Thermodynamics

1. Thermodynamics

• Study of exchange of heat energy between bodies.
• Conversion of heat into mechanical energy and vice-versa.

2. Thermodynamic System

• (a) Open system : exchanges both energy and matter with surroundings.
• (b) Closed system : exchanges only energy (not matter).
• (c) Isolated system : exchanges neither energy nor matter.

3. Extensive & Intensive Variables

• Extensive : scale with mass / size of system.
  e.g. volume, mass, moles, entropy, internal energy.
• Intensive : independent of size of system.
  e.g. temperature, pressure, density.

4. Zeroth Law of Thermodynamics

• Thermal equilibrium ⇒ same temperature.
• If A & B and A & C are in thermal equilibrium, then B & C are also in thermal equilibrium.

5. Ideal Gas Equation

• In terms of density :

ρ in kg/m³, M (molar mass) in kg
e.g. H₂ gas : M = 2 g = 2 × 10⁻³ kg

Unit conversions :

• 1 m³ = 1000 litre = 10⁶ cm³ = 10⁶ cc
• 1 litre = 10⁻³ m³ = 1000 mL = 1000 cm³
• 1 mL = 1 cm³

6. STP (Standard Temp. & Pressure)

7. Quasi-Static Process

• Process that happens infinitely slowly, so the system is always in thermal & mechanical equilibrium with surroundings.
• "Quasi" = almost ⇒ an "almost static" process.
• Piston moves very slowly ⇒ F_net = 0 on piston (even if piston is moving).
• It is a reversible process.

8. First Law of Thermodynamics

• Heat supplied to the gas is converted into internal energy of gas and work done by the gas.
• It is a statement of conservation of energy.

• Q and W are process (path) dependent.
• ΔU is path independent – depends only on initial & final state.
• Chemistry convention : W_{by gas} = −W_{on gas}
  ⇒ Q + W = ΔU (W = work done on gas)

9. Change in Internal Energy (ΔU)

• Same formula for every process (path).
• ΔU_A→B = − ΔU_B→A

10. Work Done by Gas (W)

• Force by gas on piston : F = PA
• W = ∫F dx = ∫PA dx   (A dx = dV)

• Sign of dW depends only on dV (P is always +ve).

11. Work Done in P–V Diagram

• If final vol. = initial vol., then W may be +ve, −ve or zero (not necessarily zero).
• Net work = sum of work in each part :
  W_net = W_AC + W_CD + W_DB

12. Work is Path Dependent

13. Heat Supplied to the Gas (Q)

• C → molar specific heat capacity of the process.
• C can have any value :  −∞ < C < ∞
• ΔU = nfRΔT2 = nC_VΔT,   W = ∫PdV

14. Cyclic Process

• Initial and final states are same.

15. W and Q in Cyclic Process

16. Area of Elliptical Cycle

17. Graph Conversion (Solved)

• P = nRV T ⇒ slope ∝ 1V (V more ⇒ slope less)
• T = PnR V ⇒ slope ∝ P (P more ⇒ slope more)
• ρ = massV ⇒ ρ ∝ 1V ;  P = ρRTM

18. Max. Temperature in Linear Process

• 2P₀V₀ = nR(300) ⇒ P₀V₀nR = 150
• Line : P = −P₀2V₀ V + 5P₀2
• T = PVnR,  dTdV = 0 ⇒ V = 5V₀2

19. Partition Problems (Steady State)

• P₁ = n₁RT₁V₀ = 4·R·770R = 3080 Pa
• P₂ = 2·R·1540R = 3080 Pa  ⇒ P₁ = P₂

20. Isochoric Process (V = constant)

• Fixed piston ⇒ dV = 0 ,  W = 0
• P₁T₁ = P₂T₂  (P ∝ T)
• Q = ΔU → 100% heat supplied goes into internal energy.

C_V → molar specific heat at constant volume

P–T slope = tanθ = nRV ⇒ V more ⇒ slope less.

21. Isobaric Process (P = constant)

• Movable piston :  V₁T₁ = V₂T₂  (V ∝ T)
• Piston : PA = P₀A + mg ⇒ P = P₀ + mgA

• Q = ΔU + W = nfRΔT2 + nRΔT

C_P → molar specific heat at constant pressure

• % heat converted into internal energy :

T–V slope = tanθ = PnR ⇒ P more ⇒ slope more.

22. Isothermal Process (T = constant)

• Conducting vessel, piston moving / heat supplied very slowly.
• P₁V₁ = P₂V₂ ,  ΔU = 0

• Q = nCΔT with ΔT → 0 ⇒ C → ± ∞
• PdV + VdP = 0 ⇒ slope :

23. Adiabatic Process (Q = 0)

• No heat exchange ⇒ dQ = 0, Q = 0
• Adiabatic → non-conducting (adiabatic) vessel & piston, or very fast process.

• ΔU = nC_VΔT ,  Q = 0 ⇒

• Q = nCΔT = 0 ⇒ C = 0
• C_P − C_V = R & C_PC_V = γ ⇒

• As V increases, P decreases. Slope of tangent :
• P(γV^γ−1dV) + V^γdP = 0

24. Isothermal v/s Adiabatic

25. Sign of ΔT and Q Near Curves

• From a point on an isotherm (ΔT = 0, ΔU = 0) :
  • Going to a state above it → ΔT +ve, ΔU +ve
  • Going to a state below it → ΔT −ve, ΔU −ve
• From a point on an adiabat (Q = 0) :
  • Going to a state above it → Q +ve
  • Going to a state below it → Q −ve

26. Solved Example (Compression)

Case 2, 3, 4 (adiabatic, γ = 5/3) :

• P₁V₁^5/3 = P₂(V₁/8)^5/3 ⇒ P₂ = (8)^5/3P₁ = 32 atm
• T₁V₁^2/3 = T₂(V₁/8)^2/3 ⇒ T₂ = 4T₁ = 1200 K
• Q = 0
• ΔU = nC_VΔT = 6 (3R2)(1200 − 300) = 8100 R
• W = −ΔU = −8100 R

Case 1 (isothermal) :

• P₂ = 8 atm, T₂ = 300 K, ΔU = 0
• W = Q = nRT ln18 = −5400 R ln2

27. Comparison : Expansion

n moles expand from same initial state to same final volume via AB (isobaric), AC (isothermal), AD (adiabatic).

• W = AUC = +ve :  W₁ > W₂ > W₃
• ΔU₁ > ΔU₂ > ΔU₃
• Q = ΔU + W :  Q₁ > Q₂ > Q₃
• T_B > T_C > T_D ,  P_B > P_C > P_D ,  V_B = V_C = V_D

28. Comparison : Compression

Same initial state, compressed to same final volume via AB (isobaric), AC (isothermal), AD (adiabatic). Work done by gas is −ve.

• |W₃| > |W₂| > |W₁| ⇒ W₃ < W₂ < W₁
• ΔU₃ > ΔU₂ > ΔU₁
• T₃ > T₂ > T₁
• Q₃ > Q₂ > Q₁ ,  |Q₃| < |Q₂| < |Q₁|

29. Polytropic Process (General Process)

• −∞ < m < ∞ ,  m ≠ 1
• If m = 1 ⇒ PV = const ⇒ isothermal

30. Free Expansion

• Expansion of an isolated system against no surrounding (into vacuum) in an adiabatic vessel.

• Q = 0  (no heat exchange)
• W = 0  (no work done by gas, even though gas expands)
• ΔU = 0  (temperature constant)

32. Efficiency of Cyclic Process

• For a cyclic process : ΔU_net = 0
• W_net = Q_net = ± (area enclosed)
• Q_s → total heat supplied to gas (sum of all +ve heats)
• Q_e → total heat ejected from gas (|sum of all −ve heats|)

• Q_supply → sum of only +ve heat.

33. Solved Example (Efficiency)

• AB : Q₁ = nC_VΔT = 4(5R2)(T₀) = 10RT₀ (+ve)
• BC : Q₂ = nRT lnV₂V₁ = 4R(2T₀) ln3 = 8RT₀ ln3 (+ve)
• CD : Q₃ = 4(5R2)(−T₀) = −10RT₀ (−ve)
• DA : Q₄ = 4RT₀ ln13 = −4RT₀ ln3 (−ve)

34. Second Law of Thermodynamics

(i) Kelvin–Planck Statement

• Impossible for an engine working in a cyclic process to extract heat from a reservoir and convert it completely into work.
• 100% conversion of heat into work is impossible.
• Max^m efficiency of engine → Carnot engine.

(ii) Clausius Statement

• Impossible for a self-acting machine, unaided by any external agency, to transfer heat from a cold to hot reservoir.
• Heat cannot by itself flow from a colder to a hotter body.

35. Heat Engine

• Device which converts heat into work.

W = work done by engine, Q₁ = heat supplied by source

36. Refrigerator

• Coefficient of Performance (COP) :

• COP = Q₂W = Q₁ − WW = 1η − 1

37. Carnot Engine Cycle

• Two isothermal + two adiabatic processes.
• T₁ → temp. of source,  T₂ → temp. of sink

1→2 isothermal (Q₁ in) · 2→3 adiabatic · 3→4 isothermal (Q₂ out) · 4→1 adiabatic

Derivation :

• 2 → 3 : T₁V₂^γ−1 = T₂V₃^γ−1
• 4 → 1 : T₁V₁^γ−1 = T₂V₄^γ−1  ⇒ V₂V₁ = V₃V₄
• Heat in : Q₁ = nRT₁ lnV₂V₁
• Heat out : Q₂ = nRT₂ lnV₃V₄ = nRT₂ lnV₂V₁
• η = Q₁ − Q₂Q₁ = T₁ − T₂T₁ = 1 − T₂T₁

38. Series Carnot Engine System

• Engine 1 : T₁ → T  (η₁),  Engine 2 : T → T₂  (η₂)
• η₁ = 1 − TT₁ ⇒ T = (1 − η₁)T₁
• η₂ = 1 − T₂T ⇒ T = T₂1 − η₂
• Overall : η = 1 − T₂T₁ = W₁ + W₂Q₁

39. Entropy

• Measure of randomness / disorder of a system.

• Temperature always in kelvin.
• If C = constant :

• Heat = area under T–S curve :

40. Quick Formula Revision