Heat, Work, and Chaos: Mastering the Laws of Thermodynamics

Thermodynamics is the study of the macroscopic world. It doesn’t care about individual molecules; it cares about the “Big Three”: Pressure (P), Volume (V), and Temperature (T).

It is the science that powered the Industrial Revolution and continues to define the limits of every engine, refrigerator, and even the ultimate fate of our universe.

The Core Pillars of Thermodynamics

1. The Zeroth Law: The Basis of Temperature

If two systems are each in thermal equilibrium with a third system, they are in thermal equilibrium with each other. This sounds obvious, but it’s the logical foundation that allows us to use thermometers to measure temperature.

2. The First Law: Conservation of Energy

Energy cannot be created or destroyed, only transformed. In a thermodynamic system, the heat you add (ΔQ) does two things: it increases the internal energy (ΔU) and performs work (ΔW).

• ΔQ = ΔU + ΔW
• Sign Convention: Work done by the system is positive; work done on the system is negative.

3. Thermodynamic Processes

• Isothermal: Temperature remains constant (ΔU = 0).
• Adiabatic: No heat exchange (ΔQ = 0). These processes happen very fast (like a tire bursting).
• Isobaric: Pressure remains constant.
• Isochoric: Volume remains constant (ΔW = 0).

4. The Second Law: The Arrow of Time

Heat never flows spontaneously from a colder body to a hotter body. This law introduces Entropy—the measure of disorder. It tells us that no engine can ever be 100% efficient.

The Gauntlet: 10 Challenging Aptitude Questions

Question 1: The P-V Work Integral

A gas expands from volume V₁ to V₂ following the relation P = kV². Find the work done by the gas during this expansion.

Question 2: The Adiabatic Slope

On a P-V diagram, why is the slope of an adiabatic curve always steeper than the slope of an isothermal curve at the same point? (Hint: Use the adiabatic constant γ).

Question 3: The Internal Energy Trap

An ideal gas undergoes a cyclic process (returning to its original state). If the total heat added to the gas is 500 J, what is the net work done by the gas, and what is the change in internal energy?

Question 4: The Carnot Efficiency Limit

A heat engine operates between a source at 500 K and a sink at 300 K. What is its maximum possible efficiency? If the engine performs 1000 J of work, how much heat is rejected to the sink?

Question 5: The Refrigerator Coefficient

A refrigerator’s door is left open in a closed, thermally insulated room. Will the room get cooler or warmer over time? Explain your answer using the Second Law.

Question 6: Molar Specific Heats (Cp and Cv)

For an ideal gas, prove that Cp – Cv = R (Mayer’s Relation). Why is Cp always greater than Cv?

Question 7: The Sudden Compression

A gas is suddenly compressed to 1/4th of its original volume. If the process is adiabatic and γ = 1.5, by what factor does the pressure increase?

Question 8: Work in a Cyclic Process

On a P-V graph, a cyclic process forms a perfect circle. If the cycle is clockwise, is the net work positive or negative? What does the area inside the circle represent?

Question 9: The Free Expansion Paradox

A gas undergoes “Free Expansion” into a vacuum inside an insulated container.

1. Is work done?
2. Is there a change in temperature?
3. Is there a change in entropy?

Question 10: Mixing of Gases

Two moles of Helium (monatomic) are mixed with one mole of Hydrogen (diatomic). Find the equivalent γ (ratio of specific heats) for the mixture.

Detailed Explanations & Solutions

1. Variable Work

Work W = ∫ P dV.

W = ∫ k V² dV from V₁ to V₂.

Result: W = (k/3) [V₂³ – V₁³].

2. The Slope Ratio

For Isothermal: PV = constant → dP/dV = -P/V.

For Adiabatic: PVᵞ = constant → dP/dV = -γ(P/V).

Result: Since γ > 1, the adiabatic slope is γ times steeper.

3. Cyclic Process

In a cyclic process, the system returns to its initial state, so ΔU = 0.

By the First Law: ΔQ = ΔW.

Result: ΔU = 0; Net Work = 500 J.

4. Carnot Efficiency

η = 1 – (T_sink / T_source) = 1 – (300/500) = 0.4 or 40%.

Work = 1000 J. Heat Input = 1000 / 0.4 = 2500 J.

Result: Heat Rejected = 2500 – 1000 = 1500 J.

5. The Refrigerator Trap

A refrigerator is a heat pump; it removes heat from the inside and exhausts it (plus the electrical work done) to the outside.

Result: The room gets warmer. The heat exhausted into the room is greater than the heat removed from the fridge’s interior.

6. Cp vs Cv

At constant volume (Cv), all heat goes into increasing internal energy. At constant pressure (Cp), some heat is used to do work against external pressure. Hence, more heat is required for the same temperature rise.

Result: Cp > Cv.

7. Adiabatic Compression

P₁V₁ᵞ = P₂V₂ᵞ.

P₂/P₁ = (V₁/V₂)ᵞ = (4)^1.5 = (2²)^1.5 = 2³ = 8.

Result: Pressure increases by 8 times.

8. P-V Loop Area

The area inside the loop represents the net work done.

Result: Clockwise = Positive Work; Anti-clockwise = Negative Work.

9. Free Expansion

• W = 0 (No external pressure to push against).
• ΔU = 0 (If ideal, T remains constant).
• Entropy Increases (The system becomes more disordered).

10. Mixing Gases

Use Cv_mix = (n₁Cv₁ + n₂Cv₂) / (n₁ + n₂).

For Helium: Cv = 3/2 R. For Hydrogen: Cv = 5/2 R.

Cv_mix = [2(3/2 R) + 1(5/2 R)] / 3 = [3R + 2.5R] / 3 = 1.83R.

Then find Cp_mix = Cv_mix + R and γ = Cp/Cv.

Result: γ_mix ≈ 1.54.

Pro-Tip: The “Isothermal vs. Adiabatic” Cheat

• If a process happens slowly, it’s usually Isothermal (system has time to exchange heat with surroundings).
• If a process happens suddenly, it’s usually Adiabatic (no time for heat exchange).