Thermodynamic Processes

Real-Life Hook: How Does a Diesel Engine Work Without Spark Plugs?

Unlike petrol engines that use spark plugs, diesel engines rely on extreme compression to ignite fuel. When air is compressed rapidly (adiabatically), its temperature shoots up to over 500°C—hot enough to ignite diesel fuel spontaneously!

This is a real-world adiabatic process: rapid compression prevents heat exchange, converting work directly into temperature rise. Understanding thermodynamic processes explains:

• Why diesel engines are more efficient
• How refrigerators cool
• Why pumping a bike tire heats the pump
• How atmospheric pressure changes with altitude

Let’s master all four fundamental processes!

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The Four Fundamental Processes

Quick Comparison Table

Process	Constant	Equation	Q	W	ΔU
Isothermal	Temperature (T)	$PV = \text{const}$	$nRT\ln(V_2/V_1)$	$nRT\ln(V_2/V_1)$	0
Adiabatic	Heat (Q = 0)	$PV^\gamma = \text{const}$	0	$\frac{nR(T_1-T_2)}{\gamma-1}$	$-W$
Isobaric	Pressure (P)	$V/T = \text{const}$	$nC_P\Delta T$	$P\Delta V$	$nC_V\Delta T$
Isochoric	Volume (V)	$P/T = \text{const}$	$nC_V\Delta T$	0	$nC_V\Delta T$

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1. Isothermal Process (Constant Temperature)

Definition

Isothermal process: Temperature remains constant throughout.

$$T = \text{constant} \quad \Rightarrow \quad \Delta T = 0$$

Conditions Required

1. Slow process: Allows heat exchange with surroundings
2. Thermally conducting walls: System in contact with heat reservoir
3. Constant temperature bath: Maintains T constant

For Ideal Gas

From $PV = nRT$ with T constant:

First Law Application

Since $\Delta T = 0$ for ideal gas:

$$\Delta U = nC_V\Delta T = 0$$

From First Law:

$$Q = \Delta U + W$$ $$Q = 0 + W$$

Work Done in Isothermal Process

$$W = \int_{V_1}^{V_2} P \, dV$$

Using $PV = nRT$ (constant):

$$P = \frac{nRT}{V}$$ $$W = \int_{V_1}^{V_2} \frac{nRT}{V} \, dV$$

Sign Convention

• Expansion ($V_2 > V_1$): W > 0 (work done BY gas)
• Compression ($V_2 < V_1$): W < 0 (work done ON gas)

Memory Trick

“ISO-THERMAL → ISO-ENERGY”

• Temperature constant → Internal energy constant (for ideal gas)
• Q = W (heat and work are equal and opposite)

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2. Adiabatic Process (No Heat Exchange)

Definition

Adiabatic process: No heat exchange with surroundings.

$$Q = 0$$

Conditions Required

1. Rapid process: No time for heat transfer
2. Thermally insulated walls: Prevents heat exchange
3. Examples:
   • Rapid compression/expansion
   • Sound wave propagation
   • Diesel engine compression

For Ideal Gas

From First Law with Q = 0:

$$0 = \Delta U + W$$ $$\Delta U = -W$$

This leads to three equivalent equations:

Derivation of PV^γ = Constant

Step 1: First Law for adiabatic process

$$dU = -dW$$ $$nC_VdT = -PdV$$

Step 2: From ideal gas law $PV = nRT$:

$$PdV + VdP = nRdT$$ $$dT = \frac{PdV + VdP}{nR}$$

Step 3: Substitute in step 1:

$$nC_V\frac{PdV + VdP}{nR} = -PdV$$ $$C_V(PdV + VdP) = -RPdV$$ $$C_VPdV + C_VVdP = -RPdV$$ $$(C_V + R)PdV = -C_VVdP$$

Using $C_P = C_V + R$:

$$C_PPdV = -C_VVdP$$ $$\frac{C_P}{C_V}\frac{dV}{V} = -\frac{dP}{P}$$ $$\gamma\frac{dV}{V} = -\frac{dP}{P}$$

Step 4: Integrate:

$$\gamma \ln V = -\ln P + \text{const}$$ $$\ln V^\gamma + \ln P = \text{const}$$ $$\ln(PV^\gamma) = \text{const}$$

Work Done in Adiabatic Process

Slope of Adiabatic Curve

From $PV^\gamma = K$:

$$\frac{dP}{dV} = -\gamma\frac{P}{V}$$

Slope of adiabatic = $-\gamma P/V$

For isothermal: $PV = K$

$$\frac{dP}{dV} = -\frac{P}{V}$$

Slope of isothermal = $-P/V$

Comparison:

$$\left|\frac{dP}{dV}\right|_{\text{adiabatic}} = \gamma \left|\frac{dP}{dV}\right|_{\text{isothermal}}$$

Since γ > 1, adiabatic curve is steeper than isothermal!

Memory Trick

“Adiabatic is STEEPER”

• Adiabatic slope = γ × Isothermal slope
• γ > 1 → Adiabatic steeper on P-V diagram

“No HEAT? Temperature CHANGES!”

• Q = 0 (adiabatic)
• ΔT ≠ 0 (temperature changes!)
• Opposite of isothermal

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3. Isobaric Process (Constant Pressure)

Definition

Isobaric process: Pressure remains constant.

$$P = \text{constant}$$

For Ideal Gas

From $PV = nRT$ with P constant:

Work Done

Heat Supplied

From First Law:

$$Q = \Delta U + W$$ $$Q = nC_V\Delta T + nR\Delta T$$ $$Q = n(C_V + R)\Delta T$$

P-V Diagram

Horizontal line (pressure constant, volume changes)

Real-Life Examples

• Boiling water at constant atmospheric pressure
• Heating gas in a cylinder with movable piston
• Atmospheric processes at constant altitude

Memory Trick

“Constant P → Use C_P”

• Pressure constant → Use $C_\mathbf{P}$
• $Q = nC_P\Delta T$

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4. Isochoric Process (Constant Volume)

Definition

Isochoric process: Volume remains constant.

$$V = \text{constant} \quad \Rightarrow \quad \Delta V = 0$$

Also called isometric or isovolumetric process.

For Ideal Gas

From $PV = nRT$ with V constant:

Work Done

Since $\Delta V = 0$:

Heat Supplied

From First Law:

$$Q = \Delta U + W$$ $$Q = \Delta U + 0$$

P-V Diagram

Vertical line (volume constant, pressure changes)

Real-Life Examples

• Heating gas in a rigid container
• Bomb calorimeter in chemistry
• Sealed pressure cooker heating

Memory Trick

“Constant V → Use C_V”

• Volume constant → Use $C_\mathbf{V}$
• $Q = nC_V\Delta T$

“No Volume change → No Work”

• W = PΔV = 0

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P-V Diagrams for All Processes

Visual Comparison

On the same P-V diagram starting from same point:

1. Isothermal: Rectangular hyperbola, moderate slope
2. Adiabatic: Steeper curve than isothermal
3. Isobaric: Horizontal line
4. Isochoric: Vertical line

Key Point: Adiabatic is always steeper than isothermal through the same point!

Slope Summary

Process	Slope (dP/dV)
Isothermal	$-P/V$
Adiabatic	$-\gamma P/V$
Isobaric	0 (horizontal)
Isochoric	∞ (vertical)

Steepness order: Isochoric > Adiabatic > Isothermal > Isobaric

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Common Mistakes & Tricks

Mistake 1: Confusing Isothermal and Adiabatic

Aspect	Isothermal	Adiabatic
Temperature	Constant (ΔT = 0)	Changes (ΔT ≠ 0)
Heat	Q ≠ 0	Q = 0
Process speed	Slow	Fast
Equation	$PV = \text{const}$	$PV^\gamma = \text{const}$
ΔU	0	≠ 0
Q vs W	Q = W	W = -ΔU

Mistake 2: Using Wrong Heat Capacity

❌ Isochoric: $Q = nC_P\Delta T$ (Wrong!) ✅ Isochoric: $Q = nC_V\Delta T$

❌ Isobaric: $Q = nC_V\Delta T$ (Wrong!) ✅ Isobaric: $Q = nC_P\Delta T$

Trick: Match the subscript - V with V, P with P!

Mistake 3: Sign of Work in Compression

For adiabatic compression:

• $V_2 < V_1$ (compressed)
• $T_2 > T_1$ (temperature increases)
• $W = \frac{nR(T_1 - T_2)}{\gamma - 1} < 0$ (work ON gas)

Mistake 4: Forgetting γ in Adiabatic

❌ $PV = \text{const}$ (that’s isothermal!) ✅ $PV^\gamma = \text{const}$ (adiabatic)

Memory Trick: “4 I’s”

1. Isothermal: Internal energy constant
2. Isobaric: Includes work (Q = ΔU + W with both nonzero)
3. Isochoric: Ignore work (W = 0)
4. Isolated (Adiabatic): Insulated (Q = 0)

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Comparative Analysis

Work Done (Expansion from V₁ to V₂)

Order of work done:

$$W_{\text{isobaric}} > W_{\text{isothermal}} > W_{\text{adiabatic}}$$

Why?

• Isobaric: Pressure stays high throughout
• Isothermal: Pressure decreases moderately
• Adiabatic: Pressure decreases rapidly (steeper curve)

Work = Area under curve → Higher curve → More work

Temperature Change

For expansion from V₁ to V₂:

Process	Temperature
Isothermal	No change (T constant)
Adiabatic	Decreases (gas cools)
Isobaric	Increases ($V \propto T$)
Isochoric	Not applicable (V fixed)

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Practice Problems

Level 1: JEE Main Basics

Q1. An ideal gas expands isothermally at 300 K from 1 L to 5 L. If initial pressure is 5 atm, find: (a) Final pressure (b) Work done (R = 8.314 J/mol·K, 1 atm = $10^5$ Pa)

Solution

(a) Final pressure:

Isothermal: $P_1V_1 = P_2V_2$

$$P_2 = P_1\frac{V_1}{V_2} = 5 \times \frac{1}{5} = 1 \text{ atm}$$

(b) Work done:

First find n:

$$n = \frac{P_1V_1}{RT} = \frac{5 \times 10^5 \times 10^{-3}}{8.314 \times 300}$$ $$n = \frac{500}{2494.2} = 0.2004 \text{ mol}$$ $$W = nRT\ln\frac{V_2}{V_1}$$ $$W = 0.2004 \times 8.314 \times 300 \times \ln(5)$$ $$W = 499.4 \times 1.609 = 803.6 \text{ J}$$

Alternative:

$$W = P_1V_1\ln\frac{V_2}{V_1} = 5 \times 10^5 \times 10^{-3} \times \ln(5)$$ $$W = 500 \times 1.609 = 804.5 \text{ J}$$

Q2. A gas is compressed adiabatically to 1/3 of its volume. If γ = 1.5, by what factor does pressure increase?

Solution

Adiabatic: $P_1V_1^\gamma = P_2V_2^\gamma$

$$\frac{P_2}{P_1} = \left(\frac{V_1}{V_2}\right)^\gamma$$

Given: $V_2 = V_1/3$, so $V_1/V_2 = 3$

$$\frac{P_2}{P_1} = 3^{1.5} = 3^{3/2} = 3\sqrt{3} = 3 \times 1.732 = 5.196$$

Pressure increases by factor of ~5.2

Q3. 2 moles of gas at 300 K are heated at constant volume to 450 K. If $C_V = 20$ J/mol·K, find: (a) Heat supplied (b) Work done (c) Change in internal energy

Solution

(a) Heat supplied:

$$Q = nC_V\Delta T = 2 \times 20 \times (450 - 300)$$ $$Q = 40 \times 150 = 6000 \text{ J}$$

(b) Work done: Constant volume → $W = 0$

(c) Change in internal energy:

$$\Delta U = Q - W = 6000 - 0 = 6000 \text{ J}$$

Or directly: $\Delta U = nC_V\Delta T = 6000$ J ✓

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Level 2: JEE Main/Advanced

Q4. One mole of ideal gas (γ = 1.4) at 27°C expands adiabatically from 1 L to 2 L. Find: (a) Final temperature (b) Work done (c) Change in internal energy

Solution

Given: $T_1 = 27°C = 300$ K, $V_1 = 1$ L, $V_2 = 2$ L, γ = 1.4

(a) Final temperature:

Adiabatic: $T_1V_1^{\gamma-1} = T_2V_2^{\gamma-1}$

$$T_2 = T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}$$ $$T_2 = 300 \times \left(\frac{1}{2}\right)^{1.4-1}$$ $$T_2 = 300 \times (0.5)^{0.4}$$

Calculate: $(0.5)^{0.4} = 0.758$

$$T_2 = 300 \times 0.758 = 227.4 \text{ K} = -45.6°C$$

(b) Work done:

$$W = \frac{nR(T_1 - T_2)}{\gamma - 1}$$ $$W = \frac{1 \times 8.314 \times (300 - 227.4)}{1.4 - 1}$$ $$W = \frac{8.314 \times 72.6}{0.4}$$ $$W = \frac{603.6}{0.4} = 1509 \text{ J}$$

(c) Change in internal energy: Adiabatic: $\Delta U = -W = -1509$ J

Or: $\Delta U = nC_V\Delta T$

$$C_V = \frac{R}{\gamma-1} = \frac{8.314}{0.4} = 20.785 \text{ J/mol·K}$$ $$\Delta U = 1 \times 20.785 \times (227.4 - 300)$$ $$\Delta U = 20.785 \times (-72.6) = -1509 \text{ J}$$

Gas expands, cools down, internal energy decreases.

Q5. An ideal gas undergoes process where $P \propto V^2$. If temperature changes from 300 K to 400 K, find the ratio $V_2/V_1$.

Solution

Given: $P \propto V^2$ → $P = kV^2$

From ideal gas law: $PV = nRT$

Substitute $P = kV^2$:

$$kV^2 \cdot V = nRT$$ $$kV^3 = nRT$$ $$V^3 \propto T$$

Therefore:

$$\frac{V_2^3}{V_1^3} = \frac{T_2}{T_1} = \frac{400}{300} = \frac{4}{3}$$ $$\frac{V_2}{V_1} = \left(\frac{4}{3}\right)^{1/3} = 1.10$$

Volume increases by factor of 1.10

Q6. Compare work done when 1 mole of ideal gas expands from 1 L to 2 L at 300 K: (a) Isothermally (b) Isobarically (starting at 10 atm)

Solution

(a) Isothermal work:

$$W_{iso-T} = nRT\ln\frac{V_2}{V_1}$$ $$W_{iso-T} = 1 \times 8.314 \times 300 \times \ln(2)$$ $$W_{iso-T} = 2494.2 \times 0.693 = 1728.5 \text{ J}$$

(b) Isobaric work:

From initial state:

$$P_1 = \frac{nRT_1}{V_1} = \frac{1 \times 8.314 \times 300}{10^{-3}}$$ $$P_1 = 2494200 \text{ Pa} = 24.94 \text{ atm}$$

Wait, problem says “starting at 10 atm”, so use that:

$$P = 10 \text{ atm} = 10 \times 10^5 \text{ Pa}$$ $$W_{iso-P} = P\Delta V = 10 \times 10^5 \times (2-1) \times 10^{-3}$$ $$W_{iso-P} = 10^6 \times 10^{-3} = 1000 \text{ J}$$

But this doesn’t make sense - if initial state is 1 mol at 300 K in 1 L, pressure must be:

$$P = \frac{nRT}{V} = \frac{1 \times 8.314 \times 300}{10^{-3}} = 2.49 \times 10^6 \text{ Pa} ≈ 24.9 \text{ atm}$$

Let me recalculate isobaric with correct pressure:

$$W_{iso-P} = P\Delta V = 2.49 \times 10^6 \times 1 \times 10^{-3}$$ $$W_{iso-P} = 2490 \text{ J}$$

Or using: $W = nR\Delta T$

For isobaric expansion from 1L to 2L:

$$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$ $$\frac{1}{300} = \frac{2}{T_2}$$ $$T_2 = 600 \text{ K}$$ $$W_{iso-P} = nR\Delta T = 1 \times 8.314 \times (600-300)$$ $$W_{iso-P} = 8.314 \times 300 = 2494.2 \text{ J}$$

Comparison:

$$W_{iso-P} = 2494.2 \text{ J} > W_{iso-T} = 1728.5 \text{ J}$$

Isobaric work is greater (as expected - pressure stays higher).

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Level 3: JEE Advanced

Q7. An ideal gas undergoes cycle ABCA where:

• AB: Isothermal expansion at 400 K from 1 L to 4 L
• BC: Isobaric compression
• CA: Isochoric heating back to initial state

Find: (a) Pressure and volume at each point (b) Work done in each process (c) Net work in cycle

Assume 1 mole, initial pressure 10 atm.

Solution

State A: $T_A = 400$ K, $V_A = 1$ L, $P_A = 10$ atm

(a) Find all states:

State B (after isothermal expansion):

• $T_B = T_A = 400$ K (isothermal)
• $V_B = 4$ L (given)
• $P_AV_A = P_BV_B$ $$P_B = P_A\frac{V_A}{V_B} = 10 \times \frac{1}{4} = 2.5 \text{ atm}$$

State C (after isobaric compression to original volume):

• $V_C = V_A = 1$ L (isochoric CA brings back to A)
• $P_C = P_B = 2.5$ atm (isobaric BC)
• From ideal gas: $\frac{V_B}{T_B} = \frac{V_C}{T_C}$ $$T_C = T_B\frac{V_C}{V_B} = 400 \times \frac{1}{4} = 100 \text{ K}$$

Verification: In process CA (isochoric), volume stays at 1 L:

• Initial: C at 1 L, 2.5 atm, 100 K
• Final: A at 1 L, 10 atm, 400 K
• Check: $P/T$ constant? $2.5/100 = 10/400$ → $0.025 = 0.025$ ✓

(b) Work in each process:

AB (Isothermal):

$$W_{AB} = nRT\ln\frac{V_B}{V_A}$$ $$W_{AB} = 1 \times 8.314 \times 400 \times \ln(4)$$ $$W_{AB} = 3325.6 \times 1.386 = 4609.4 \text{ J}$$

BC (Isobaric compression):

$$W_{BC} = P_B(V_C - V_B)$$

Convert pressure: $P_B = 2.5 \times 10^5$ Pa

$$W_{BC} = 2.5 \times 10^5 \times (1 - 4) \times 10^{-3}$$ $$W_{BC} = 2.5 \times 10^5 \times (-3) \times 10^{-3}$$ $$W_{BC} = -750 \text{ J}$$

CA (Isochoric):

$$W_{CA} = 0$$

(no volume change)

(c) Net work:

$$W_{net} = W_{AB} + W_{BC} + W_{CA}$$ $$W_{net} = 4609.4 + (-750) + 0$$ $$W_{net} = 3859.4 \text{ J}$$

Also equals net heat absorbed (cyclic process: ΔU = 0).

Q8. Prove that for an adiabatic process of ideal gas:

$$TV^{\gamma-1} = \text{constant}$$

Solution

Given: $PV^\gamma = K_1$ (constant)

From ideal gas law: $PV = nRT$

$$P = \frac{nRT}{V}$$

Substitute in adiabatic equation:

$$\frac{nRT}{V} \cdot V^\gamma = K_1$$ $$nRT \cdot V^{\gamma-1} = K_1$$

Since n and R are constants:

$$T \cdot V^{\gamma-1} = \frac{K_1}{nR} = K_2 \text{ (constant)}$$

$$\boxed{TV^{\gamma-1} = \text{constant}}$$

Similarly, can derive:

From $TV^{\gamma-1} = K_2$ and $PV = nRT$:

$$T = \frac{PV}{nR}$$

Substitute:

$$\frac{PV}{nR} \cdot V^{\gamma-1} = K_2$$ $$PV^\gamma = nRK_2 = K_1$$

Which gives back $PV^\gamma = \text{const}$ ✓

For T-P relation:

From $PV = nRT$: $V = nRT/P$

Substitute in $TV^{\gamma-1} = K_2$:

$$T \cdot \left(\frac{nRT}{P}\right)^{\gamma-1} = K_2$$ $$T \cdot \frac{(nRT)^{\gamma-1}}{P^{\gamma-1}} = K_2$$ $$T \cdot T^{\gamma-1} \cdot \frac{(nR)^{\gamma-1}}{P^{\gamma-1}} = K_2$$ $$T^\gamma \cdot \frac{(nR)^{\gamma-1}}{P^{\gamma-1}} = K_2$$ $$T^\gamma = K_2 \cdot \frac{P^{\gamma-1}}{(nR)^{\gamma-1}}$$ $$T^\gamma \propto P^{\gamma-1}$$

$$\boxed{TP^{(\gamma-1)/\gamma} = \text{constant}}$$

Or:

$$\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{(\gamma-1)/\gamma}$$

All three forms proven!

Q9. Show that slope of adiabatic curve is γ times the slope of isothermal curve at the same point.

Solution

Isothermal curve: $PV = K_1$

Differentiate:

$$\frac{d(PV)}{dV} = 0$$ $$P + V\frac{dP}{dV} = 0$$ $$\frac{dP}{dV}\bigg|_{iso-T} = -\frac{P}{V}$$

Adiabatic curve: $PV^\gamma = K_2$

Differentiate:

$$\frac{d(PV^\gamma)}{dV} = 0$$ $$V^\gamma \frac{dP}{dV} + P \cdot \gamma V^{\gamma-1} = 0$$ $$V^\gamma \frac{dP}{dV} = -\gamma P V^{\gamma-1}$$ $$\frac{dP}{dV}\bigg|_{adiabatic} = -\gamma \frac{P}{V}$$

Ratio:

$$\frac{\left|\frac{dP}{dV}\right|_{adiabatic}}{\left|\frac{dP}{dV}\right|_{iso-T}} = \frac{\gamma P/V}{P/V} = \gamma$$

$$\boxed{\text{Slope}_{adiabatic} = \gamma \times \text{Slope}_{isothermal}}$$

Since γ > 1, adiabatic curve is steeper!

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Connection to Other Topics

→ PV Diagrams

All processes visualized on P-V diagrams, work as area: PV Diagrams and Cycles →

→ First Law of Thermodynamics

Each process applies First Law differently: First Law →

→ Heat Engines

Carnot cycle uses isothermal and adiabatic processes: Heat Engines →

→ Kinetic Theory

Adiabatic exponent γ from degrees of freedom: Kinetic Theory →

→ Sound Waves

Sound propagates adiabatically:

$$v = \sqrt{\frac{\gamma P}{\rho}}$$

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Quick Revision Table

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JEE Strategy Tips

1. Identify the process from given conditions (constant T, P, V, or Q)
2. Write the governing equation (PV, PV^γ, etc.)
3. Use First Law to connect Q, W, ΔU
4. For work: Use specific formula for each process
5. Slopes: Remember adiabatic is steeper than isothermal
6. Sign conventions: Expansion → W positive, Compression → W negative

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Next Topic: PV Diagrams and Cyclic Processes →

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Last updated: February 2025