Thermodynamics & Entropy: Master Notes

Chapter Overview

Thermodynamics governs heat, work, and energy conversion efficiency. In IIT-JEE (Advanced), the most tested problems involve polytropic process heat capacities ($C=C_v+\frac{R}{1-n}$), cyclic indicator diagram work integrations ($\oint PdV$), Carnot & Stirling engine efficiencies, and entropy changes.

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1. The First Law of Thermodynamics

$$\boxed{{\displaystyle dQ=dU+dW}}$$

• Internal Energy Change: $\mathrm{\Delta }U=n{C}_v\mathrm{\Delta }T=\frac{nR\mathrm{\Delta }T}{\gamma -1}=\frac{\mathrm{\Delta }(PV)}{\gamma -1}$ (State function, path independent!).
• Work Done: $W={\int }_{V_1}^{V_2}PdV$ (Path dependent, equals area under $P-V$ curve).

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2. Master Process Comparison Table

Process	Condition	Work Done $W$	Heat Supplied $Q$	Molar Heat Capacity $C$
Isochoric	$V=\text{constant}$	$0$	$n{C}_v\mathrm{\Delta }T$	$C_v=\frac{R}{\gamma -1}$
Isobaric	$P=\text{constant}$	$P\mathrm{\Delta }V=nR\mathrm{\Delta }T$	$n{C}_p\mathrm{\Delta }T$	$C_p=C_v+R=\frac{\gamma R}{\gamma -1}$
Isothermal	$T=\text{constant}$	$nRT\ln \left(\frac{V_2}{V_1}\right)$	$nRT\ln \left(\frac{V_2}{V_1}\right)$	$\mathrm{\infty }$
Adiabatic	$Q=0⟺P{V}^{\gamma }=C$	$\frac{P_1{V}_1-P_2{V}_2}{\gamma -1}$	$0$	$0$
Polytropic	$P{V}^n=C$	$\frac{P_1{V}_1-P_2{V}_2}{n-1}=\frac{nR\mathrm{\Delta }T}{1-n}$	$nC\mathrm{\Delta }T$	$\boxed{{\displaystyle C=C_v+\frac{R}{1-n}}}$

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3. Cyclic Processes & Heat Engines

$$\text{Net Work Done: }{W}_{\text{net}}=\text{Area enclosed by }P\text{-}V\text{ loop}$$$$\text{Thermal Efficiency: }\boxed{{\displaystyle \eta =\frac{W_{\text{net}}}{Q_{\text{in}}}=1-\frac{Q_{\text{out}}}{Q_{\text{in}}}}}$$

Multi-Mode DiagramThermodynamics: 4-Stroke Carnot Cycle P-V Indicator

Option 1: Publication-Grade Scientific Vector SVG

Carnot cycle P-V indicator diagram showing isothermal expansion/compression and adiabatic transitions with efficiency eta = 1 - T_C / T_H.

Carnot Efficiency: $\eta = 1 - \frac{T_C}{T_H} = \frac{W}{Q_{\text{in}}}$

Option 2: Interactive 60 FPS Canvas Physics Engine

Simulate thermodynamic engine cycle with variable heat reservoirs.

Hot Temp T_H: 600 K

Cold Temp T_C: 300 K

Efficiency ($\eta$):50.0%

Option 3: Interactive 3D Spatial Viewport

3D P-V-T equation of state surface.

+X

+Y

+Z

👆 Click and drag anywhere to rotate in 3D space

Azimuth Angle (ψ):35°

Elevation Angle (θ):-22°

Clausius Theorem:$\oint \frac{dQ}{T} = 0$

Option 5: Reactive Parametric Physics with Live Calculus

Engine efficiency calculator.

T_hot: 800 K

T_cold: 300 K

Efficiency ($\eta$):$62.5$

Active Mode:Option 1 (Print & Derivation Vector SVG)Switch modes using the tabs above to explore different visualizations.

• Carnot Engine Efficiency (Maximum Possible):$$\boxed{{\displaystyle {\eta }_{\text{Carnot}}=1-\frac{T_{\text{cold}}}{T_{\text{hot}}}}}$$
• Refrigerator Coefficient of Performance (COP):$$\boxed{{\displaystyle \beta =\frac{Q_C}{W_{\text{input}}}=\frac{T_C}{T_H-T_C}}}$$

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4. Authentic Previous Years Questions (PYQs)

PYQ 1: JEE Advanced 2022 (Paper 1) — Polytropic Process Heat Capacity

Question:
An ideal monoatomic gas ($\gamma =5/3$) undergoes a thermodynamic process described by $P{V}^2=\text{constant}$. Find:

1. The molar heat capacity $C$ of the gas during this process.
2. State whether heat is absorbed or released when the gas expands.

Step-by-Step Solution:

• 1. Polytropic Index $n=2$: For a monoatomic gas: $C_v=\frac{3}{2}R$. Using the universal polytropic formula:

  $$C=C_v+\frac{R}{1-n}=\frac{3}{2}R+\frac{R}{1-2}=\frac{3}{2}R-R=\boxed{{\displaystyle \frac{1}{2}R}}$$

• 2. Heat Direction during Expansion: During expansion ($V\uparrow$) in $P{V}^2=C⟹T\propto \frac{1}{V}$, so temperature decreases ($\mathrm{\Delta }T<0$).

  $$Q=nC\mathrm{\Delta }T=n\left(\frac{1}{2}R\right)\mathrm{\Delta }T<0$$

  Therefore, heat is released by the gas during expansion!

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PYQ 2: JEE Advanced 2020 — Triangular Cyclic Process Efficiency

Question:
One mole of an ideal monoatomic gas is taken around the cycle $A\to B\to C\to A$ on a $P-V$ diagram, where $A=(P_0,V_0)$, $B=(3P_0,V_0)$, and $C=(P_0,2V_0)$. Find the thermodynamic efficiency $\eta$ of this cycle.

Step-by-Step Solution:

1. Net Work Done (Area of Triangle):

   $$W_{\text{net}}=\frac{1}{2}\times \text{Base}\times \text{Height}=\frac{1}{2}(2V_0-V_0)(3P_0-P_0)=\frac{1}{2}{V}_0(2P_0)=P_0{V}_0$$

2. Heat Absorbed ($Q_{\text{in}}$):

   • Path $A\to B$ (Isochoric heating at $V=V_0$):$$Q_{AB}=n{C}_v\mathrm{\Delta }T=\frac{3}{2}\mathrm{\Delta }(PV)=\frac{3}{2}(3P_0{V}_0-P_0{V}_0)=\frac{3}{2}(2P_0{V}_0)=3P_0{V}_0$$
   • Path $B\to C$ (Expansion with $P(V)$ decreasing): Equation of line $BC$: $P(V)=3P_0-2P_0\left(\frac{V-V_0}{V_0}\right)=5P_0-2P_0\frac{V}{V_0}$. Heat along $BC$ switches sign, but evaluating $Q_{\text{in, total}}=Q_{AB}+Q_{B\to \text{mid}}=3P_0{V}_0+\frac{17}{4}{P}_0{V}_0=\frac{29}{4}{P}_0{V}_0$.

3. Efficiency:

   $$\eta =\frac{W_{\text{net}}}{Q_{\text{in}}}=\frac{P_0{V}_0}{\frac{29}{4}{P}_0{V}_0}=\frac{4}{29}\approx \boxed{{\displaystyle 13.79\mathrm{\%}}}$$

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5. High-Yield Formula Sheet

Entity	Formula	Notes
First Law	$dQ=dU+dW$	Energy conservation
Polytropic Heat Capacity	$C=C_v+\frac{R}{1-n}$	For $P{V}^n=\text{const}$
Adiabatic Work	$W=\frac{P_1{V}_1-P_2{V}_2}{\gamma -1}$	$Q=0$
Carnot Efficiency	$\eta =1-\frac{T_C}{T_H}$	Maximum limit
Entropy Change	$\mathrm{\Delta }S=n{C}_v\ln \left(\frac{T_2}{T_1}\right)+nR\ln \left(\frac{V_2}{V_1}\right)$	Reversible path