GATE Chemistry (CY): Statistical Thermodynamics Basics
GATE CY physical chemistry note on statistical thermodynamics: partition functions, population ratios, U, A and S, with two solved practice questions and a st
This note explains the core ideas of statistical thermodynamics that GATE CY expects you to handle.
You get the concept, the exam angle, two worked practice questions and a short study plan.
The big idea: counting how molecules spread over levels
Classical thermodynamics deals with bulk quantities such as pressure and temperature. Statistical thermodynamics asks a deeper question. Given the energy levels of one molecule, how do about 1023 molecules share those levels at a temperature T?
The answer is the population law. For a level of energy Îľi and degeneracy gi (the number of states with that same energy), the population is proportional to gi eâÎľi/kT. Here k is the gas constant per molecule, k = R/NA. Three points follow from this form:
- Lower levels are more populated than higher levels, when the degeneracies are equal.
- A higher temperature makes the populations more even.
- Degeneracy multiplies the population. A level with g = 3 holds three times as much as a single-state level at the same energy.
Entropy is linked to counting through S = k ln W, where W is the number of microstates (distinct arrangements) consistent with the macroscopic state. Keep this in mind. It explains why entropy rises when more arrangements become available.
The partition function and what it gives you
The molecular partition function is q = ÎŁ gi eâÎľi/kT. Think of it as a count of the levels that are thermally reachable. At very low T, only the ground level counts, so q is close to g0. At high T, many levels count, so q is large.
For a whole system we use the canonical partition function Q. It describes a system at fixed N, V and T. The two are related as follows:
- Distinguishable particles (for example, atoms fixed at lattice sites): Q = qN.
- Indistinguishable particles (a gas): Q = qN/N!.
The N! corrects for counting the same state many times when identical molecules swap places. Forgetting it is a common mistake in entropy questions.
Once you have Q, the thermodynamic functions follow:
- Internal energy: U = kT2 (â ln Q/âT)V
- Helmholtz energy (the free energy at constant volume): A = âkT ln Q
- Entropy: S = U/T + k ln Q
Do not just memorise these. Check that S follows from U and A, because A = U â TS. A question may give you ln Q as a function of T and ask for the heat capacity. Then you differentiate once more: CV = (âU/âT)V.
Partition functions for each type of motion
For a molecule, the energy is treated as a sum of independent parts. So q is a product: q = qtrans Ă qrot Ă qvib Ă qelec. This is why each part can be worked out separately.
| Motion | High-temperature form | What to remember |
|---|---|---|
| Translation (3D) | q = V/Î3, with Î = h/(2ĎmkT)1/2 | Depends on V, m and T. Gives the ideal-gas behaviour. |
| Rotation (linear molecule) | q = T/(ĎÎrot) | Ď is the symmetry number: 1 for a heteronuclear diatomic, 2 for a homonuclear one. |
| Vibration (one mode) | q = 1/(1 â eâÎvib/T), energy zero at the ground level | Îvib = hν/k. Usually large, so few vibrational levels are populated near room temperature. |
| Electronic | q â g0 | Excited states are normally too high in energy to matter. |
Îrot and Îvib are characteristic temperatures. They compare the spacing of levels with kT. If T is much larger than Î, that motion is "fully excited" and behaves classically. This leads to the equipartition theorem: each quadratic term in the energy contributes ½kT per molecule. So a linear molecule has translational energy (3/2)kT and rotational energy kT.
Three kinds of statistics
You must know when to use which distribution.
- Classical statistics: distinguishable or dilute particles. Levels are sparsely occupied, as in an ordinary gas at normal conditions.
- Boson statistics: identical particles with integer spin (bosons). Any number can share a state.
- Fermion statistics: identical particles with half-integer spin (fermions). The exclusion principle allows at most one per state.
At high temperature and low density, both quantum distributions reduce to the classical form.
The exam angle
GATE CY questions on this topic are usually short and calculation-based. Statistical thermodynamics sits within the thermodynamics part of physical chemistry. Always check the current official syllabus for the exact list. The usual question types are:
- Population ratio of two levels, with or without degeneracy.
- Computing q for a two-level or few-level system.
- Deriving U, heat capacity or entropy from a given Q.
- Using characteristic temperatures to decide which motions contribute classically.
- Choosing the correct statistics for a described particle.
Speed matters. Practise writing eâÎľ/kT in terms of the ratio Îľ/kT, because numbers in such questions are often set up so that this ratio is a simple logarithm.
Two solved practice questions
These are practice questions written for this note. They are not past-paper questions.
Practice question 1. A system has a non-degenerate ground level at energy 0 and a triply degenerate excited level at energy Îľ. At temperature T, Îľ = kT ln 3. Find (a) q and (b) the ratio of the number of molecules in the excited level to that in the ground level.
Solution.
- The factor for the excited level is eâÎľ/kT = eâln 3 = 1/3.
- q = g0 + g1 eâÎľ/kT = 1 + 3 Ă (1/3) = 2.
- N1/N0 = (g1/g0) eâÎľ/kT = 3 Ă (1/3) = 1.
Answer: q = 2 and N1/N0 = 1. The excited level lies higher in energy, yet it holds as many molecules as the ground level. Degeneracy balances the exponential factor. Many students answer "less than 1" because they forget g.
Practice question 2. A hypothetical heteronuclear diatomic molecule has Îrot = 15 K. At 300 K, find qrot using the high-temperature form. Then find the rotational contribution to the molar internal energy. Finally, state how qrot changes if the molecule were homonuclear with the same Îrot.
Solution.
- T is much larger than Îrot (300 K against 15 K), so the high-temperature form is valid.
- qrot = T/(ĎÎrot) = 300/(1 Ă 15) = 20.
- For rotation, ln q = ln T + constant. Then the energy per molecule = kT2 (d ln q/dT) = kT2 Ă (1/T) = kT.
- Multiply by NA: the molar rotational energy is RT. This agrees with equipartition for a linear molecule.
- For a homonuclear molecule, Ď = 2, so qrot = 10. The energy is still RT, because Ď is a constant and disappears on differentiation.
Answer: qrot = 20 and the molar rotational energy is RT. For the homonuclear case qrot = 10, with the same energy. The symmetry number changes the entropy but not the energy.
A short study plan
Do this over three weeks, alongside your other physical chemistry topics.
- Days 1â3: Learn the population law and write q for a two-level system. Solve ten population-ratio problems, half of them with degeneracy.
- Days 4â7: Derive U, A and S from Q on paper. Repeat until you can write them without notes.
- Week 2: Learn the table of q for each motion. For each one, find U and CV by differentiating ln q. Check that your answers match equipartition.
- Week 3: Revise the three kinds of statistics, then attempt timed mixed sets. Review every wrong answer and note whether the error was in g, in N!, or in Ď.
Practise mixed sets from a chemistry question bank, and keep an error log. Do not move on until the log shows the same mistake has stopped repeating.
Key takeaways
- The population of a level is proportional to g eâÎľ/kT. Always include the degeneracy.
- Q = qN for distinguishable particles and qN/N! for indistinguishable ones.
- U = kT2(â ln Q/âT)V, A = âkT ln Q, and S = U/T + k ln Q.
- Compare T with Îrot and Îvib before choosing a high-temperature formula.
- The symmetry number Ď affects qrot and entropy, but not the rotational energy.
Frequently asked questions
What is the difference between q and Q?
q is the partition function of one molecule. Q is the partition function of the whole system of N molecules. For indistinguishable molecules, Q = qN/N!.
Why do we divide by N! for a gas?
Identical gas molecules cannot be told apart, so swapping two of them does not give a new state. Dividing by N! removes this overcounting.
Is the vibrational contribution significant at room temperature?
Usually not, for light diatomic molecules. Îvib is typically much larger than room temperature, so most molecules stay in the lowest vibrational level. Check Îvib against T in each question.
When can I use classical statistics for quantum particles?
Use it when the temperature is high and the density is low, so that the number of available states is far larger than the number of particles. Then boson and fermion statistics give almost the same result as the classical form.
Do I need to memorise the derivations?
Know the derivation of U, A and S from Q well enough to rebuild it. For the q of each motion, remember the final forms and the conditions under which they hold.
