Sackur-Tetrode Equation: Ideal Gas Entropy Calculations
Calculating the absolute entropy of a substance without integrating heat capacity data from absolute zero requires statistical thermodynamics. For an ideal monatomic gas, Otto Sackur and Hugo Tetrode derived the fundamental equation linking microscopic atomic parameters to macroscopic entropy. In competitive examinations including JEE Advanced, IIT-JAM, GATE, CSIR-NET, and CUET, the Sackur-Tetrode equation evaluates understanding of partition functions, quantum states, and gas laws.
Physical Basis and Derivation Outline
Consider an ideal gas of N indistinguishable atoms of mass m confined in volume V at temperature T. Indistinguishability requires dividing the canonical partition function by N! to prevent Gibbs paradox:
Q = (qN) / N!
For a monatomic gas at moderate temperatures, energy is purely translational. The particle-in-a-box translational partition function is:
qtrans = V × (2πmkBT / h2)3/2 = V / Λ3
Where Λ = h / √(2πmkBT) is the thermal de Broglie wavelength.
Using S = kB ln Q + U / T with U = (3/2)NkBT and applying Stirling's approximation ln(N!) ≈ N ln N − N yields the Sackur-Tetrode equation:
S = NkB [ ln( (V / N) × (2πmkBT / h2)3/2 ) + 5/2 ]
The constant 5/2 combines 3/2 from internal kinetic energy with 1 from Stirling's approximation.
Key Formula and Thermodynamic Conventions
For 1 mole of gas (N = NA), setting V / N = kBT / P gives the standard molar entropy formula:
Or using thermal wavelength: Sm = R · [ ln( Vm / (NA · Λ3) ) + 5/2 ]
Where: Sm = molar entropy (J·mol−1·K−1); R = 8.314 J·mol−1·K−1; kB = 1.381 × 10−23 J·K−1; h = 6.626 × 10−34 J·s; m = atomic mass in kg (m = M / NA); T = temperature in K; P = pressure in Pa (1 bar = 105 Pa; 1 atm = 101,325 Pa); Vm = molar volume in m3/mol; Λ = thermal wavelength in metres.
Textbooks differ on two conventions: IUPAC defines standard pressure as P° = 1 bar (105 Pa; 22.71 L/mol at 273.15 K), while older texts use P° = 1 atm (101,325 Pa; 22.41 L/mol), differing by ΔS = R ln(1.01325) ≈ 0.109 J·mol−1·K−1. Also, IUPAC uses ΔU = q + w, whereas older physics texts write ΔU = q − w.
Exam details
- Source: practice problem written for this article (not a past-paper question)
- Relevant exams: JEE Advanced, IIT-JAM, GATE, CSIR-NET, CUET PG
- Topic: Statistical Thermodynamics and Ideal Gas Entropy
- Difficulty: Advanced
Worked Numerical Examples
Question: Calculate the absolute standard molar entropy of helium gas (monatomic, M = 4.003 g/mol = 0.004003 kg/mol) at T = 298.15 K and P = 1.000 × 105 Pa (1 bar) using the Sackur-Tetrode equation. Use kB = 1.381 × 10−23 J/K, h = 6.626 × 10−34 J·s, NA = 6.022 × 1023 mol−1, and R = 8.314 J/(mol·K).
Solution:
Step 1: Calculate mass of a single helium atom.
m = M / NA = 0.004003 / (6.022 × 1023) = 6.647 × 10−26 kg.
Step 2: Compute thermal wavelength Λ.
2πmkBT = 2 × 3.1416 × (6.647 × 10−26) × (1.381 × 10−23) × 298.15 = 1.720 × 10−46.
√(2πmkBT) = 4.147 × 10−23 kg·m·s−1.
Λ = (6.626 × 10−34) / (4.147 × 10−23) = 5.054 × 10−11 m.
Λ3 = 1.291 × 10−31 m3.
Step 3: Calculate volume per particle v.
v = kBT / P = (1.381 × 10−23 × 298.15) / 105 = 4.117 × 10−26 m3.
Step 4: Evaluate argument of logarithm and molar entropy.
Argument = v / Λ3 = (4.117 × 10−26) / (1.291 × 10−31) = 3.189 × 105.
ln(3.189 × 105) = 12.673.
Sm = R × [12.673 + 2.500] = 8.314 × 15.173 = 126.15 J/(mol·K).
Double check by second route:
Using molar volume:
Vm = RT / P = (8.314 × 298.15) / 105 = 0.02479 m3/mol.
Vm / (NA · Λ3) = 0.02479 / [(6.022 × 1023) × (1.291 × 10−31)] = 3.189 × 105.
Sm = 8.314 × [ln(3.189 × 105) + 2.5] = 126.15 J/(mol·K). Both routes match.
Answer: 126.15 J/(mol·K)
Question: At 298.15 K and 1.000 bar, calculate the molar entropy difference ΔSm = Sm(Ar) − Sm(He) between argon gas (M = 39.95 g/mol) and helium gas (M = 4.003 g/mol), treating both as ideal monatomic gases. Use R = 8.314 J/(mol·K).
Solution:
Step 1: Simplify Sackur-Tetrode formula at constant T and P.
All terms cancel except atomic mass: Sm = (3/2)R ln(M) + Constant(T, P).
Step 2: Set up entropy difference expression.
ΔSm = Sm(Ar) − Sm(He) = (3/2)R ln(MAr / MHe).
Step 3: Evaluate mass ratio and natural logarithm.
Ratio = 39.95 / 4.003 = 9.980.
ln(9.980) = 2.3006.
Step 4: Compute final entropy difference.
ΔSm = 1.5 × 8.314 J/(mol·K) × 2.3006 = 12.471 × 2.3006 = 28.69 J/(mol·K).
Double check by second route:
Using base-10 logarithm:
ΔSm = 1.5 × 2.3026 × R × log10(9.980) = 1.5 × 2.3026 × 8.314 × 0.99913 = 28.69 J/(mol·K). Both routes agree.
Answer: 28.69 J/(mol·K)
Common Mistakes to Avoid
- Using Molar Mass in Grams: In (2πmkBT / h2), particle mass m must be in kilograms (kg), not grams.
- Confusing ln with log10: Sackur-Tetrode uses natural logarithm (base e). Using base-10 without converting leads to severe errors.
- Forgetting the 5/2 Term: Omitting the constant 5/2 neglects kinetic energy and indistinguishability contributions.
- Applying to Polyatomic Gases: Sackur-Tetrode accounts only for translational entropy. Polyatomic gases require rotational and vibrational terms.
- Incorrect Pressure Units: When using V / N = kBT / P, pressure P must be expressed in Pascals (N/m2), not bar or atm.
Exam Relevance and Question Patterns
| Examination | Question Focus | Key Skills Tested |
|---|---|---|
| JEE Advanced | Dependence of S on molar mass, temperature, and volume | Applying S ∝ (3/2)R ln M and ΔS for expansion/heating |
| IIT-JAM Chemistry | Translational partition function and absolute entropy numericals | Evaluating thermal de Broglie wavelength and molar entropy |
| GATE & CSIR-NET | Statistical thermodynamics, indistinguishability, Gibbs paradox | Deriving S from Q, comparing noble gas standard entropies |
| CUET PG | Direct formula substitution and ratio-based numericals | Connecting macroscopic thermodynamic S to microscopic variables |
Turn this concept into marks: solve previous-year CSIR-NET questions and take a timed quiz.
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