Two rulebooks for odds in physics

Science · Thermal & Statistical Physics · ages 21-22

Name ______________________ Date ____________

At positive temperature, how is a canonical microstate's probability assigned?
- Divide its energy by the total of all energies
- Divide its Boltzmann factor by the partition function
- Give every state the same probability regardless of energy
Which quantities are fixed in the microcanonical ensemble?
- Energy, volume and particle number
- Temperature, volume and particle number
- Temperature, pressure and particle number
Dividing each canonical Boltzmann factor by the partition function makes the state probabilities sum to 1.
Circle one: True False
A sealed, rigid box is perfectly insulated and has no work input or other energy exchange. Which ensemble describes its equilibrium gas?
- Canonical, because insulation fixes the temperature
- Microcanonical, because E,V,N remain fixed
- Fixed-pressure ensemble, because sealing fixes pressure
A sealed, rigid flask exchanges heat with a large room and has equilibrated at its steady temperature. Which ensemble models the gas?
- Microcanonical: sealing prevents every energy exchange
- Fixed-pressure ensemble: the room fixes the gas pressure
- Canonical: T,V,N are fixed while energy can fluctuate
For an ordinary short-range system away from singular regimes, why can matched canonical and microcanonical ensembles approach the same local thermodynamic averages at large N?
- Canonical energy becomes sharply concentrated relative to its mean
- Canonical energy becomes exactly fixed for every finite particle count
- Canonical temperature must approach absolute zero as particle count grows
A five-particle system is in a heat bath. A friend claims ensemble equivalence guarantees every average matches a microcanonical calculation. Which reply is justified?
- Heat-bath contact makes the system perfectly isolated
- Five particles do not justify a large-system equivalence claim
- Every observable must differ whenever a system has five particles
A canonical system at positive finite T has two states of different energies. Rafi assigns them equal probabilities without an approximation. What is the problem?
- Equal probabilities follow whenever particle number is fixed
- Different energies require different volume constraints instead
- Canonical probabilities depend on energy through Boltzmann factors

Answer key

For grown-ups. Fold this page away before handing over the rest.

Two rulebooks for odds in physics W1-mt_NSCYLK-brt-s1

Divide its Boltzmann factor by the partition function · The canonical probability is exp(-E/(k_B T))/Z.
Energy, volume and particle number · The microcanonical constraints are E,V,N. The allowed states share the specified energy shell.
True · Z is the total weight. Summing the weights divided by Z gives Z/Z=1.
Microcanonical, because E,V,N remain fixed · All energy-transfer routes are excluded, and volume and particle number are fixed. These are microcanonical constraints.
Canonical: T,V,N are fixed while energy can fluctuate · The seal fixes N, rigidity fixes V, and the large room sets T through thermal contact.
Canonical energy becomes sharply concentrated relative to its mean · The relative root-mean-square energy fluctuation usually falls as 1/sqrt(N). This supports agreement of local averages, not equality of energy variances.
Five particles do not justify a large-system equivalence claim · The large-system approximation is not justified by five particles. Some averages could still agree, so universal disagreement is also unsupported.
Canonical probabilities depend on energy through Boltzmann factors · For different energies at positive finite T, the Boltzmann factors differ. Equal probabilities within an energy shell belong to the microcanonical description.