PhysicsJTHERM

Thermodynamics: Laws, Processes & Engines

Apply concepts from Thermodynamics: Laws, Processes & Engines to problem-solving. Focus on numerical practice, shortcuts, and real-world applications.

3%50 minPhase 1 · APPLICATIONMCQ + Numerical

Concept Core

Thermodynamic System and State Variables

A thermodynamic system is a definite quantity of matter bounded by a surface. State variables (P, V, T, U) describe the system's condition. For an ideal gas: PV=nRTPV = nRT.

Internal energy (UU): total kinetic energy of molecules. For an ideal gas, UU depends only on temperature: U=nCvT=f2nRTU = nC_vT = \frac{f}{2}nRT where ff is degrees of freedom (3 for monatomic, 5 for diatomic at moderate T, 6 for polyatomic).

Zeroth Law of Thermodynamics

If systems A and B are each in thermal equilibrium with system C, then A and B are in thermal equilibrium with each other. This establishes the concept of temperature.

First Law of Thermodynamics

Energy conservation for thermodynamic systems: ΔQ=ΔU+ΔW\Delta Q = \Delta U + \Delta W

  • ΔQ\Delta Q: heat added to the system (positive if absorbed)
  • ΔU\Delta U: change in internal energy (positive if increased)
  • ΔW\Delta W: work done by the system (positive if gas expands)

Work Done by a Gas

W=V1V2PdVW = \int_{V_1}^{V_2} P \, dV

Work equals the area under the P-V curve. Work is path-dependent (not a state function).

Thermodynamic Processes

1. Isothermal Process (TT = constant, ΔU=0\Delta U = 0): PV=constantW=nRTlnV2V1=nRTlnP1P2PV = \text{constant} \quad \Rightarrow \quad W = nRT \ln\frac{V_2}{V_1} = nRT \ln\frac{P_1}{P_2} ΔQ=ΔW=nRTlnV2V1\Delta Q = \Delta W = nRT \ln\frac{V_2}{V_1}

2. Adiabatic Process (ΔQ=0\Delta Q = 0): PVγ=constant,TVγ1=constant,TPγ/(γ1)=constantPV^\gamma = \text{constant}, \quad TV^{\gamma-1} = \text{constant}, \quad TP^{-\gamma/(\gamma-1)} = \text{constant} W=P1V1P2V2γ1=nR(T1T2)γ1W = \frac{P_1V_1 - P_2V_2}{\gamma - 1} = \frac{nR(T_1 - T_2)}{\gamma - 1} ΔU=W=nCv(T2T1)\Delta U = -W = nC_v(T_2 - T_1)

3. Isobaric Process (PP = constant): W=PΔV=P(V2V1)=nRΔTW = P\Delta V = P(V_2 - V_1) = nR\Delta T ΔQ=nCpΔT,ΔU=nCvΔT\Delta Q = nC_p\Delta T, \quad \Delta U = nC_v\Delta T

4. Isochoric Process (VV = constant, W=0W = 0): ΔQ=ΔU=nCvΔT\Delta Q = \Delta U = nC_v\Delta T

5. Polytropic Process (PVn=constantPV^n = \text{constant}): W=nR(T1T2)n1,Cpoly=Cvγn1nW = \frac{nR(T_1 - T_2)}{n - 1}, \quad C_{\text{poly}} = C_v\frac{\gamma - n}{1 - n}

P-V diagram comparing isothermal, adiabatic, isobaric, and isochoric processes

Specific Heats of Ideal Gas

Cv=f2R,Cp=Cv+R=f+22RC_v = \frac{f}{2}R, \quad C_p = C_v + R = \frac{f+2}{2}R γ=CpCv=f+2f\gamma = \frac{C_p}{C_v} = \frac{f+2}{f}

Gas TypeffCvC_vCpC_pγ\gamma
Monatomic332R\frac{3}{2}R52R\frac{5}{2}R53=1.67\frac{5}{3} = 1.67
Diatomic552R\frac{5}{2}R72R\frac{7}{2}R75=1.4\frac{7}{5} = 1.4
Polyatomic63R3R4R4R43=1.33\frac{4}{3} = 1.33

Second Law of Thermodynamics

Kelvin-Planck statement: No engine can convert heat entirely into work in a cyclic process. Some heat must be rejected to a cold reservoir.

Clausius statement: Heat cannot spontaneously flow from a colder body to a hotter body without external work.

Heat Engines

A heat engine absorbs heat QHQ_H from a hot reservoir, converts part to work WW, and rejects QCQ_C to a cold reservoir.

Efficiency: η=WQH=1QCQH\eta = \frac{W}{Q_H} = 1 - \frac{Q_C}{Q_H}

Carnot Engine (Maximum Efficiency)

Operating between temperatures THT_H and TCT_C (Kelvin): ηCarnot=1TCTH\eta_{\text{Carnot}} = 1 - \frac{T_C}{T_H}

Carnot cycle: isothermal expansion → adiabatic expansion → isothermal compression → adiabatic compression.

Carnot cycle on P-V diagram showing four processes

Refrigerator (Reverse Heat Engine)

Coefficient of Performance: COP=QCW=QCQHQC=TCTHTC\text{COP} = \frac{Q_C}{W} = \frac{Q_C}{Q_H - Q_C} = \frac{T_C}{T_H - T_C}


Key Testable Concept

---

Study Materials

Available in the NoteTube app — start studying for free.

100 Flashcards

SM-2 spaced repetition flashcards with hints and explanations

100 Quiz Questions

Foundation and PYQ-style questions with AI feedback

15 Study Notes

Structured notes across 10 scientifically grounded formats

10 Summaries

Progressive summaries from comprehensive guides to cheat sheets

Continue studying in NoteTube

Frequently Asked Questions

Common questions about studying Thermodynamics: Laws, Processes & Engines for JEE Main 2027.

Thermodynamics: Laws, Processes & Engines — JEE Main 2027 Physics | NoteTube