Thermodynamics By Sahender Sir Handwritten Pdf

Thermodynamics Handwritten Notes For IIT-JAM/CSIR-NET /GATE by ChemistryABC.com
Ad
ChemistryABC Banner
Sponsored / Ad
Thermodynamics Handwritten Notes PDF Preview by Sahendra Sir
Physical Chemistry Master Series • Topper Notes

Chemical & Statistical Thermodynamics Handwritten Notes PDF (By Sahendra Sir)

Complete, high-yield classroom notes compiled by top rankers from Sahendra Sir’s coaching sessions. Covers classical laws of thermodynamics, Maxwell relations, Joule-Thomson expansion, statistical partition functions, and residual entropy calculations for CSIR NET & GATE.

📌 Why Thermodynamics is a Scoring Heavyweight

In CSIR UGC NET Chemical Sciences and GATE Chemistry, Classical and Statistical Thermodynamics collectively carry 16 to 24 marks. Unlike highly conceptual descriptive subjects, thermodynamics questions are strictly equation-governed, making them highly predictable and rewarding for prepared candidates.

  • Direct Maxwell Relations: 4-mark Part C questions regularly ask for thermodynamic equations of state, internal pressure derivations, and Joule-Thomson inversion temperatures.
  • Statistical Mechanics Shortcut Formulas: Questions on molecular partition functions (qtrans, qrot, qvib) and residual entropy follow standard algebraic rules.
  • Zero Ambiguity: Numerical and conceptual problems have definitive mathematical answers, ensuring maximum scoring efficiency.

🔬 Core Topics Covered in These Notes

1. First Law of Thermodynamics & Thermochemistry

  • State Functions vs. Path Functions: Exact vs. inexact differentials, Euler’s reciprocity condition, cyclic rule: (∂P/∂V)T (∂V/∂T)P (∂T/∂P)V = −1.
  • Internal Energy (U) & Enthalpy (H): Heat capacity relations for ideal and real gases:
    Cp − Cv = T (∂P/∂T)V (∂V/∂T)P
    For an ideal gas: Cp − Cv = R. For a van der Waals gas: Cp − Cv ≈ R [ 1 + (2a / RTV) ].
  • Reversible vs. Irreversible Expansions: Work done in isothermal reversible expansion (w = −nRT ln(V2/V1)) versus irreversible expansion against constant external pressure; adiabatic reversible expansion (PVγ = constant, TVγ−1 = constant).
  • Thermochemistry: Kirchhoff’s law of heat of reaction with temperature (ΔH2 − ΔH1 = ∫ ΔCp dT), Hess’s law, and bond enthalpy calculations.

2. Second Law, Carnot Engine & Entropy Calculations

  • Carnot Cycle Efficiency (η):
    η = (−wnet / qh) = (Th − Tc) / Th = 1 − (Tc / Th)
  • Clausius Inequality: ∮ (dq / T) ≤ 0 (equality holds for reversible processes, strict inequality for spontaneous irreversible processes).
  • Entropy Changes for Ideal Gases:
    • In terms of T and V: ΔS = Cv ln(T2 / T1) + R ln(V2 / V1)
    • In terms of T and P: ΔS = Cp ln(T2 / T1) − R ln(P2 / P1)
    • Phase transitions: ΔStrans = ΔHtrans / Ttrans (Trouton’s rule: ΔSvap ≈ 88 J K−1 mol−1 for non-associated liquids).
    • Entropy of mixing of ideal gases: ΔSmix = −R ∑ ni ln xi (always positive, ΔGmix = −T ΔSmix < 0).

3. Maxwell Relations & Thermodynamic Equations of State

The four fundamental differential equations and their corresponding Maxwell relations:

  • dU = TdS − PdV ⇒ (∂T/∂V)S = −(∂P/∂S)V
  • dH = TdS + VdP ⇒ (∂T/∂P)S = (∂V/∂S)P
  • dA = −SdT − PdV ⇒ (∂S/∂V)T = (∂P/∂T)V
  • dG = −SdT + VdP ⇒ (∂S/∂P)T = −(∂V/∂T)P
Internal Pressure (πT):
πT = (∂U/∂V)T = T (∂P/∂T)V − P
For an ideal gas, πT = 0. For a van der Waals gas, πT = a / Vm2.
Joule-Thomson Coefficient (μJT):
μJT = (∂T/∂P)H = (1 / Cp) [ T (∂V/∂T)P − V ]
Inversion Temperature: Ti = 2a / (Rb). At T < Ti, gas cools upon expansion (μJT > 0).

4. Phase Equilibria & Chemical Potential

  • Clausius-Clapeyron Equation: ln(P2 / P1) = (ΔHvap / R) [ (1 / T1) − (1 / T2) ].
  • Gibbs-Helmholtz Equation: [ ∂(ΔG / T) / ∂T ]P = −ΔH / T2.
  • Gibbs-Duhem Equation: ∑ ni dμi = −SdT + VdP (at constant T and P: ∑ ni dμi = 0).

5. Statistical Thermodynamics & Residual Entropy

  • Molecular Partition Functions (q):
    • Translational: qtrans = [ (2πmkT)3/2 / h3 ] × V ∝ T3/2 V.
    • Rotational: Linear molecules: qrot = kT / (σBhc) ∝ T. Non-linear molecules: qrot ∝ T3/2.
    • Vibrational: qvib = 1 / [ 1 − exp(−hν / kT) ].
    • Electronic: qelec = g0 + g1 exp(−Δε1 / kT).
  • Canonical Partition Function (Q): For indistinguishable particles: Q = qN / N!. For distinguishable particles: Q = qN.
  • Residual Entropy (Third Law of Thermodynamics): Calculated using Boltzmann’s formula: S0 = k ln W.
    • Carbon monoxide (CO): 2 orientations ⇒ W = 2N ⇒ S0 = R ln 2 ≈ 5.76 J K−1 mol−1.
    • Nitrous oxide (N2O): 2 orientations ⇒ W = 2N ⇒ S0 = R ln 2 ≈ 5.76 J K−1 mol−1.
    • Ice (H2O): Pauling’s model with W = (3/2)N ⇒ S0 = R ln(1.5) ≈ 3.37 J K−1 mol−1.

📊 Exam Weightage: Thermodynamics

Mark distribution across competitive chemistry examinations:

ExaminationType of QuestionAverage QuestionsTotal Marks
CSIR UGC NET (Part C)4-Mark Analytical (Maxwell relations, Statistical Thermo)3 – 5 Questions12 – 20 Marks
CSIR UGC NET (Part B)2-Mark Factual (Carnot η, Residual entropy, Work done)2 – 3 Questions4 – 6 Marks
GATE Chemistry1-Mark & 2-Mark MCQs / Numerical Answer Types (NATs)4 – 6 Questions7 – 10 Marks
BARC / TIFRClassical Equations of State & Ensembles3 – 4 Questions9 – 12 Marks
  1. Draw the Thermodynamic Square (Mnemonic): Practice writing all four Maxwell relations and internal energy relations in under 30 seconds using the standard thermodynamic square mnemonic.
  2. Memorize Partition Function Dependencies on T: Remember that qtrans ∝ T3/2, qrot(linear) ∝ T, and qrot(non-linear) ∝ T3/2. These temperature dependencies appear frequently in GATE NAT questions.
  3. Differentiate Between Reversible and Irreversible Work: Always check if expansion occurs against constant external pressure (irreversible) or slowly in equilibrium (reversible) before choosing formulas.

❓ Frequently Asked Questions (FAQ)

Q1. How do I calculate the residual entropy of ice (H2O) at absolute zero?

Using Pauling’s hydrogen-bonding model, a water molecule in ice has 6 possible orientations but only 1/4 of them satisfy hydrogen-bonding requirements per pair, giving W = (3/2)N. The residual entropy is S0 = R ln(3/2) = R ln(1.5) ≈ 3.37 J K−1 mol−1.

Q2. What is the value of the internal pressure for an ideal gas vs. a van der Waals gas?

For an ideal gas, the internal pressure πT = (∂U/∂V)T = 0 because there are no intermolecular attractive forces. For a van der Waals gas, πT = a / Vm2, reflecting intermolecular attractive interactions.

Q3. What is the Joule-Thomson inversion temperature formula?

The inversion temperature (Ti) is given by Ti = 2a / (Rb), where a and b are van der Waals constants. When a gas expands below its inversion temperature (T < Ti), it experiences cooling (μJT > 0).

Disclaimer & Educational Fair Use: ChemistryABC.com is an open educational platform dedicated to assisting chemistry students and competitive exam aspirants. These study notes are student classroom summaries prepared for academic reference and self-study purposes. We encourage learners to support original authors and publishers by purchasing authorized textbooks.

Ad
ChemistryABC Banner
Sponsored / Ad

“ChemistryABC.com has always been very supportive of the Science community. Membership feels like a huge effort towards a strong, diverse community of people who care about their Education. Highly recommended for every Student & Tutor who wants to learn and connect — even those who don’t like Chemistry.”

19 Comments

  1. Due to thousands of students downloading so speed of this page may down or some time you will see the database errors.
    Keep calm and revisit after some time if the page unavailable.
    Sorry for the inconvenience caused.
    Chemistry ABC team.

Leave a Reply

*