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

Definition Symbols used on this page

$U$ = internal energy — the total random kinetic energy of the gas molecules (joules).
$n$ = number of moles of gas. One mole is $6.022 \times 1 0^{23}$ molecules; $n$ counts how many such packets of molecules you have. It is a pure count, no units beyond "mol."
$f$ = number of active degrees of freedom per molecule — the independent ways one molecule stores energy (see Degrees of freedom).
$T$ = absolute temperature (kelvin); $R$ = the gas constant.
$Q$ = heat — energy that flows into the gas from outside because of a temperature difference (joules). $Q > 0$ means heat is absorbed by the gas.
$W$ = work done by the gas — energy the gas gives out by pushing on its surroundings, e.g. shoving a piston outward (joules). $W > 0$ means the gas expends energy on the outside world.
$P$ = pressure — force per unit area the gas exerts on its container walls (units: pascals, $Pa = N m^{- 2}$ ).
$V$ = volume — the space the gas occupies (units: cubic metres, $m^{3}$ ).
$k_{B}$ = the Boltzmann constant, $k_{B} = 1.381 \times 1 0^{- 23} J K^{- 1}$ — the energy-per-kelvin bookkeeping constant for a single molecule (whereas $R$ does the same job per mole; indeed $R = N_{A} k_{B}$ ).

These are tied together by the First law of thermodynamics: $Δ U = Q - W$ — "the change in internal energy equals heat in minus work out."

The key result we exercise everywhere below is $U = \frac{f}{2} n R T$

Everything here rests on the parent result Internal energy of an ideal gas, U = (f/2)nRT and its supporting ideas: Degrees of freedom, the Equipartition theorem, Kinetic theory of gases, the First law of thermodynamics, the Ideal gas law, and Molar specific heats Cv and Cp.

Before we start, here is the one master picture the whole page leans on.

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

Reading the figure: the horizontal axis names three gas types; the vertical axis is the internal energy per mole, measured in units of $R T$ , i.e. its height is $\frac{f}{2} R T$ . A monatomic gas ( $f = 3$ ) is the short burnt-orange bar; a diatomic gas ( $f = 5$ ) is the taller teal bar; a nonlinear polyatomic gas ( $f = 6$ ) is the tallest plum bar. Adding a pair of storage modes ( $f$ jumps by 2) simply stacks another slab of energy. Keep this picture in mind — most problems are just "read the bar" or "compare two bars."

Level 1 — Recognition

L1·Q1 — Read off the formula

Recall Solution

WHAT we need: the formula $U = \frac{f}{2} n R T$ with $n = 1$ mole. WHY $f = 3$ : a single atom is a point mass, so it can only move in three independent directions ( $x, y, z$ ) — three translational Degrees of freedom, no rotation, no vibration. $U = \frac{f}{2} n R T = \frac{3}{2} (1) (8.314) (300)$ $U = 1.5 \times 8.314 \times 300 \approx 3741 J \approx 3.74 kJ$

L1·Q2 — Identify $f$

Recall Solution

(a) Argon is monatomic. WHY $f = 3$ : a lone atom is essentially a point, so the only way to store energy is to move it along the three space directions $x, y, z$ → $f = 3$ .
(b) $O_{2}$ is diatomic (a dumbbell). WHY $f = 5$ : 3 translational modes for moving the whole molecule, plus 2 rotational modes for tumbling end-over-end about the two axes perpendicular to the bond. Spinning about the bond axis is not counted — see the note below on why.
(c) $H_{2} O$ is a nonlinear polyatomic molecule. WHY $f = 6$ : 3 translational, plus 3 rotational, because a bent (non-linear) molecule has real, comparable moments of inertia about all three axes, so all three tumbling modes store energy.

On the "frozen" third rotation of a diatomic: the honest reason is quantum-mechanical, not just "small moment of inertia." Rotational energy comes in discrete steps, and the smaller the moment of inertia about an axis, the larger the energy gap to the first excited rotational state. For spinning about the bond axis that gap is huge — far bigger than the thermal energy $k_{B} T$ available at moderate temperature (recall $k_{B} = 1.381 \times 1 0^{- 23} J K^{- 1}$ is the per-molecule energy-per-kelvin constant defined in the symbol list) — so collisions cannot supply enough energy to excite it. That mode is therefore "frozen out" and contributes nothing, leaving $f = 5$ . (The classical "negligible moment of inertia" phrasing is a shorthand for this; taken literally it would be misleading.)

Level 2 — Application

L2·Q1 — Change in internal energy

Recall Solution

WHAT: because $U$ depends only on $T$ , the change is $Δ U = \frac{f}{2} n R Δ T$ with $n = 2$ moles. WHY subtract temperatures: $U (T)$ is a straight line through the origin, so $U (T_{2}) - U (T_{1})$ depends only on $Δ T = 450 - 250 = 200 K$ . $Δ U = \frac{5}{2} (2) (8.314) (200) = 5 \times 8.314 \times 200 \approx 8314 J \approx 8.31 kJ$

L2·Q2 — Two gases, same energy

Recall Solution

WHAT: write $U = \frac{f}{2} n R T$ for each gas and set them equal. WHY set them equal: the question demands the same internal energy, so their two expressions must be numerically identical. WHY the $R T$ cancels: both gases share the same $T$ , and $R$ is a universal constant, so the common factor $R T$ divides out of both sides, leaving only the mole– $f$ balance. $\frac{3}{2} n_{He} R T = \frac{5}{2} (3) R T \Rightarrow 3 n_{He} = 15 \Rightarrow n_{He} = 5 mol$ Look at the bar picture: helium's bar is shorter (per mole), so you need more moles of it to reach the same total energy — $5$ vs $3$ , exactly the ratio $5/3$ of the two $f$ values.

L2·Q3 — Isothermal expansion

Recall Solution

WHAT: $Δ U = \frac{f}{2} n R Δ T$ . WHY it is zero: isothermal means $Δ T = 0$ , so $Δ U = 0$ regardless of how much the volume $V$ or pressure $P$ changed. $Δ U = \frac{f}{2} n R (0) = 0$ The gas still does work $W$ and absorbs heat $Q$ — but by the First law of thermodynamics ( $Δ U = Q - W$ ), $Δ U = 0$ forces $Q = W$ , and the internal energy never budges.

Level 3 — Analysis

L3·Q1 — Same heat, different processes

Recall Solution

$Δ T = 60 K$ , $n = 1$ mole.

Internal energy — same for both paths. WHY the path does not matter: the master formula $U = \frac{f}{2} n R T$ shows $U$ depends only on $T$ , so its change $Δ U = \frac{f}{2} n R Δ T$ is fixed by $Δ T = 60 K$ alone — whether we hold volume or pressure fixed is irrelevant to $Δ U$ . $Δ U = \frac{3}{2} (1) (8.314) (60) = 1.5 \times 8.314 \times 60 \approx 748.3 J$

(a) Constant volume: no work is done ( $W = 0$ ), so by the First law of thermodynamics $Q = Δ U$ : $Q_{V} = n C_{v} Δ T = Δ U \approx 748.3 J$

(b) Constant pressure: now the gas also pushes the piston out, doing work $W = n R Δ T$ , so more heat is needed. Using $C_{p} = C_{v} + R = \frac{3}{2} R + R = \frac{5}{2} R$ : $Q_{P} = n C_{p} Δ T = \frac{5}{2} (1) (8.314) (60) = 2.5 \times 8.314 \times 60 \approx 1247.1 J$

The lesson: $Δ U$ is identical ( $\approx 748 J$ ) both times, but $Q$ differs because at constant pressure part of the heat becomes work. See Molar specific heats Cv and Cp.

L3·Q2 — Extracting temperature from energy

Recall Solution

Rearrange $U = \frac{f}{2} n R T$ for $T$ (here $n = 0.5$ mole): $T = \frac{2 U}{f n R} = \frac{2 ( 3120 )}{5 ( 0.5 ) ( 8.314 )} = \frac{6240}{20.785} \approx 300.2 K$

Level 4 — Synthesis

L4·Q1 — Freezing out modes

Recall Solution

WHY vibration adds two DOF: a vibrating bond stores energy both as kinetic energy of stretching and as potential energy in the "spring." By the Equipartition theorem each quadratic term gets $\frac{1}{2} k_{B} T$ , so two terms → $f$ rises by 2. $U_{f = 5} = \frac{5}{2} (2) (8.314) (2000) = 5 \times 8.314 \times 2000 = 83140 J$ $U_{f = 7} = \frac{7}{2} (2) (8.314) (2000) = 7 \times 8.314 \times 2000 = 116396 J$ $Δ U = U_{f = 7} - U_{f = 5} = 116396 - 83140 = 33256 J \approx 33.3 kJ$ Equivalently, each extra pair of DOF adds $n R T = 2 \times 8.314 \times 2000 = 33256 J$ — one slab on the bar chart.

L4·Q2 — Mixture of gases

Recall Solution

WHAT: compute each gas's internal energy with its own $f$ and its own number of moles $n$ , then add. WHY add separately: internal energy is a total — each molecule contributes independently, so the helium molecules ( $3$ mol, $f = 3$ ) and the nitrogen molecules ( $2$ mol, $f = 5$ ) each carry their own share; there is no single $f$ that describes both species at once. WHY the $R T$ factors out: both gases sit at the same $T$ , so $R T$ is common and can be pulled out of the sum. $U = [\frac{3}{2} (3) + \frac{5}{2} (2)] R T = [4.5 + 5] (8.314) (400)$ $U = 9.5 \times 8.314 \times 400 \approx 31593 J \approx 31.6 kJ$ Note: the effective $f$ here is $\frac{9.5 \times 2}{5} = 3.8$ , between 3 and 5, weighted by moles — but you must never assume that value up front; it only emerges after adding the species.

Level 5 — Mastery

L5·Q1 — Full first-law cycle leg

Recall Solution

(a) First law: $Δ U = Q - W = 2000 - 800 = 1200 J$ . (b) Since $Δ U = \frac{f}{2} n R Δ T$ (with $n = 1$ mole), solve for $Δ T$ : $Δ T = \frac{2 Δ U}{f n R} = \frac{2 ( 1200 )}{3 ( 1 ) ( 8.314 )} = \frac{2400}{24.942} \approx 96.2 K$ Only the internal $1200 J$ raises the temperature; the $800 J$ that left as work does not.

L5·Q2 — Reasoning about a $P$ – $V$ triangle cycle

Reading the figure: the horizontal axis is volume $V$ , the vertical axis is pressure $P$ . The gas starts at corner $A$ , is heated at constant pressure along $A \to B$ (moving right, volume increases), then along $B \to C$ pressure rises while volume falls, and finally $C \to A$ closes the loop. The shaded burnt-orange area enclosed by the triangle equals the net work $W_{cycle}$ done by the gas over one loop.

Recall Solution

WHAT: internal energy is a state function — it depends only on the current $(P, V, T)$ , not on the path taken to get there. WHY the cycle change is zero: the gas returns to its starting state $A$ , so its temperature returns to the start, hence $Δ U_{cycle} = \frac{f}{2} n R Δ T_{cycle} = \frac{f}{2} n R (0) = 0$ Consequence (first law over the cycle): $0 = Q_{cycle} - W_{cycle}$ , therefore $Q_{cycle} = W_{cycle}$ All the net heat absorbed over the loop comes out as net work — that is exactly how a heat engine runs. The enclosed area of the triangle is that net work $W_{cycle}$ : geometrically, the " $P$ times change in $V$ " summed around the loop leaves precisely the triangle's area, so reading off that shaded area gives you both $W_{cycle}$ and (since $Δ U_{cycle} = 0$ ) the net heat $Q_{cycle}$ as well.

L5·Q3 — Path independence, numerically

Recall Solution

$Δ U$ depends only on the endpoints' temperatures, so only $T_{A}$ and $T_{C}$ matter — the intermediate $T_{B}$ and the specific nature of each leg (isochoric, isobaric, …) are irrelevant. $Δ U_{A \to C} = \frac{3}{2} (1) (8.314) (500 - 300) = 1.5 \times 8.314 \times 200 \approx 2494.2 J$ Via the detour, add the two legs: $Δ U_{A \to B} + Δ U_{B \to C} = \frac{3}{2} (8.314) (250 - 300) + \frac{3}{2} (8.314) (500 - 250)$ $= 1.5 (8.314) (- 50) + 1.5 (8.314) (250) = - 623.55 + 3117.75 = 2494.2 J$ Identical — because $U$ is a state function; the $T_{B}$ dip on the cooling leg is exactly repaid on the heating leg, so it cancels out.

Active recall

Recall One-line summary of every level

L1: plug into $U = \frac{f}{2} n R T$ ; know $f$ = 3 / 5 / 6.
L2: $Δ U = \frac{f}{2} n R Δ T$ ; isothermal $\Rightarrow Δ U = 0$ .
L3: $Δ U$ same for all paths, but $Q$ differs; $Q_{V} = Δ U$ , $Q_{P} = n C_{p} Δ T$ .
L4: add species / modes separately; vibration adds 2 to $f$ .
L5: cycle $\Rightarrow Δ U = 0 \Rightarrow Q = W$ ; $U$ is a state function.