[PDF] Chemical & Statistical Thermodynamics Notes (CSIR NET & GATE)

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Chemical & Statistical Thermodynamics Handwritten Notes PDF

Master the most scoring mathematical unit of physical chemistry. High-yield handwritten classroom notes covering Maxwell relations, partition functions, residual entropy, and solved CSIR NET & GATE questions.

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💡 Recommended Companion Material: Check out our Thermodynamics Notes by Sahendra Sir and revise key operators with Quantum Chemistry Notes PDF.
📑 Table of Contents (Quick Navigation)
  1. 1. File Specifications & Syllabus Coverage
  2. 2. Classical Thermodynamics: Maxwell Relations & Equations of State
  3. 3. Thermodynamic Square Mnemonic (Maxborn Relations)
  4. 4. Statistical Thermodynamics: Partition Functions & Ensembles
  5. 5. Hierarchy of Energy Level Spacings & Partition Functions
  6. 6. Solved Benchmark CSIR NET Part-C Questions
  7. 7. Classroom Notes Sample Page Preview
  8. 8. Frequently Asked Questions (FAQs)
  9. 9. Direct High-Speed Download Mirror

📚 1. File Specifications & Exam Weightage

Thermodynamics (Classical + Statistical) represents 25 to 35 marks in CSIR NET Chemical Science and 8 to 12 marks in GATE (CY). Because the questions are direct application of fundamental formulas, this unit yields nearly 100% accuracy with systematic revision.

Module NameChemical and Statistical Thermodynamics (Career Notes)
Target ExamsCSIR UGC NET (JRF/LS), GATE Chemistry (CY), BARC, TIFR, SET, M.Sc./B.Sc.
File Size & Format7.4 MB • High-Resolution PDF (High-contrast, clean printer scan)
Authors / SourceCurated by Top Rankers of CSIR NET & GATE Chemical Sciences
Access Cost100% Free Direct Download via Google Drive

⚖️ 2. Classical Thermodynamics: Fundamental Equations

A. Four Fundamental Thermodynamic Relations
Internal Energy: dU = T dS − P dV
Enthalpy: dH = T dS + V dP
Helmholtz Energy: dA = −S dT − P dV
Gibbs Free Energy: dG = −S dT + V dP
B. The Four Maxwell Relations
Origin PotentialMaxwell RelationPhysical Significance
Internal Energy (U)(∂T/∂V)S = −(∂P/∂S)VAdiabatic reversible volume changes
Enthalpy (H)(∂T/∂P)S = (∂V/∂S)PAdiabatic reversible pressure changes
Helmholtz Energy (A)(∂S/∂V)T = (∂P/∂T)VDetermines internal pressure (πT)
Gibbs Energy (G)(∂S/∂P)T = −(∂V/∂T)PRelates entropy change to thermal expansivity (α)
C. Thermodynamic Equations of State
Internal Pressure: πT = (∂U/∂V)T = T(∂P/∂T)V − P
• For an Ideal Gas: πT = 0
• For a van der Waals Gas: πT = a / Vm2

🧩 3. Maxborn Thermodynamic Square Derivations

Use the mnemonic “Vampires And Trolls Go Prowling Here” to easily write down any Maxwell relation or differential equation:

Corner VariablesEdge PotentialsSign RuleShortcut Derivation
V (Volume) & T (Temperature)A (Helmholtz Energy)Both incoming arrows are negativedA = −P dV − S dT
T (Temperature) & P (Pressure)G (Gibbs Energy)T is negative, P is positivedG = −S dT + V dP
P (Pressure) & S (Entropy)H (Enthalpy)Both incoming arrows are positivedH = V dP + T dS
S (Entropy) & V (Volume)U (Internal Energy)S is positive, V is negativedU = T dS − P dV

🔬 4. Statistical Thermodynamics: Partition Functions

The molecular partition function (q) connects microscopic quantum states to macroscopic thermodynamic properties (U, H, S, G):

Total Molecular Partition Function: q = qtrans × qrot × qvib × qelec
Canonical Ensemble (Indistinguishable): Q = qN / N!   (Distinguishable: Q = qN)
ModePartition Function FormulaTemperature DependenceSymmetry Factor (σ)
Translational (3D)qtrans = (2πm kBT / h2)3/2 × Vqtrans ∝ T3/2 · M3/2—
Rotational (Linear)qrot = kBT / (σ h c B) = 8π2I kBT / (σ h2)qrot ∝ THomonuclear (σ=2), Heteronuclear (σ=1)
Rotational (Non-Linear)qrot = (π1/2 / σ) × (kBT / hc)3/2 × (1 / √(IAIBIC))qrot ∝ T3/2H2O (σ=2), NH3 (σ=3), CH4 (σ=12)
Vibrationalqvib = 1 / [1 − e−hν/kBT] ≈ kBT / hν (High T)qvib ∝ T0 (Low T) → T1 (High T)—
Thermodynamic Properties from Partition Function Q:
Internal Energy: U = kBT2 (∂ ln Q / ∂T)V
Entropy: S = kB ln Q + U / T
Helmholtz Energy: A = −kBT ln Q
Residual Entropy: Sres = kB ln W

📊 5. Hierarchy of Energy Level Spacings & Partition Functions

At standard room temperature (300 K), the energy gap (Δε) relative to thermal energy (kBT) dictates the number of populated quantum states:

1. Translational
Δε << kBT
qtrans ≈ 1028 – 1030
Extremely tiny energy gaps in macroscopic volumes. Behavior approaches a continuous spectrum. Highly populated at all temperatures.
Continuum of States
2. Rotational
Δε < kBT
qrot ≈ 10 – 100
Level spacing corresponds to the microwave region. Many rotational J-states are thermally populated at 300 K.
Dozens of Levels Populated
3. Vibrational
Δε > kBT
qvib ≈ 1.00 – 1.50
Spacing corresponds to the infrared region. Over 95% of molecules remain in the ground vibrational level (v = 0).
Mostly Ground State (v = 0)
4. Electronic
Δε >> kBT
qelec ≈ g0 (usually 1.0)
Spacing corresponds to UV-Visible radiation. Excited electronic states are completely unpopulated at ordinary room temperatures.
Ground State Degeneracy Only

💡 6. Solved Benchmark CSIR NET Exam Problems

CSIR NET Chemical Science (Part C • 4 Marks)

Problem 1: Calculate the molar residual entropy of crystalline Carbon Monoxide (CO) at absolute zero (0 K).

Step-by-Step Solution:

  • In crystalline solid CO, each molecule has two nearly degenerate orientations in the crystal lattice (C≡O vs. O≡C) due to a very small dipole moment (0.1 Debye).
  • Total number of microstates for 1 mole (NA molecules): W = 2NA.
  • Applying Boltzmann’s entropy formula:
    Sres = kB ln W = kB ln(2NA) = NA kB ln 2 = R ln 2.
  • Numerically: Sres = 8.314 × 0.693 = 5.76 J K−1 mol−1.
CSIR NET Chemical Science (Part C • 4 Marks)

Problem 2: What is the ratio of the rotational partition functions of HD and D2 at the same temperature T? (Assume bond lengths are identical: rHD = rD2).

Step-by-Step Solution:

  • Rotational partition function: qrot = 8π2I kBT / (σh2) ∝ μ / σ (since I = μr2).
  • Reduced masses:
    μHD = (1 × 2) / (1 + 2) = 2/3 amu
    μD2 = (2 × 2) / (2 + 2) = 1.0 amu
  • Symmetry numbers (σ):
    HD is heteronuclear → σHD = 1
    D2 is homonuclear → σD2 = 2
  • Taking the ratio:
    qrot(HD) / qrot(D2) = (μHD / σHD) / (μD2 / σD2) = (2/3 / 1) / (1 / 2) = (2/3) × 2 = 4/3 = 1.33.

🖼️ 7. Classroom Notes Sample Page Preview

Below is a sample preview from the high-resolution scanned PDF notes:

Chemical and Statistical Thermodynamics Notes PDF Preview
Figure 1: High-contrast sample page from Chemical & Statistical Thermodynamics notes.

❓ 8. Frequently Asked Questions (FAQs)

Q1: Are these thermodynamics notes sufficient for CSIR NET & GATE?

Yes. These classroom notes cover both classical thermodynamics (laws, Maxwell relations, chemical potential, non-ideal solutions) and statistical mechanics (ensembles, partition functions, residual entropy). Paired with previous year questions, it provides complete preparation.

Q2: Why is the symmetry number (σ) 2 for homonuclear diatomics?

In homonuclear diatomics like N2 or O2, a 180° rotation around the perpendicular axis leaves the molecule in an indistinguishable orientation. To avoid double-counting quantum microstates in phase space, the rotational partition function is divided by σ = 2.

Q3: How do I download the PDF file?

Click the download button below. The link directs to an open Google Drive mirror where you can read online or download the 7.4 MB PDF file directly without registration.

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