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PhysicsMediumClass 11

First Law of Thermodynamics — ΔU = Q - W

Thermodynamics & KTG

21

JEE Qs

8%

Hard

75

min

Master the sign conventions for Q and W, understand that ΔU depends only on initial/final temperatures for ideal gases, and correctly apply formulas for work in different processes.

🧮 Key Formulas

ΔU = Q - W
W = ∫P dV (work done BY the system)
ΔU = nC_vΔT (for ideal gas, always)
W_isothermal = nRT ln(V_final / V_initial) = nRT ln(P_initial / P_final)
W_adiabatic = (P_initial V_initial - P_final V_final) / (γ - 1) = nR(T_initial - T_final) / (γ - 1)
C_v = R / (γ - 1)
C_p = γR / (γ - 1)
C_p - C_v = R

✅ Key Points for JEE

  • 1The First Law of Thermodynamics is a statement of energy conservation for a thermodynamic system, defining the relationship between internal energy change (ΔU), heat exchanged (Q), and work done (W).
  • 2Internal energy (ΔU) is a state function, depending only on the initial and final states (primarily temperature for an ideal gas). Q and W are path functions, dependent on the process.
  • 3Correct sign convention is crucial: Q is positive when heat is supplied TO the system; W is positive when work is done BY the system (expansion). ΔU is positive when internal energy increases.
  • 4For a cyclic process, ΔU = 0, which implies Q = W, meaning net heat absorbed equals net work done by the system.
  • 5ΔU = nC_vΔT is universally applicable for an ideal gas undergoing *any* process, as internal energy depends only on temperature.

⚠️ Common Mistakes

  • Using incorrect sign conventions for Q and W, leading to errors in the First Law equation.
  • Confusing work done *by* the system (W = ∫PdV) with work done *on* the system (W_on = -W). JEE problems typically refer to work done by the system unless specified.
  • Incorrectly assuming ΔU = 0 for processes other than isothermal or cyclic processes, especially if temperature change is not explicitly zero.
  • Applying nC_pΔT for ΔU for an ideal gas. ΔU is *always* nC_vΔT for an ideal gas, regardless of the process type.

📝 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