Heat and internal energy — microscopic vs macroscopic: Worked examples

This is the "roll up your sleeves" companion to Heat and Internal Energy — Microscopic vs Macroscopic. The parent note built the ideas; here we hit every kind of scenario the topic can throw at you, one worked example per cell.

Before we begin, a promise: every symbol below was defined in the parent, except a few we add here. We use

$U$ = internal energy (energy stored inside), $Δ U$ = its change,
$Q$ = heat (energy crossing the boundary because of a temperature difference),
$W$ = work done by the gas (energy leaving as ordered pushing),
$n$ = number of moles, $R = 8.314 J mol^{- 1} K^{- 1}$ (gas constant), $T$ = temperature in kelvin,
$f$ = degrees of freedom, and the two master formulas $Δ U = Q - W and U = \frac{f}{2} n R T .$

Every problem below is just bookkeeping with these lines.

The scenario matrix

Think of "every scenario" as choosing a value for three knobs plus a few special cases. The table lists each cell and the example that lands on it.

Cell	What is special about it	$Q$	$W$	Sign of $Δ U$	Covered by
A Heat in, gas does work	classic first-law bookkeeping	$> 0$	$> 0$	$+$	Ex 1
B Constant volume ( $W = 0$ ), heating	all heat $\to$ internal energy	$> 0$	$0$	$+$	Ex 2
C Isothermal ( $Δ U = 0$ )	heat in = work out exactly	$> 0$	$> 0$	$0$	Ex 3
D Adiabatic ( $Q = 0$ )	no heat; work changes $U$	$0$	$> 0$	$-$ (expansion)	Ex 4
E Sign flip: work done on gas, heat leaks out	both negative	$< 0$	$< 0$	$+$	Ex 5
F Heat out, gas compressed	both negative	$< 0$	$< 0$	either sign	Ex 6
G Degenerate: free expansion into vacuum	$Q = 0, W = 0$	$0$	$0$	$0$	Ex 7
H Degrees of freedom change ( $f$ : 3 vs 5)	diatomic gas	$> 0$	$0$	$+$	Ex 8
I Real-world word problem	tyre / pump framing	context	context	context	Ex 9
J Exam twist: cyclic process	$Δ U = 0$ over a full loop	net $> 0$	net $> 0$	$0$	Ex 10
K Heat in and gas compressed	both add to $U$	$> 0$	$< 0$	$+$	Ex 11
L Heat out while gas still does work	opposite signs	$< 0$	$> 0$	$-$	Ex 12
M Constant-volume cooling	heat removed, no work	$< 0$	$0$	$-$	Ex 13
N Heat in but too much work out	net cooling despite heating	$> 0$	$> 0$ (with $W > Q$ )	$-$	Ex 14

Ex 1 — Cell A: heat in, gas does work

Ex 2 — Cell B: constant volume, all heat becomes internal energy

Worked example Statement

Heat $2 mol$ of monatomic gas from $300 K$ to $330 K$ inside a rigid sealed can. How much heat was needed?

Forecast: a rigid can can't move — so where can the heat possibly go?

Rigid $\Rightarrow$ volume fixed $\Rightarrow W = 0$ . Why this step? Work is force through a distance; a wall that cannot move means zero distance, zero work.
Monatomic $\Rightarrow f = 3 \Rightarrow U = \frac{3}{2} n R T$ , so $Δ U = \frac{3}{2} n R Δ T = \frac{3}{2} (2) (8.314) (30) = 748.26 J .$ Why this step? $Δ U$ needs only the start and end temperatures — it's a state function, so the path is irrelevant.
First law with $W = 0$ : $Q = Δ U = 748.26 J$ . Why this step? With the work term gone, $Δ U = Q - 0$ collapses to $Q = Δ U$ — the heat has nowhere to go but into storage.

Verify: All heat went to internal energy because no work escaped — this is precisely why $C_{V} = \frac{3}{2} R$ (defined above; see Specific Heats Cv and Cp). Sanity: heating raised $T$ , so $Q > 0$ . ✓

Ex 3 — Cell C: isothermal, $Δ U = 0$

Worked example Statement

$1 mol$ of ideal gas expands isothermally at $T = 350 K$ from volume $V_{1}$ to $V_{2} = 3 V_{1}$ . Find $Δ U$ , $W$ , and $Q$ .

Forecast: heat is clearly flowing in — so surely $U$ goes up? (It doesn't. Watch.)

Isothermal means $T$ constant. For an ideal gas $U = \frac{f}{2} n R T$ depends only on $T$ , so $Δ U = 0$ . Why this step? No temperature change $\Rightarrow$ average molecular speed unchanged $\Rightarrow$ stored energy unchanged.
Isothermal work of an ideal gas (from Isothermal and Adiabatic Processes): $W = n R T ln \frac{V _{2}}{V _{1}} = (1) (8.314) (350) ln 3 = 3196.87 J .$ Why this step? At constant $T$ the pressure $P$ falls as the gas spreads out; adding up the little pushes $P Δ V$ with $P = n R T / V$ (from $P V = n R T$ ) gives the logarithm.
First law: $Q = Δ U + W = 0 + 3196.87 = 3196.87 J$ . Why this step? Rearranging $Δ U = Q - W$ gives $Q = Δ U + W$ ; with $Δ U = 0$ the heat needed equals exactly the work performed.

Verify: $Q = W$ exactly — every joule of heat entering leaves immediately as work, keeping $⟨ v^{2} ⟩$ (hence $T$ ) fixed. This is the parent note's "heat in ≠ internal energy up" made concrete. ✓

Ex 4 — Cell D: adiabatic, $Q = 0$

Look at the figure: a gas pushing a piston out while wrapped in perfect insulation. No heat can cross the wall, yet the gas cools. Why? Because the only energy account left is $U$ , and the gas spends it doing work.

Figure — Heat and internal energy — microscopic vs macroscopic

Worked example Statement

$1 mol$ of monatomic gas expands adiabatically ( $Q = 0$ ) and does $500 J$ of work. Find $Δ U$ and the temperature drop $Δ T$ .

Forecast: with no heat allowed in, and the gas spending energy on work, which way does $T$ move?

Adiabatic $\Rightarrow Q = 0$ . Why this step? "Insulated / fast" processes exchange no heat; this kills the $Q$ term.
First law: $Δ U = Q - W = 0 - 500 = - 500 J$ . Why this step? The only place the $500 J$ of work could come from is the stored energy, so $U$ must drop.
Convert to temperature via $Δ U = \frac{3}{2} n R Δ T$ : $Δ T = \frac{Δ U}{\frac{3}{2} n R} = \frac{- 500}{\frac{3}{2} ( 1 ) ( 8.314 )} = - 40.09 K .$ Why this step? $U$ maps one-to-one to $T$ for an ideal gas, so a known $Δ U$ gives a known $Δ T$ .

Verify: $Δ T < 0$ — the gas cooled without losing any heat, purely from doing work. This is why bike pumps warm on compression and aerosol cans chill on release. ✓

Ex 5 — Cell E: sign flip, work done ON the gas

Ex 6 — Cell F: heat out AND compression together

Ex 7 — Cell G: degenerate case, free expansion into vacuum

The figure shows the degenerate scenario: gas on the left, vacuum on the right, a valve between. Open the valve and the gas rushes into the empty half. There is nothing pushing back, and the box is insulated.

Worked example Statement

A thermally insulated box is split by a valve. Gas fills the left half; the right half is a perfect vacuum. The valve opens. Find $Q$ , $W$ , $Δ U$ , and $Δ T$ .

Forecast: the gas expands to double the volume — surely it cools, like Ex 4?

Insulated $\Rightarrow Q = 0$ . Why this step? No heat can cross the wall.
The gas pushes against vacuum — no opposing force, no piston, so $W = 0$ . Why this step? Work is force through distance; expanding into nothing means zero opposing force, hence zero work. This is the degenerate cell — both terms vanish.
$Δ U = Q - W = 0 - 0 = 0$ , so $Δ T = 0$ : the temperature does not change. Why this step? With $U$ unchanged and $U$ set only by $T$ , $T$ is frozen even though $V$ doubled.

Verify: Contrast with Ex 4 (adiabatic against a piston, which cooled). Same $Q = 0$ , but here $W = 0$ too — no work means no cooling. The lone difference is whether something pushes back. ✓

Ex 8 — Cell H: degrees of freedom change (diatomic gas)

Worked example Statement

Heat $1 mol$ of diatomic gas ( $f = 5$ : three translational + two rotational) at constant volume from $400 K$ to $420 K$ . Compare the heat needed to that for a monatomic gas over the same rise.

Forecast: more ways to store energy — should a diatomic need more or less heat for the same $Δ T$ ?

Constant volume $\Rightarrow W = 0 \Rightarrow Q = Δ U$ . Why this step? A fixed volume does no work (no distance moved), so the first law $Δ U = Q - W$ collapses to $Q = Δ U$ — every joule of heat is stored.
Diatomic: $U = \frac{5}{2} n R T$ , so $Q_{di} = \frac{5}{2} n R Δ T = \frac{5}{2} (1) (8.314) (20) = 415.70 J .$ Why this step? Each of the $5$ degrees of freedom carries $\frac{1}{2} k_{B} T$ per molecule (with $k_{B}$ defined above; equipartition, see Equipartition Theorem), so more channels $\Rightarrow$ more energy per kelvin.
Monatomic same rise: $Q_{mono} = \frac{3}{2} (1) (8.314) (20) = 249.42 J$ . Why this step? Repeat with $f = 3$ instead of $5$ so we can compare like for like — same $n$ , $Δ T$ , only the number of storage channels differs.

Verify: Ratio $415.70/249.42 = 5/3 \approx 1.667$ — exactly $f_{di} / f_{mono} = 5/3$ . The extra rotational storage demands proportionally more heat. See Degrees of Freedom and Molecular Structure. ✓

Ex 9 — Cell I: real-world word problem

Worked example Statement

A hand pump seals off $0.02 mol$ of air (treat as diatomic, $f = 5$ ) and compresses it quickly enough that no heat escapes. You do $18 J$ of work on the air. By how much does its temperature rise?

Forecast: you've felt a pump barrel get hot. Will the rise be a few degrees or a hundred?

"No heat escapes" $\Rightarrow Q = 0$ ; you do work on the gas $\Rightarrow W = - 18 J$ . Why this step? Fast pumping is adiabatic; your effort is negative "work by gas."
$Δ U = Q - W = 0 - (- 18) = + 18 J$ . Why this step? All your effort is stored, since none leaks out as heat.
$Δ U = \frac{5}{2} n R Δ T \Rightarrow Δ T = \frac{18}{\frac{5}{2} ( 0.02 ) ( 8.314 )} = 43.30 K$ . Why this step? Convert stored energy to a temperature rise using the diatomic $U$ – $T$ link.

Verify: A $\sim 43 K$ jump from just $18 J$ into a tiny $0.02 mol$ of gas — that's why the barrel genuinely feels hot. Units: J / (J K $^{- 1}$ ) $=$ K. ✓

Ex 10 — Cell J: exam twist, a full cycle

The figure sketches a closed loop on a pressure–volume diagram: the gas is taken through several steps and returns to its exact starting point (red dot). The key fact is where it ends.

Worked example Statement

A gas is carried through a complete thermodynamic cycle, returning to its starting state. Over the whole cycle it absorbs $600 J$ of heat and rejects $370 J$ . Find $Δ U_{cycle}$ and the net work done by the gas.

Forecast: heat came in and went out — is there any leftover stored energy after a round trip?

A cycle returns to the same state, and $U$ is a state function, so $Δ U_{cycle} = 0$ . Why this step? Same start and end state $\Rightarrow$ same $U$ $\Rightarrow$ zero net change, no matter how loopy the path.
Net heat: $Q_{net} = 600 - 370 = 230 J$ . Why this step? Heat absorbed counts positive, heat rejected counts negative; their sum is the net heat crossing the boundary over the loop.
First law over the cycle: $0 = Q_{net} - W_{net} \Rightarrow W_{net} = 230 J$ . Why this step? With $Δ U = 0$ , whatever net heat entered must have left entirely as net work — this is how an engine produces work from a cycle.

Verify: $W_{net} = 230 J > 0$ : the gas did net work, so on the $P$ – $V$ loop the path went clockwise, enclosing an area of $230 J$ . Round-trip stored change is zero, as any state function demands. ✓

Ex 11 — Cell K: heat in AND gas compressed (both add to $U$ )

Ex 12 — Cell L: heat OUT while the gas still does work

Ex 13 — Cell M: constant-volume cooling

Ex 14 — Cell N: heat in, but so much work out that the gas cools

Worked example Statement

A gas absorbs $80 J$ of heat while expanding and doing $200 J$ of work on its piston. Find $Δ U$ and say whether the gas warms or cools.

Forecast: heat is flowing in — but the gas is working hard. Can $T$ still fall?

Signs: heat in $\Rightarrow Q = + 80 J$ ; gas does work $\Rightarrow W = + 200 J$ . Why this step? Both positive, but this time the withdrawal ( $W$ ) is bigger than the deposit ( $Q$ ).
$Δ U = Q - W = 80 - 200 = - 120 J$ . Why this step? The gas paid out $200 J$ of work but only took in $80 J$ of heat, so it had to dip $120 J$ into its own store.
Since $Δ U < 0$ and $U = \frac{f}{2} n R T$ , the temperature falls. Why this step? Lower internal energy means lower average molecular KE, hence lower $T$ — despite heat coming in.

Verify: $Δ U = - 120 < 0$ : heat was added yet the gas cooled, because work out exceeded heat in. This shatters the intuition "heat in $\Rightarrow$ warmer" and completes the $Q > 0, W > 0, W > Q$ cell. ✓

Recall Quick self-test across the matrix

Constant volume means $W = ?$ ::: $W = 0$ , so $Q = Δ U$ . Isothermal (ideal gas) means $Δ U = ?$ ::: $Δ U = 0$ , so $Q = W$ . Adiabatic means $Q = ?$ ::: $Q = 0$ , so $Δ U = - W$ . Free expansion into vacuum: $Q, W, Δ T$ ? ::: All zero — no heat, no work, no temperature change. Heat in and gas compressed: sign of $Δ U$ ? ::: Positive and large — both $Q > 0$ and $- W > 0$ deposit energy. Heat out while gas does work: sign of $Δ U$ ? ::: Negative — both effects withdraw energy ( $Q < 0, W > 0$ ). Heat in but work out exceeds it ( $W > Q > 0$ ): warms or cools? ::: Cools — $Δ U = Q - W < 0$ despite heat entering. Over one full cycle, $Δ U = ?$ ::: $0$ , because $U$ is a state function; then $W_{net} = Q_{net}$ . Work done on the gas enters the first law as ::: negative $W$ (since $W$ = work done by gas). Relation between $k_{B}$ and $R$ ::: $k_{B} = R / N_{A}$ — $R$ per mole equals $k_{B}$ per molecule.

Connections

Heat and Internal Energy — Microscopic vs Macroscopic
First Law of Thermodynamics
Isothermal and Adiabatic Processes
Specific Heats Cv and Cp
Equipartition Theorem
Degrees of Freedom and Molecular Structure

Worked examples — Heat and internal energy — microscopic vs macroscopic

The scenario matrix

Ex 1 — Cell A: heat in, gas does work

Ex 2 — Cell B: constant volume, all heat becomes internal energy

Ex 3 — Cell C: isothermal, $Δ U = 0$

Ex 4 — Cell D: adiabatic, $Q = 0$

Ex 5 — Cell E: sign flip, work done ON the gas

Ex 6 — Cell F: heat out AND compression together

Ex 7 — Cell G: degenerate case, free expansion into vacuum

Ex 8 — Cell H: degrees of freedom change (diatomic gas)

Ex 9 — Cell I: real-world word problem

Ex 10 — Cell J: exam twist, a full cycle

Ex 11 — Cell K: heat in AND gas compressed (both add to $U$ )

Ex 12 — Cell L: heat OUT while the gas still does work

Ex 13 — Cell M: constant-volume cooling

Ex 14 — Cell N: heat in, but so much work out that the gas cools

Connections

Concept Map