1.7.10 · D5Thermodynamics
Question bank — Internal energy of ideal gas U = (f - 2)nRT
True or false — justify
Each statement is either true or false. Say which, and — this is the whole point — say why in one line of real physics.
The internal energy of an ideal gas depends on its volume.
False. For an ideal gas there are no intermolecular forces, so no potential energy hides in the spacing between molecules — is pure kinetic energy and kinetic energy tracks only , giving .
If you double the volume of an ideal gas at constant temperature, its internal energy doubles.
False. Constant means constant molecular speeds; the molecules just have more room to roam. no matter what happens to .
Two gas samples at the same temperature always have the same internal energy.
False. also scales with the number of moles and the degrees of freedom . Same but different or different molecular structure gives different .
Heating a gas always increases its internal energy.
False. depends on , not on heat . In an isothermal expansion you supply heat yet , so — the heat leaves as work (see First law of thermodynamics).
For an ideal gas, is the same for any process connecting the same two temperatures.
True. is a state function of alone, so depends only on the endpoints, not the path taken between them.
The equipartition rule applies per molecule, not per mole.
True. Each quadratic degree of freedom carries per molecule; multiplying by molecules and using gives the per-mole form (see Equipartition theorem).
A monatomic and a diatomic gas at the same have the same average translational kinetic energy per molecule.
True. Translational KE per molecule is for both — it depends only on . The diatomic's extra energy lives in its 2 rotational modes, not in faster translation.
Internal energy includes the kinetic energy of the container sliding across a table.
False. is strictly microscopic random energy; the ordered bulk motion of the whole container is external kinetic energy, deliberately excluded from .
Spot the error
Each line contains a flawed statement or reasoning step. Find and fix it.
"For all gases, because translation has 3 degrees of freedom."
The error is using universally. That is only monatomic. Diatomic and polyatomic gases add rotational (and at high vibrational) modes, so can be 5, 6, or more — .
"In an isothermal expansion the gas does work, so its internal energy must fall."
For an ideal gas depends only on , and is constant, so . The work done is paid for by heat absorbed (), not by dipping into internal energy.
"I added of heat, so rose by ."
Only true at constant volume where . In general the first law says , so some of that heat may have become work — can be less than .
"A diatomic molecule spins about 3 axes, so it has 3 rotational degrees of freedom."
Rotation about the bond axis has negligible moment of inertia (mass sits on the line), so that mode is frozen out. Only the 2 perpendicular axes count, giving .
"Since depends only on , and is the same, an ideal gas and a real gas at the same have equal internal energy."
The " only" statement is an ideal-gas result. A real gas has intermolecular forces, hence potential energy that varies with volume, so its depends on as well.
"Equipartition gives every molecule exactly of energy."
It gives the average energy per molecule. Individual molecules have a wide spread of speeds; only the ensemble average equals .
"To find for a nitrogen gas process I need to know whether it was isobaric or isochoric."
You don't — needs only . The path matters for and separately, but not for (see Molar specific heats Cv and Cp).
Why questions
Answer the "why" with mechanism, not restatement.
Why does the internal energy of an ideal gas depend on temperature but not on pressure?
Temperature is a direct measure of average molecular kinetic energy, which is the internal energy for an ideal gas; pressure is just how often molecules hit the walls and carries no separate stored energy since there are no intermolecular forces.
Why do we multiply energy-per-molecule by and then swap for ?
Multiplying by turns the per-molecule average into the total energy of the whole sample; swapping converts the microscopic molecule count into the macroscopic, measurable quantity of moles (see Ideal gas law).
Why does each degree of freedom get exactly and not more?
Equipartition follows from statistical mechanics: any energy term that is quadratic in a coordinate or velocity independently thermalizes to the same average , because energy is shared democratically among all independent quadratic modes.
Why is called a "state function"?
Because its value is fixed entirely by the current state (for an ideal gas, by and ) — it has no memory of how the gas got there, so between two states is path-independent.
Why does adding rotational degrees of freedom make a gas harder to heat up (higher heat capacity)?
The supplied energy must be shared among more modes, so a given input raises the translational (temperature-setting) energy less — more of each joule goes into rotation, so more heat is needed per degree of temperature rise.
Why can a diatomic gas behave like a monatomic gas () at very low temperature?
Rotational modes require a minimum quantum of energy to activate; at low collisions are too gentle to excite them, so those modes are "frozen out" and only the 3 translational DOF respond.
Edge cases
Boundary and degenerate scenarios — the ones exams hide in.
What is for a free (Joule) expansion of an ideal gas into vacuum?
Zero. No heat is exchanged and no work is done against a vacuum, so , and since the temperature is unchanged too.
What is of an ideal gas at absolute zero, ?
Classically — all molecular motion has ceased. (Quantum zero-point energy is a real-gas refinement beyond the ideal model.)
For a cyclic process that returns to the starting state, what is the total ?
Exactly zero, because is a state function: same start and end state means same , so over the full cycle regardless of the loop's shape.
If a gas is heated but simultaneously does exactly as much work as the heat supplied, what happens to ?
It stays constant: . This is precisely the isothermal case, where all incoming heat converts to work.
What happens to (and hence to for the same ) as a diatomic gas is heated to very high temperature?
Vibrational modes switch on, adding 2 more DOF (one kinetic, one potential), pushing toward 7, so grows for the same once vibrations activate.
For a monatomic gas, can ever exceed 3 by heating?
No — a point-like single atom has no rotational or vibrational structure to excite, so stays 3 at all attainable temperatures.
Connections
- Kinetic theory of gases — where per molecule comes from
- Equipartition theorem — the -per-mode rule these traps lean on
- First law of thermodynamics — , root of the heat-vs-energy traps
- Degrees of freedom — why changes with structure and temperature
- Molar specific heats Cv and Cp — path dependence of vs path independence of
- Ideal gas law — the conversion