πŸ”₯ JEE Main 2027 Session 1 is in January β€” start the JEE Mains Physics Test Series β†’

CLASS 11 Β· CHAPTER 11 Β· HEAT

QUICK REVISION

Thermodynamics

Every key result from this chapter, boxed and ready for a last look before the exam. No derivations here, just what to recall and when to use it β€” for the full explanation, see the detailed notes.

1. Zeroth Law & Thermal Equilibrium

Two systems each in thermal equilibrium with a third are in thermal equilibrium with each other. This is what makes a thermometer valid β€” touch each object to the same thermometer in turn, never to each other.

2. First Law of Thermodynamics

First law of thermodynamics
Ξ”U=Qβˆ’W\Delta U = Q - W

Q positive = heat absorbed by the system; W positive = work done by the system on its surroundings. Get a sign backwards and every later result flips.

W=∫V1V2P dVW = \int_{V_1}^{V_2} P\,dV

= area under the P-V curve, path-dependent. Ξ”U\Delta U is a state variable (endpoints only); Q absorbs whatever path-dependence is left over.

3. Isothermal vs. Adiabatic β€” Side by Side

  1. Isothermal (Ξ”T = 0, reservoir contact): Boyle’s law PV=nRT=constantPV = nRT = \text{constant}; since Ξ”U=0\Delta U = 0, first law gives Q=WQ = W.
  2. Adiabatic (Q = 0, insulated or fast): PVΞ³=constantPV^{\gamma} = \text{constant} (also TVΞ³βˆ’1=constantTV^{\gamma-1} = \text{constant}); Ξ”U=βˆ’W\Delta U = -W β€” compression does work ON the gas with zero heat input, so T rises with no heater anywhere (diesel-engine effect).
Isothermal work
W=nRTln⁑ ⁣(V2V1)W = nRT\ln\!\left(\dfrac{V_2}{V_1}\right)
Adiabatic work
W=P1V1βˆ’P2V2Ξ³βˆ’1W = \dfrac{P_1V_1 - P_2V_2}{\gamma - 1}

Ξ³=Cp/Cv\gamma = C_p/C_v is always >1> 1, so the adiabatic curve through any point is always steeper than the isothermal curve through that same point.

4. Specific Heats: Cp and Cv

Constant volume (W = 0)
Q=nCvΞ”T=Ξ”UQ = nC_v\Delta T = \Delta U
Constant pressure
Q=nCpΞ”T=Ξ”U+PΞ”VQ = nC_p\Delta T = \Delta U + P\Delta V
Specific heats of an ideal gas
Cpβˆ’Cv=RC_p - C_v = R

Cp always exceeds Cv: the same Ξ”T costs more heat at constant pressure because some of it leaks out as expansion work instead of staying behind as Ξ”U.

5. Heat Engines & the Second Law

W=Q1βˆ’Q2W = Q_1 - Q_2
Heat engine efficiency
Ξ·=WQ1=1βˆ’Q2Q1\eta = \dfrac{W}{Q_1} = 1 - \dfrac{Q_2}{Q_1}

Kelvin-Planck statement (second law): no process is possible whose sole result is absorbing heat from a reservoir and converting it completely into work β€” Ξ·=1\eta = 1 is forbidden; every real engine rejects some heat to a cold sink.

6. The Carnot Engine β€” the Efficiency Ceiling

Carnot efficiency
Ξ·Carnot=1βˆ’T2T1\eta_{Carnot} = 1 - \dfrac{T_2}{T_1}

T always in Kelvin. Four reversible steps between hot reservoir T₁ and cold reservoir Tβ‚‚:

  1. Isothermal expansion at T₁ β€” absorbs Q₁ from the hot reservoir.
  2. Adiabatic expansion β€” cools T₁ β†’ Tβ‚‚, no heat exchange.
  3. Isothermal compression at Tβ‚‚ β€” rejects Qβ‚‚ to the cold reservoir.
  4. Adiabatic compression β€” returns Tβ‚‚ β†’ T₁, closing the cycle.

These give Q2/Q1=T2/T1Q_2/Q_1 = T_2/T_1. No engine beats Ξ·Carnot\eta_{Carnot} between the same two temperatures; it reaches 1 only if T2=0Β KT_2 = 0\text{ K} β€” unreachable (third law).

7. Quick Numbers & Conversions

  • 1 cal = 4.186 J
  • 1 kPa Γ— 1 L = 1 J β€” convenient shortcut for P-V work
  • Ξ³=Cp/Cv\gamma = C_p/C_v, always greater than 1 for an ideal gas
  • T must be in Kelvin in every engine/Carnot formula

Full derivations and worked examples: detailed notes β†’