Heat and internal energy — microscopic vs macroscopic: Question bank

True or false — justify

Every one of these is a sentence you might hear in class. Decide true/false, then say why in one line.

A hot cup of coffee "contains a lot of heat."

False. It contains a large internal energy

U

; heat only exists while energy crosses a boundary due to a

Δ T

. "Heat in a body" is like "rain in a river."

If two bodies have the same temperature, they have the same internal energy.

False. Same

T

means same average molecular KE, but

U

also scales with how many molecules there are (

U = \frac{f}{2} n R T

, with

n

the mole count) — a bathtub and a teaspoon at

4 0^{\circ}

C differ hugely in

U

Adding heat to a system always raises its temperature.

False. In isothermal expansion heat enters yet

T

is unchanged because the gas does equal work; during a phase change heat goes into breaking bonds, not raising

T

Internal energy of an ideal gas can change even if its temperature stays fixed.

False. For an ideal gas

U

depends only on

T

, so

Δ T = 0 \Rightarrow Δ U = 0

, regardless of what

P

V

do.

Δ U

depends on the path taken between two states.

False.

U

is a state function, so

Δ U

depends only on start and end states. It is

Q

and

W

individually that are path-dependent.

Doing work on a gas is a different physical process from heating it, but both can raise

U

by the same amount.

True.

Q

(disordered energy from a

Δ T

) and

W

(ordered force through distance) are distinct transfers, yet

Δ U = Q - W

counts only their net effect on the stored energy.

For a real gas with intermolecular forces,

U

still depends only on temperature.

False. Real molecules attract/repel, so

U

includes a potential-energy term that changes with separation, hence with

V

. Only the ideal gas drops this term.

Heat always flows from the body with more internal energy to the body with less.

False. Heat flows from higher to lower temperature, not higher

U

. A large lukewarm object can have more

U

than a small hot one, yet still receive heat from the hot one.

Spot the error

Each line contains one specific mistake. Name it and correct it.

Δ U = Q + W

where

W

is the work done by the gas."

Sign error against our convention. With

W

= work done by the gas, the first law is

Δ U = Q - W

; the plus sign only applies when the symbol means work done on the gas.

"In a free expansion into vacuum, the gas does work, so its temperature drops."

The gas pushes against nothing, so

W = 0

; also

Q = 0

, hence

Δ U = 0

and (ideal gas)

T

is unchanged. No piston, no work.

"Because

C_{V} = \frac{3}{2} R

, all heat added to any process goes into internal energy."

Only true at constant volume (where

W = 0

). In general some heat leaves as work, so

Q \neq = Δ U

. See Specific Heats Cv and Cp.

"Temperature is the total kinetic energy of the molecules."

It is the average translational KE per molecule:

\frac{1}{2} m ⟨ v^{2} ⟩ = \frac{3}{2} k_{B} T

(here

m

is one molecule's mass). Total KE is

U

, which also depends on molecule count.

"A diatomic and a monatomic gas at the same

T

have the same internal energy per mole."

False — different degrees of freedom

f

. Diatomic has

f = 5

(adds rotation), so

U = \frac{5}{2} n R T

versus

\frac{3}{2} n R T

"Since

P V = \frac{1}{3} N m ⟨ v^{2} ⟩

, doubling the volume doubles

⟨ v^{2} ⟩

No —

⟨ v^{2} ⟩

is fixed by

T

alone. At fixed

T

, doubling

V

instead halves

P

, so the product

P V

(and hence

⟨ v^{2} ⟩

) stays constant.

"Heat is a form of matter that flows between bodies."

Heat is energy in transit, not a substance. This is the discredited caloric picture the topic warns against.

Why questions

Force yourself to give the mechanism, not just the label.

Why does the ideal-gas internal energy depend only on

T

and not on

V

Ideal molecules have no intermolecular forces, so there is no potential-energy term to change as they spread out — only KE remains, and KE is fixed by

T

Why does the first law carry a minus sign in front of

W

With

W

= work done by the gas: if the gas pushes a piston out, it spends its own stored energy, so

U

falls. Energy leaving as work must be subtracted.

Why can we use averages like

⟨ v^{2} ⟩

instead of tracking each molecule?

There are

\sim 1 0^{23}

molecules; individual paths are hopelessly chaotic, but their statistical averages are stable and reproducible — that is exactly what macroscopic variables measure. See Kinetic Theory of Gases.

Why does each degree of freedom get

\frac{1}{2} k_{B} T

of energy?

The equipartition theorem: energy shares equally among independent quadratic modes of motion in thermal equilibrium. Three translational directions give

3 \times \frac{1}{2} k_{B} T = \frac{3}{2} k_{B} T

per molecule. See Equipartition Theorem.

Why is heat not a state function like internal energy?

Heat only exists during a transfer driven by

Δ T

; once inside, the energy is just

U

. You cannot point at a state and say "this much heat is in it," so it has no state value.

Why must two bodies at the same temperature exchange no net heat?

With no temperature difference there is no thermal driving; molecular collisions transfer energy both ways equally. This is the basis of Temperature and the Zeroth Law.

Why does isothermal expansion require heat input even though

T

doesn't change?

The gas does work on the piston, which would drain its KE and cool it; heat flows in at exactly the rate needed to keep

⟨ v^{2} ⟩

(hence

T

) constant, so

Q = W

Edge cases

The scenarios that break naive rules — cover them all.

Constant-volume heating: what are

W

Q

Δ U

W = 0

(no volume change, no piston motion), so

Q = Δ U = \frac{f}{2} n R Δ T

. All heat becomes internal energy.

Isothermal ideal-gas process: what are

Δ U

and

Q

Δ U = 0

(since

T

fixed), so the first law gives

Q = W

— every joule in leaves as work.

Adiabatic process (

Q = 0

): how does

U

change when the gas expands?

Δ U = - W

: with no heat exchange the gas can only draw on its own energy to do work, so

U

(and

T

) drops. Compression does the reverse. See Isothermal and Adiabatic Processes.

Free expansion of an ideal gas into vacuum:

Q

W

Δ U

Δ T

All four are zero: no heat crosses, no work is done against vacuum, so

Δ U = 0

and

T

is unchanged despite the volume doubling.

Zero-Kelvin limit: what happens to

⟨ v^{2} ⟩

and

U

for an ideal gas?

From

\frac{1}{2} m ⟨ v^{2} ⟩ = \frac{3}{2} k_{B} T

, as

T \to 0

the average KE

\to 0

, so ideal-gas

U \to 0

. (Real matter hits quantum limits before this.)

A single molecule bouncing in a box — does "temperature" mean anything?

Not really: temperature is a statistical average over many molecules. One molecule has a KE but no well-defined thermodynamic

T

; the concept only sharpens as

N

grows large.

Two identical gas samples, one heated at constant

P

, one at constant

V

, to the same

Δ T

: which absorbs more heat?

The constant-

P

one, because it also does expansion work, so

Q_{P} = Δ U + W > Q_{V} = Δ U

. This is exactly why

C_{P} > C_{V}

. See Specific Heats Cv and Cp.

Recall One-line summary to lock in

Store vs transfer, path-independence, and $U (T)$ alone for the ideal gas — nearly every trap here is one of those three being confused.

The trap that catches most people

Confusing "heat added" with "internal energy raised" — they are equal only when no work is done (

V

constant).

Connections

Heat and internal energy — microscopic vs macroscopic
First Law of Thermodynamics
Kinetic Theory of Gases
Equipartition Theorem
Specific Heats Cv and Cp
Isothermal and Adiabatic Processes
Temperature and the Zeroth Law
Degrees of Freedom and Molecular Structure