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PhysicsMediumClass 11

Thermodynamic Processes + PV Diagrams

Thermodynamics & KTG

95

JEE Qs

18%

Hard

75

min

Master the P-V diagram interpretation for different processes and carefully apply sign conventions for heat, work, and internal energy changes.

🧮 Key Formulas

First Law of Thermodynamics: ΔQ = ΔU + ΔW
Work done by gas: ΔW = ∫P dV
Work done for isobaric process: W = PΔV = nRΔT
Work done for isothermal process: W = nRT ln(Vf/Vi) = nRT ln(Pi/Pf)
Work done for adiabatic process: W = (P2V2 - P1V1)/(1-γ) = nR(T2 - T1)/(1-γ)
Internal energy change for ideal gas: ΔU = nCvΔT = n(f/2)RΔT
Mayer's relation: Cp - Cv = R
Adiabatic relations: PV^γ = constant, TV^(γ-1) = constant, P^(1-γ) T^γ = constant
Ratio of specific heats: γ = Cp/Cv = (f+2)/f
Polytropic process: PV^x = constant
Work done for polytropic process: W = (P2V2 - P1V1)/(1-x)
Molar heat capacity for polytropic process: C = Cv + R/(1-x)

✅ Key Points for JEE

  • 1Work done by a gas during a process on a P-V diagram is the area under the curve. For expansion, W > 0; for compression, W < 0. For a cyclic process, work done is the area enclosed by the cycle.
  • 2Internal energy (ΔU) of an ideal gas depends only on its temperature change (ΔU = nCvΔT), making it a state function. It is independent of the path taken between two states.
  • 3Heat (ΔQ) and Work (ΔW) are path functions; their values depend on the specific process connecting initial and final states.
  • 4The slope of an adiabatic process (PV^γ = const) on a P-V diagram is steeper (more negative) than that of an isothermal process (PV = const) at any given point: |slope_adiabatic| = γ |slope_isothermal|.
  • 5Memorize the characteristics and corresponding formulas for the four basic processes: Isochoric (V=const, W=0), Isobaric (P=const), Isothermal (T=const, ΔU=0), and Adiabatic (ΔQ=0).

⚠️ Common Mistakes

  • Sign errors for work done: Forgetting that work is positive for expansion and negative for compression, or using work done *on* the gas instead of *by* the gas.
  • Confusing the specific heat formulas (Cv, Cp, or polytropic C) or the constant relations for different thermodynamic processes (e.g., using PV=const for adiabatic process).
  • Incorrectly calculating the area under the P-V curve for complex or cyclic processes, especially when the path is not a simple rectangle or trapezoid.

📝 Practice Questions

See all

Q29.Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : With the increase in the pressure of an ideal gas, the volume falls off more rapidly in an isothermal process in comparison to the adiabatic process. Reason (R) : In isothermal process, PV = constant, while in adiabatic process PVγ = constant. Here γ is the ratio of specific heats, P is the pressure and V is the volume of the ideal gas. In the light of the above statements, choose the correct answer from the options given below : (1) Both (A) and (R) are true and (R) is the correct (2) Both (A) and (R) are true but (R) is NOT the explanation of (A) correct explanation of (A) (3) (A) is true but (R) is false (4) (A) is false but (R) is true 2025 (29 Jan Shift 2) JEE Main Previous Year Paper

2025·Assertion ReasoningMedium

Q40.A cup of coffee cools from 90∘C to 80∘C in t minutes when the room temperature is 20∘C. The time taken by the similar cup of coffee to cool from 80∘C to 60∘C at the same room temperature is : (1) 13 10 t (2) 1013 t (3) 13 5 t (4) 135 t

2025·NumericalMedium

Q42. A poly-atomic molecule ( CV = 3R, CP = 4R, where R is gas constant) goes from phase space point A (PA = 105 Pa, VA = 4 × 10−6 m3) to point B (PB = 5 × 104 Pa, VB = 6 × 10−6 m3) to point 2025 (29 Jan Shift 2) JEE Main Previous Year Paper C (PC = 104 Pa, VC = 8 × 10−6 m3). A to B is an adiabatic path and B to C is an isothermal path. The net heat absorbed per unit mole by the system is : (1) 500R(ln 3 + ln 4) (2) 450R(ln 4 −ln 3) (3) 500R ln 2 (4) 400R ln 4

2025·MCQHard

Q29.An amount of ice of mass 10−3 kg and temperature −10∘C is transformed to vapour of temperature 110∘C by applying heat. The total amount of work required for this conversion is, (Take, specific heat of ice = 2100Jkg−1 K−1 , specific heat of water = 4180Jkg−1 K−1 , specific heat of steam = 1920Jkg−1 K−1 , Latent heat of ice = 3.35 × 105Jkg−1 and Latent heat of steam = 2.25 × 106 Jkg−1 ) (1) 3043 J (2) 3024 J (3) 3003 J (4) 3022 J

2025·MCQMedium

Q41.Two spherical bodies of same materials having radii 0.2 m and 0.8 m are placed in same atmosphere. The temperature of the smaller body is 800 K and temperature of the bigger body is 400 K . If the energy radiated from the smaller body is E, the energy radiated from the bigger body is (assume, effect of the surrounding temperature to be negligible), (1) 16 E (2) E (3) 64 E (4) 256 E

2025·MCQMedium

Q49.Three conductors of same length having thermal conductivity k1, k2 and k3 are connected as shown in figure. Area of cross sections of 1st and 2nd conductor are same and for 3rd conductor it is double of the 1st conductor. The temperatures are given in the figure. In steady state condition, the value of θ is _______ ∘C. (Given : k1 = 60Js−1 m−1 K−1, k2 = 120Js−1 m−1 K−1, k3 = 135Js−1 m−1 K−1 ) ∣ ∣∣ 2025 (22 Jan Shift 1) JEE Main Previous Year Paper

2025·NumericalMedium

NCERT Chapters

  • Class 11 Physics Ch 12: Thermodynamics