 4 - 16 \times 1}{1.5 - 1}$$ $$W_{BC} = \frac{8 - 16}{0.5} = \frac{-8}{0.5} = -16 \text{ atm·L}$$ $$W_{BC} = -1600 \text{ J}$$CA (Isochoric):
$$W_{CA} = 0$$(d) Net work:
$$W_{net} = 1109 + (-1600) + 0 = -491 \text{ J}$$Negative → Counter-clockwise → Refrigerator
Note: Without additional information, we assumed γ = 1.5. In actual JEE problems, γ would be given or derivable from additional constraints.
Q9. Prove that for any cyclic process, the net heat absorbed equals the area enclosed on the PV diagram.
Solution
Given: Cyclic process (returns to initial state)
From First Law:
$$\Delta Q = \Delta U + \Delta W$$For cyclic process:
$$\Delta U_{cycle} = 0$$(state function, returns to same state)
Therefore:
$$Q_{net} = W_{net}$$Work in thermodynamics:
$$W = \int P \, dV$$For a complete cycle:
$$W_{cycle} = \oint P \, dV$$On PV diagram:
The integral $\oint P \, dV$ represents:
- Going right (expansion): Positive contribution (area under upper curve)
- Going left (compression): Negative contribution (area under lower curve)
Net work = Area under upper path - Area under lower path
= Area enclosed by the loop
Therefore:
$$Q_{net} = W_{net} = \text{Area enclosed on PV diagram}$$Direction matters:
- Clockwise: Net work positive (area counted positive)
- Counter-clockwise: Net work negative (area counted negative)
Proved!
This fundamental result makes PV diagrams so powerful—the area directly tells us energy conversion!
Connection to Other Topics
→ Thermodynamic Processes
Each process has characteristic PV curve: Processes →
→ Heat Engines
Carnot, Otto, Diesel cycles on PV diagrams: Heat Engines →
→ First Law
Q, W, ΔU related through First Law: First Law →
→ Real Engines
- Otto cycle: Petrol engines
- Diesel cycle: Diesel engines
- Refrigeration cycle: ACs, refrigerators
Quick Revision Points
Key Formulas:
Work:
$$W = \int_{V_1}^{V_2} P \, dV = \text{Area under curve}$$Cyclic Process:
$$\Delta U = 0$$ $$Q_{net} = W_{net} = \text{Area enclosed}$$Slopes:
- Isothermal: $-P/V$
- Adiabatic: $-\gamma P/V$ (steeper)
- Isobaric: 0 (horizontal)
- Isochoric: ∞ (vertical)
Direction:
- Clockwise → Heat Engine (W > 0)
- Counter-clockwise → Refrigerator (W < 0)
JEE Strategy Tips
- Draw the diagram - visualize the process
- Identify each leg - isothermal, adiabatic, etc.
- Calculate work process-by-process - use appropriate formula
- Check direction - clockwise or counter-clockwise
- Use geometry - rectangles, triangles, trapezoids
- Remember: Area = Work (fundamental principle)
- For cycles: ΔU = 0, so Q = W
Next Topic: Second Law of Thermodynamics →
Last updated: February 2025