Internal energy of ideal gas U = (f - 2)nRT: Worked examples

Everything here rests on the parent: the parent topic — internal energy of an ideal gas. Prerequisites we lean on: Kinetic theory of gases, Equipartition theorem, Degrees of freedom, First law of thermodynamics, Molar specific heats Cv and Cp, Ideal gas law.

The scenario matrix

Before any symbols mean anything, let us name them, then list the knobs a problem can turn.

Two quantities decide $U$ : the number of energy-storing modes $f$ (set by the gas) and the temperature $T$ (set by the process). So every problem lives in a grid of "which gas" × "which question about $T$ ".

Cell	Case class	What makes it different	Worked in
A	Monatomic, absolute $U$	$f = 3$ , want $U$ itself	Ex 1
B	Diatomic, change $Δ U$	$f = 5$ , want $Δ U$ from $Δ T$	Ex 2
C	Polyatomic (nonlinear)	$f = 6$ , bigger $U$	Ex 3
D	Isothermal (degenerate $Δ T = 0$ )	$Δ U = 0$ regardless of $P, V$	Ex 4
E	Path-independence check	same $Δ U$ via two different paths	Ex 5
F	Cooling / negative $Δ T$	sign of $Δ U$ is negative	Ex 6
G	Mixture of two gases	add $U$ of parts, different $f$	Ex 7
H	Limiting: high- $T$ vibration "unfreezes"	$f$ jumps from 5 to 7	Ex 8
I	Real-world word problem + first law	combine $Δ U = Q - W$	Ex 9
J	Exam twist: given $U$ , find $T$	invert the formula	Ex 10
K	Adiabatic ( $Q = 0$ )	$Δ U$ comes wholly from work	Ex 11

The grid below is the same table drawn as a picture: the vertical axis is which gas (it fixes $f$ ), the horizontal axis is what the problem asks about $T$ . Each shaded box is a lettered cell (A–K) matching the table above and the example that solves it — trace any exam problem to one box.

Figure — Internal energy of ideal gas U = (f - 2)nRT

Worked examples

Cell A — Absolute internal energy of a monatomic gas

Worked example Example 1 — Argon in a tank

$3 mol$ of argon (monatomic, so $f = 3$ ) sits at $T = 350 K$ . Find the total internal energy $U$ .

Forecast: guess the order of magnitude before computing — kilojoules? tens of kJ? Jot it down.

Step 1. Identify $f$ . Argon is a single atom → only 3 translational modes. Why this step? $f$ is set by molecular shape (see Degrees of freedom); a lone atom cannot store rotation (no "arms" to spin) or vibration (no bond).

Step 2. Apply $U = \frac{f}{2} n R T$ . Why this step? This is the master formula; for an absolute energy we use $T$ itself (not a change). $U = \frac{3}{2} (3) (8.314) (350)$

Step 3. Multiply. Why this step? Just arithmetic; group the constants to get a single number in Joules. $U = 1.5 \times 3 \times 8.314 \times 350 \approx 13, 095 J \approx 13.1 kJ$

Verify: Units: $mol \cdot (J mol^{- 1} K^{- 1}) \cdot K = J$ . ✓ And $13 kJ$ for a few moles at room-ish temperature is the right ballpark (energy per mole $\approx 4.4 kJ$ ).

Cell B — Change in internal energy of a diatomic gas

Cell C — Polyatomic gas (more storage bins)

Worked example Example 3 — Water vapour

$0.5 mol$ of steam (H₂O is nonlinear polyatomic, $f = 6$ ) at $500 K$ . Find $U$ .

Forecast: More degrees of freedom than diatomic → more energy per mole at the same $T$ .

Step 1. Set $f = 6$ : 3 translation + 3 rotation (a bent, non-linear molecule can spin about all three axes and each stores energy). Why this step? Unlike a straight dumbbell, a bent molecule has real moment of inertia about all three axes → the third rotation is not frozen.

Step 2. Apply the master formula $U = \frac{f}{2} n R T$ with $f = 6$ . Why this step? We want an absolute $U$ , so we feed in $T$ itself; substituting fixes every symbol to a number. $U = \frac{6}{2} (0.5) (8.314) (500) = 3 \times 0.5 \times 8.314 \times 500$

Step 3. Multiply. Why this step? Arithmetic only — reduce the product to a single energy in Joules. $U = 6235.5 J \approx 6.24 kJ$

Verify: $f /2 = 3$ , so $U = 3 n R T$ — the polyatomic case in the parent table. Per mole $= 12, 471 J$ , larger than argon's per-mole $\frac{3}{2} R T$ at the same $T$ . ✓

Cell D — Degenerate case: isothermal ( $Δ T = 0$ )

Worked example Example 4 — Isothermal expansion trap

An ideal gas expands isothermally; its volume triples and pressure drops. Find $Δ U$ .

Forecast: The gas does work and its $P, V$ change dramatically — surely $U$ changes? (This is the trap.)

Step 1. Read the word isothermal: temperature constant, so $Δ T = 0$ . Why this step? "Iso-thermal" literally = same temperature; that is the only quantity $U$ cares about.

Step 2. Apply $Δ U = \frac{f}{2} n R Δ T$ with $Δ T = 0$ . Why this step? Substituting the one fact we know ( $Δ T = 0$ ) makes the whole product vanish — no other quantity can revive it, because $U$ ignores $P$ and $V$ . $Δ U = \frac{f}{2} n R (0) = 0$

Verify: By the First law of thermodynamics, $Δ U = Q - W$ . Here the gas absorbs heat $Q$ and does equal work $W$ , so $Q = W$ and $Δ U = 0$ . ✓ $P$ and $V$ changed, but $U$ did not — because there are no intermolecular forces.

Cell E — Same $Δ U$ by two different paths

Worked example Example 5 — Path independence

$1 mol$ of a monatomic gas ( $f = 3$ ) goes from $300 K$ to $400 K$ by (a) constant volume, then (b) constant pressure. Show $Δ U$ is identical.

Forecast: The heat supplied will differ (constant- $P$ needs more, to also do work), but $Δ U$ should match.

Step 1. Both paths share the same endpoints $Δ T = 100 K$ . Why this step? Because $U$ is a state function (defined just above), only the endpoints matter, not the route.

Step 2. Compute $Δ U$ once. Why this step? Because $Δ U$ only sees $Δ T$ , one calculation covers both paths at once. $Δ U = \frac{3}{2} (1) (8.314) (100) = 1247.1 J$

Step 3. Cross-check with specific heats. At constant volume the heat is $Q_{V} = n C_{V} Δ T$ with $C_{V} = \frac{f}{2} R = \frac{3}{2} R$ (see Molar specific heats Cv and Cp). Why this step? $Q_{V}$ equals $Δ U$ (no work), so it must give the same number. $Q_{V} = (1) (\frac{3}{2} \cdot 8.314) (100) = 1247.1 J$

Verify: Both routes give $Δ U = 1247.1 J$ . ✓ The constant-pressure path needs more heat ( $Q_{P} = n C_{P} Δ T$ with $C_{P} = \frac{5}{2} R$ ), but the extra goes into work, not $U$ .

The $P$ – $V$ diagram below draws these two paths so you can see the path independence: the blue route (rise pressure at fixed volume, then expand at fixed pressure) and the yellow route (expand first, then raise pressure) start at the red dot (State 1, $300 K$ ) and end at the green dot (State 2, $400 K$ ). The two coloured tracks enclose different areas — so they involve different work and different heat — yet both connect the same two temperatures, so $Δ U$ is the same $1247 J$ for each.

Cell F — Negative $Δ T$ (cooling)

Cell G — Mixture of two gases

Worked example Example 7 — A mix of helium and nitrogen

A vessel holds $2 mol$ He ( $f = 3$ ) and $3 mol$ N₂ ( $f = 5$ ) in thermal equilibrium at $300 K$ . Find the total internal energy.

Forecast: Each gas keeps its own $f$ ; total $U$ = sum of parts.

Step 1. Internal energy is additive: $U_{tot} = U_{He} + U_{N_{2}}$ . Why this step? Energy is a bulk sum over all molecules; the two species don't share degrees of freedom.

Step 2. Compute each part at the same $T = 300 K$ . Why this step? They are in thermal equilibrium, so a single $T$ feeds both terms — we apply the master formula once per gas with its own $f$ . $U_{He} = \frac{3}{2} (2) (8.314) (300) = 7482.6 J$ $U_{N_{2}} = \frac{5}{2} (3) (8.314) (300) = 18706.5 J$

Step 3. Add. Why this step? Additivity from Step 1 means the total is just the sum of the two energies. $U_{tot} = 7482.6 + 18706.5 = 26189.1 J \approx 26.2 kJ$

Verify: Units all Joules; both terms positive. ✓ An "effective" mixture $f$ would be $\frac{2 ( 3 ) + 3 ( 5 )}{5} = 4.2$ , and $\frac{4.2}{2} (5) (8.314) (300) = 26189 J$ matches. ✓

Cell H — Limiting behaviour: vibration unfreezes at high $T$

Worked example Example 8 — Diatomic gas at very high temperature

$1 mol$ of a diatomic gas is heated so hot that its vibrational mode activates, raising $f$ from $5$ to $7$ . Compare $U$ at $T = 2000 K$ for $f = 5$ vs $f = 7$ .

Forecast: More active modes → more stored energy at the same $T$ .

Step 1. Why $f = 7$ ? A vibrating bond stores energy two ways: kinetic (atoms moving) and potential (spring stretch). Each is a quadratic term → $+ 2$ degrees of freedom. Why this step? Equipartition theorem counts every quadratic energy term, and vibration contributes two of them.

Step 2. Compute both with the master formula $U = \frac{f}{2} n R T$ . Why this step? Only $f$ differs between the two scenarios; feeding both values into the same formula isolates the effect of the extra modes. $U_{f = 5} = \frac{5}{2} (1) (8.314) (2000) = 41570 J$ $U_{f = 7} = \frac{7}{2} (1) (8.314) (2000) = 58198 J$

Step 3. Difference. Why this step? Subtracting isolates exactly the energy the two new vibrational bins carry. $Δ U = 58198 - 41570 = 16628 J \approx 16.6 kJ$

Verify: The extra $= \frac{2}{2} n R T = n R T = (1) (8.314) (2000) = 16628 J$ — exactly the two vibrational bins' worth. ✓ At low $T$ these bins are frozen (quantum gap too big), so $f = 5$ ; the parent's " $f$ at moderate $T$ " caveat is exactly this limiting case.

Cell I — Real-world word problem with the first law

Worked example Example 9 — Piston does work while heating

A cylinder of $1 mol$ monatomic gas ( $f = 3$ ) absorbs $Q = + 2000 J$ of heat while doing $W = + 500 J$ of work pushing a piston (positive because the gas does work on its surroundings). Find $Δ U$ and the temperature rise $Δ T$ .

Forecast: Not all the heat becomes internal energy — some leaks out as work, so $Δ U < 2000 J$ .

Step 1. First law: $Δ U = Q - W$ . Why this step? First law of thermodynamics — heat in minus work done by the gas equals the energy that stays inside (using the sign convention boxed above). $Δ U = 2000 - 500 = 1500 J$

Step 2. Invert $Δ U = \frac{f}{2} n R Δ T$ to get $Δ T$ . Why this step? We know $Δ U$ ; solve for the only unknown, $Δ T$ . $Δ T = \frac{2 Δ U}{f n R} = \frac{2 ( 1500 )}{3 ( 1 ) ( 8.314 )}$

Step 3. Compute. Why this step? Arithmetic only — divide out to get the temperature rise in kelvin. $Δ T = \frac{3000}{24.942} \approx 120.3 K$

Verify: Plug back: $\frac{3}{2} (1) (8.314) (120.3) = 1500 J = Δ U$ . ✓ And $Δ U < Q$ as forecast, because $500 J$ went into work. ✓

Cell J — Exam twist: given $U$ , find $T$

Worked example Example 10 — Reverse-engineer the temperature

A sample of $0.25 mol$ diatomic gas ( $f = 5$ ) has internal energy $U = 1300 J$ . Find its temperature $T$ .

Forecast: Rearranging the formula — expect a few hundred kelvin.

Step 1. Solve $U = \frac{f}{2} n R T$ for $T$ . Why this step? Same formula, unknown moved to the subject; pure algebra, no new physics. $T = \frac{2 U}{f n R}$

Step 2. Substitute the known numbers. Why this step? Replacing each symbol with its value turns the algebra into a single arithmetic fraction. $T = \frac{2 ( 1300 )}{5 ( 0.25 ) ( 8.314 )} = \frac{2600}{10.3925}$

Step 3. Compute. Why this step? Arithmetic only — divide to get the temperature in kelvin. $T \approx 250.2 K$

Verify: Forward-check: $\frac{5}{2} (0.25) (8.314) (250.2) = 1300 J$ . ✓ Units: $\frac{J}{mol \cdot J mol ^{- 1} K ^{- 1}} = K$ . ✓ The answer is a sensible few-hundred kelvin, as forecast. ✓

Cell K — Adiabatic process ( $Q = 0$ )

Worked example Example 11 — Insulated compression

A perfectly insulated cylinder holds $2 mol$ of a monatomic gas ( $f = 3$ ). A piston compresses the gas, so work is done on it: in our sign convention this is $W = - 900 J$ (work done by the gas is negative). No heat enters or leaves ( $Q = 0$ ). Find $Δ U$ and the temperature change $Δ T$ .

Forecast: With no heat flow, all the compression work becomes internal energy → $T$ must rise, so $Δ U > 0$ .

Step 1. First law with $Q = 0$ : $Δ U = Q - W = 0 - (- 900) = + 900 J$ . Why this step? First law of thermodynamics — "adiabatic" means the insulation blocks all heat ( $Q = 0$ ), so the only source of energy change is work. Here $W = - 900 J$ (work done on the gas), which raises $U$ .

Step 2. Convert $Δ U$ to $Δ T$ via $Δ U = \frac{f}{2} n R Δ T$ , solving for $Δ T$ . Why this step? $Δ U$ is now known; the master formula links it to the temperature change we want. $Δ T = \frac{2 Δ U}{f n R} = \frac{2 ( 900 )}{3 ( 2 ) ( 8.314 )}$

Step 3. Compute. Why this step? Arithmetic only — divide out to get the rise in kelvin. $Δ T = \frac{1800}{49.884} \approx 36.1 K$

Verify: Positive $Δ T$ , as forecast (compression heats the gas). ✓ Back-check: $\frac{3}{2} (2) (8.314) (36.1) = 900 J = Δ U$ . ✓ Contrast with Cell D (isothermal): there $Q$ and $W$ balanced; here $Q = 0$ so work alone drives $U$ .

Recall

Recall Which cell does each cue belong to?

"Volume triples, $T$ fixed, find $Δ U$ " ::: Cell D — isothermal, $Δ U = 0$ . "He + N₂ in one box, find total $U$ " ::: Cell G — sum the parts with their own $f$ . "Bond starts vibrating at 2000 K" ::: Cell H — $f$ jumps $5 \to 7$ . "Given $U$ , asked for $T$ " ::: Cell J — invert to $T = 2 U / (f n R)$ . "Cooled from 320 K to 250 K" ::: Cell F — keep the negative $Δ T$ . "Insulated, no heat, piston compresses gas" ::: Cell K — adiabatic, $Δ U = - W$ .

Connections

Parent topic — the derivation of the master formula
Degrees of freedom — how each cell's $f$ is chosen
Equipartition theorem — why vibration adds 2 (Cell H)
First law of thermodynamics — $Δ U = Q - W$ (Cells I, D, K)
Molar specific heats Cv and Cp — cross-check in Cell E
Ideal gas law — underpins path independence
Kinetic theory of gases — the microscopic root of it all