Physics · Thermodynamics

Internal energy of a monatomic ideal gas

Q: What does Internal energy of a monatomic ideal gas describe physically?

$U=\frac32nRT$ describes internal energy of a classical monatomic ideal gas sample at equilibrium. It sums random translational energy of all atoms; motion of the whole sample is excluded. A constant atomic ground-state contribution is omitted through the chosen energy zero. At fixed n, U in this model depends only on absolute temperature. For example, 0.50 mol at 300 K gives about 1870.754 J. Heat and work instead describe transfers during a process, not separate contributions stored in a state. For temperature changes, $\Delta U=\frac32nR\Delta T$ at fixed n is particularly useful.

Q: How can I check a calculation involving Internal energy of a monatomic ideal gas?

For 0.80 mol between 290 and 410 K, $U_1\approx2893.433\,\mathrm J$, $U_2\approx4090.716\,\mathrm J$ and $\Delta U\approx1197.283\,\mathrm J$. Independently check the difference with $nC_{V,m}\Delta T$ and $C_{V,m}=\frac32R$. Both methods must agree. In a rigid closed vessel without other work, $Q=\Delta U$; do not transfer that equality unchecked to a moving piston. An isothermal process at fixed n instead must give $\Delta U=0$. Despite constant U, heat and work can be nonzero. Therefore also check $\Delta U=Q-W_{\mathrm{by}}$, with work done outward by the gas defined as positive. Initial and final states must also use the same energy zero. These process balances assume a closed system with unchanged or negligible bulk kinetic and external potential energy.

Q: What assumptions and limits apply to Internal energy of a monatomic ideal gas?

The model assumes a dilute classical monatomic ideal gas with negligible interaction energy and no relevant electronic excitation. Initial and final states must admit thermodynamic description. Polyatomic molecules can store rotational and vibrational energy; ionisation and chemical conversion also require additional contributions. Classical equipartition can fail at low temperature too. $\Delta U=\frac32nR\Delta T$ requires unchanged amount of substance. If amount changes, $U_2-U_1=\frac32R(n_2T_2-n_1T_1)$ is the appropriate state difference within this model. An open installation additionally needs material and energy transport balances and cannot be treated as a closed vessel.

Q: How do I check units and scale for Internal energy of a monatomic ideal gas?

With n in mol, R in J/(mol·K) and T in K, nRT has units joule. Absolute temperature matters: 20 °C corresponds to 293.15 K. For ΔT, however, kelvin and Celsius differences have the same numerical value. A useful scale is $\frac Un\approx3741.508\,\mathrm{J\,mol^{-1}}$ at 300 K. Doubling n at equal T doubles U; doubling T at fixed n does the same. For an independent unit check, use $U=\frac32pV$ when the ideal gas law applies and p is absolute pressure. Pa·m³ also gives joules, not force.

Q: When should I combine Internal energy of a monatomic ideal gas with another relation?

Combine the state formula with $\Delta U=Q-W_{\mathrm{by}}$ when a process specifies heat or work. For a rigid closed vessel without additional work, $Q=\frac32nR\Delta T$. In reversible isothermal expansion, $p=\frac{nRT}{V}$ gives $W_{\mathrm{by}}=nRT\ln\frac{V_2}{V_1}$; because $\Delta U=0$, $Q=W_{\mathrm{by}}$. For example, 0.50 mol at 300 K and doubled volume gives about 864.472 J positive work and equal heat input. Another process needs its own relation or a specified work value. The U formula alone does not establish the path or Q, although it determines the state change. These process balances assume a closed system with unchanged or negligible bulk kinetic and external potential energy.

The internal energy of a monatomic ideal gas is the sum of the translational kinetic energies of its particles.

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Formula

Calculate

U = \frac{3}{2} n RT

U in J; n in mol; R in J/(mol·K); T in K.

Variables & units

Variables & units
Symbol	Meaning	Unit
$U$	Internal energy	J
$n$	Amount of substance	mol
$R$	Universal gas constant	J/(mol·K)
$T$	Absolute temperature	K

LaTeX

U=\frac{3}{2}nRT

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Energy–temperature line for 0.80 mole. The 290 and 410 kelvin states are marked, with a difference triangle showing energy change 1.1973 kilojoules. — 0.80 mol of a classical monatomic ideal gas: between 290 and 410 K, U rises from 2.8934 to 4.0907 kJ. In a rigid closed vessel without other work or external energy changes, Q = ΔU ≈ 1.1973 kJ. The vertical scale begins at 2 kJ.

Monatomic ideal gas internal energy calculator

Find the internal energy, amount or temperature of a monatomic ideal gas such as helium or argon, plus the energy change and heat when it warms up.

Calculation model

Find

Amount of substance n

Amount of gas, e.g. 1 mol of helium, which is about 4 g.

Absolute temperature T

Temperature of the gas, e.g. 300 K; °C is converted to kelvin for you.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for U, substitute, evaluate

1. Rearrange for U
$U=\frac32nRT$
2. Substitute values (in SI units)
$\frac{3\cdot 1\cdot 8.31446\cdot 300}2=3741.51$
3. Result
U = 3.742 kJ

Result: Internal energy U

U = 3.742 kJ

With n = 1 mol and T = 300 K, U = 3.742 kJ.

Quick tips

Estimate: 3/2 · R ≈ 12.5 J per mole and kelvin. 1 mol at 300 K, which is 27 °C, stores about 3.7 kJ; always use T in kelvin.
Same warming, different heat: at constant pressure the gas needs 5/3 as much heat as at constant volume, because it also does work by expanding.

Assumptions and limits

Classical dilute monatomic ideal gas in thermal equilibrium with n>0,T>0 and only three translational contributions; electronic excitation, interactions, quantum degeneracy and ionisation neglected. Fixed model zero excludes ground-state contribution; bulk motion and external potential energy are separate. Not a real quantum model extended to T=0.

Derivation & background

The relation sums translational energies in a classical monatomic ideal gas relative to shared gas motion. Constant atomic ground-state energy is omitted through the energy zero; electronic excitation and interaction energy are negligible. At fixed n, $\Delta U=\frac32nR\Delta T$ is path independent, but $Q=\Delta U$ only without net work. Polyatomic gases need additional degrees of freedom. These process balances assume a closed system with unchanged or negligible bulk kinetic and external potential energy.

Exam blueprint

Validity range

Classical monatomic ideal gas in equilibrium states without relevant interaction energy or electronic excitation. U uses a fixed energy zero; centre-of-mass motion and external potential energy are separate. ΔU=3nRΔT/2 requires fixed n. These process balances assume a closed system with unchanged or negligible bulk kinetic and external potential energy.

Derivation steps

Sum mean translational atomic energy and convert to amount using NA.

1Consider a classical monatomic ideal gas sample at equilibrium; neglect electronic excitation.
2Three translational degrees each contribute $\frac12k_BT$ , giving $\frac32k_BT$ per atom.
3Sum over all N atoms: $U=\frac32Nk_BT$ . The constant atomic ground-state contribution is omitted through the energy reference.
4Using $N=nN_A$ and $R=N_Ak_B$ gives $U=\frac32nRT$ . Shared gas motion remains separate.
5At fixed n, subtract: $\Delta U=\frac32nR(T_2-T_1)$ .
6Then add the appropriate process balance: $\Delta U=Q-W_{\mathrm{by}}$ for a closed system without external-energy change; $Q=\Delta U$ only at zero net work.

Rearrangements

Absolute temperature

T=\frac{2U}{3nR}

This form applies to monatomic ideal gases.

Task variant

Calculate the quantity from the worked example for “Internal energy of a monatomic ideal gas”.

For $n=1.0\,\mathrm{mol}$ of a classical monatomic ideal gas at $T=300\,\mathrm K$ , $U=\frac32nRT\approx3.74\,\mathrm{kJ}$ , using the stated thermal-energy reference.

Common mistakes

Ignoring units or assumptions stated for the formula: U in J; n in mol; R in J/(mol·K); T in K.

Convert units before substitution and verify the validity range; round only at the end.

Exam context

Calculate temperature changes and energy balances for monatomic ideal gases.

Formula cluster

Thermodynamics

Related relations: erster-hauptsatz-thermodynamik, mittlere-kinetische-energie-molekuel, wirkungsgrad-carnot.

ΔU = Q − W

First Law of Thermodynamics

⟨Ekin⟩ = 3/2 kBT

Mean translational kinetic energy per gas molecule

η = 1 − T_k/T_w

Carnot Efficiency

Worked example

For $n=1.0\,\mathrm{mol}$ of a classical monatomic ideal gas at $T=300\,\mathrm K$ , $U=\frac32nRT\approx3.74\,\mathrm{kJ}$ , using the stated thermal-energy reference.

Applications

Calculate temperature changes and energy balances for monatomic ideal gases.

How do you calculate with Internal energy of a monatomic ideal gas?

Here is how to work through a typical Internal energy of a monatomic ideal gas (U = 3/2 nRT) task step by step:

1
Task
Calculate the quantity from the worked example for “Internal energy of a monatomic ideal gas”.
Solution path
For $n=1.0\,\mathrm{mol}$ of a classical monatomic ideal gas at $T=300\,\mathrm K$ , $U=\frac32nRT\approx3.74\,\mathrm{kJ}$ , using the stated thermal-energy reference.

Understand and apply the equation

How much energy does a gas sample store, and what temperature change results from heat or work? A classical monatomic ideal gas needs the sum of three translational contributions. The relation for U describes stored energy; a process balance explains how it changes.

Energy for the random motion of atoms

A hotter gas has more strongly moving atoms. A larger amount contains more atoms. Thermal internal energy accounts for both effects: energy per atom and atom count. This page applies to a classical monatomic ideal gas with three active translational degrees of freedom.[2]

Random motion is described relative to the gas mean motion. If a sealed vessel moves faster as a whole, its external kinetic energy increases, but its temperature need not rise. Conversely, a stationary vessel can contain hot gas with substantial internal energy.[1][3]

The model energy reference excludes the temperature-independent atomic ground-state contribution. The linear thermal-energy expression is not a valid quantum model down to absolute zero. For changes within its valid range, any constant reference contribution cancels anyway.[4][2]

When does the formula describe internal energy of an entire gas sample?

U=\frac32Nk_{\mathrm B}T=\frac32nRT

N counts atoms; n measures amount of substance. The chosen energy zero omits constant atomic ground-state contributions.[1]

U is an extensive state variable: twice the amount at equal temperature has twice the modelled energy store. Here this includes random atomic motion. Motion of the sample as a whole and its external potential energy are accounted for separately. Thermal equilibrium does not require every atom to carry the same energy.[3]

From an atom to amount of substance

First count independent directions of motion: x, y and z. At classical equilibrium, each quadratic translational contribution has the same mean energy. Three directions supply the factor three; each degree of freedom contributes the factor one half. Then sum over all atoms.[2]

Equipartition assigns $\frac12k_BT$ to each classical quadratic translational degree of freedom in equilibrium. Three spatial directions give $\frac32k_BT$ per atom. The monatomic model has no molecular rotation or vibration; electronic excitation is neglected. N such contributions can be added. Mean translational energy per atom is the individual statistical contribution before summing over the sample.[2]

U=N\langle E_{\mathrm{tr}}\rangle=\frac32nN_{\mathrm A}k_{\mathrm B}T=\frac32nRT

$R=N_Ak_B$ is the molar gas constant. The sum assumes negligible interaction energy.[7]

This derivation is a statistical model with stated premises, not a universal equation for every material. Molecular gases may have additional internal motions. Even a monatomic gas can develop electronic excitation and ionisation at sufficient temperature; the factor 3/2 then becomes insufficient.[4]

Changes in U and heat capacity

\begin{aligned}&\Delta U=\frac32nR(T_2-T_1)\\&C_{V,m}=\frac32R\end{aligned}

The first equation applies at fixed amount even when volume changes.[2]

$R=8.31446261815324\,\mathrm{J\,mol^{-1}\,K^{-1}}$ gives $C_{V,m}\approx12.47169393\,\mathrm{J\,mol^{-1}\,K^{-1}}$ . The product $nR\Delta T$ has units of joules. Temperature differences have equal numerical values in Celsius and kelvin; energy from a single temperature requires absolute kelvin temperature. A constant energy reference does not change an energy difference.[7]

\begin{aligned}&T=\frac{2U}{3nR}\\&n=\frac{2U}{3RT}\\&U=\frac32pV\end{aligned}

The first rearrangements use the same energy zero. $U=\frac32pV$ additionally uses $pV=nRT$ ; p is absolute pressure.[1]

Temperature and amount set thermal energy

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At equal temperature, twice the amount has twice the thermal internal energy. At fixed amount, the slope gives heat capacity at constant volume. Only a classically appropriate temperature interval is shown, not an extrapolation to absolute zero.[2]

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Is internal-energy change always equal to supplied heat?

\Delta U=Q-W_{\mathrm{by}}

Balance for a closed system without changes in external kinetic or potential energy. Heat is positive inward; all work by the system is positive outward. Other energy changes require the complete total-energy balance.[5]

A rigid closed vessel performs no volume work. If other work modes are negligible too, $Q=\Delta U$ . A piston can require a different heat input for the same temperature change because work is involved. An isothermal process at fixed n in this ideal gas has $\Delta U=0$ even though heat and work can each be nonzero. The first law of thermodynamics supplies the signed heat–work balance for this distinction.[6]

Three examples: energy, heat and process

Heating in a rigid vessel

Given: $n=0.80\,\mathrm{mol}$ , $T_1=290\,\mathrm K$ , $T_2=410\,\mathrm K$ . Classical monatomic ideal gas in a rigid closed vessel, with no other work or external energy changes.
Find: Find U1, U2, ΔU and Q.

Approach: For the classical monatomic ideal gas use $U=\frac32nRT$ . In a rigid vessel with no other work, $Q=\Delta U$ . Determine initial state energy.
$U_1=\tfrac32(0.80)(8.31446261815324)(290)=2893.43299\,\mathrm J$
Calculate final state at unchanged n.
$U_2=4090.71561\,\mathrm J$
Connect difference and energy balance.
$\begin{aligned}\Delta U&=1197.28262\,\mathrm J\\{} W_{\mathrm{by}}&=0\\{} Q&=\Delta U\end{aligned}$

Result: U rises from about 2.893 to 4.091 kJ; required heat is about 1.197 kJ.

Check: Direct calculation: $\Delta U=nC_{V,m}\Delta T=0.80\cdot12.47169393\cdot120\,\mathrm J=1197.28262\,\mathrm J$ . $\frac{U_2}{U_1}=\frac{410}{290}$ confirms temperature proportionality.

Isothermal expansion supplied by heat

Given: $n=0.50\,\mathrm{mol}$ , $T=300\,\mathrm K$ , $V_2=2V_1$ . Closed classical monatomic ideal gas in a reversible isothermal piston process; only volume work, external energies unchanged.
Find: Find thermal internal energy, its change, work by the gas and heat.

Approach: Isothermally $\Delta U=0$ ; reversible work is $W_{\mathrm{by}}=nRT\ln\!\left(\frac{V_2}{V_1}\right)$ , and $Q=W_{\mathrm{by}}$ . Temperature remains constant, so U does too.
$\begin{aligned}U_1&=U_2=\tfrac32nRT=1870.75409\,\mathrm J\\{} \Delta U&=0\end{aligned}$
Work integration uses $p=\frac{nRT}{V}$ .
$W_{\mathrm{by}}=\int_{V_1}^{2V_1}\frac{nRT}{V}\,dV=nRT\ln2\approx864.47195\,\mathrm J$
The first law requires equal heat input.
$Q=\Delta U+W_{\mathrm{by}}=864.47195\,\mathrm J$

Result: About 864.47 J of heat enters and leaves the sample as work; stored energy U remains unchanged.

Check: Final volume is doubled and absolute pressure halved. Thus pV and $\frac32pV$ remain unchanged. Positive $W_{\mathrm{by}}$ during expansion matches the sign of Q.

The same temperature change determines the same internal energy change, but not equal heat. At constant pressure, supplied heat must additionally cover volume work.[2][5][6]

Equal warming, different heat demand

Given: $n=0.300\,\mathrm{mol}$ of a classical monatomic ideal gas, $T_1=300\,\mathrm K$ , $T_2=340\,\mathrm K$ . A: rigid and closed, no other work. B: closed, reversible constant-pressure heating, only volume work. External energies unchanged.
Find: Find the common internal energy change and heat along each path.

Approach: internal energy change depends only on the temperatures. At constant pressure, the gas equation gives the additional work.
$\begin{aligned}\Delta U&=\frac32nR\Delta T\\W_A&=0\\W_B&=p\Delta V=nR\Delta T\\Q_A&=\Delta U\\Q_B&=\Delta U+W_B\end{aligned}$
Substitute: temperature change is 40 K.
$\begin{aligned}\Delta U&=\frac32(0.300)(8.314462618)(40)\,\mathrm J\\&\approx149.660\,\mathrm J\\W_B&=(0.300)(8.314462618)(40)\,\mathrm J\approx99.774\,\mathrm J\\Q_A&\approx149.660\,\mathrm J\\Q_B&\approx249.434\,\mathrm J\end{aligned}$

Result: Both paths raise internal energy by about 150 J. Path B requires about 249 J of heat because about 99.8 J additionally leaves as work.

Check: Check: $Q_B-W_B=Q_A$ . Heat capacities differ: $C_{V,m}=\frac32R$ , $C_{p,m}=\frac52R$ . This heat statement requires the specified work conditions and does not describe every heating process of arbitrary systems.

What temperature alone reveals

For two equilibrium states at the same n, ΔT suffices to determine ΔU. The path remains relevant for Q and W. The formula can therefore be used in isochoric, isobaric, isothermal or adiabatic exercises; it does not replace the process relation needed in each case.[5]

A sample of 1 mol at 300 K has about 3.74 kJ. This is not heat “contained in the gas”, but modelled internal energy relative to the selected zero. Heat denotes energy transfer; internal energy denotes a state variable.[5]

Five exercises with solutions

Calculate · Foundation

Find U for 0.25 mol of an appropriate monatomic ideal gas at 400 K.

Show hint

Insert n and T.

Learning goal: Calculate thermal energy of an appropriate monatomic ideal gas.

Show solution and calculation

Approach: multiply energy per mole of an appropriate classical monatomic ideal gas by amount.
Retain the selected thermal-energy reference.

Check: twice the amount at the same temperature would have twice this energy; units are joules.

\begin{aligned}U&=\frac32nRT\\&=\frac32(0.25)(8.314462618)(400)\,\mathrm J\\&\approx1247.16939\,\mathrm J\end{aligned}

Calculate · Application

1.20 mol cools by 50 K. Find ΔU without assuming a process for Q.

Show hint

ΔT is negative; ΔU is path independent.

Learning goal: Find energy change from temperature difference without assuming heat transfer.

Show solution and calculation

Approach: apply the same caloric equation to both states at fixed amount, then subtract.
Temperature change is −50 K, and internal energy decreases by about 748.3 J.

Check: cooling agrees with the negative sign. Heat is unspecified without work information; only in a rigid vessel with no other work would $Q=\Delta U$ .

\begin{aligned}\Delta U&=\frac32nR\Delta T\\&=\frac32(1.20)(8.314462618)(-50)\,\mathrm J\\&\approx-748.301636\,\mathrm J\end{aligned}

Interpret · Application

Volume is halved at constant n and T. What happens to p and U?

Show hint

Combine the ideal gas law and U(T).

Learning goal: Distinguish pressure change from unchanged internal energy in isothermal volume change.

Show solution and calculation

Approach: use the equation of state for pressure and the monatomic caloric model for internal energy.
Halving volume doubles pressure.
Internal energy stays unchanged at equal n and T.

Check: in reversible compression, work by the gas is negative; $Q=W_{\mathrm{by}}<0$ maintains constant temperature. Increased pressure alone does not establish increased internal energy.

\begin{aligned}\frac{p_2}{p_1}&=\frac{V_1}{V_2}=2\\U_2&=\frac32nRT=U_1\\\Delta U&=0\end{aligned}

Interpret · Application

Two identical vessels each contain 0.50 mol of the same classical monatomic ideal gas at 320 K. One rests; the other moves uniformly as a whole at 20 m/s. Compare thermal internal energies of the gas fillings. Must their total energies in the laboratory frame also agree? Use $R=8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}$ .

Show hint

Temperature describes random motion relative to mean flow. A common drift is not a temperature increase.

Learning goal: Separate thermal internal energy from bulk motion.

Show solution and calculation

Approach: use the same thermal expression for U for both fillings.
Both have 1995.36 J of thermal internal energy.
In the laboratory, the moving gas additionally has bulk kinetic energy of its total mass; gas mass would be needed to calculate it.

Check: a mere inertial-frame change must not alter thermal energy at unchanged temperature. Vessel energy is outside this gas balance.

\begin{aligned}U_1=U_2&=\frac32nRT=\frac32(0.50)(8.314)(320)\,\mathrm J\\&=1995.36\,\mathrm J\\E_{\mathrm{bulk}}&=\frac12M_{\mathrm{gas}}u^2\end{aligned}

[1][2]

Error analysis · Application

A model gas contains rigid diatomic molecules. Translation and two rotational degrees of freedom are classically active; vibration is frozen and the gas is ideal. For 0.20 mol heated by 40 K, someone uses $\Delta U=\frac32nR\Delta T$ . Correct the coefficient and ΔU using $R=8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}$ . Why is “ideal” insufficient justification for the factor three halves?

Show hint

Each classically active quadratic degree of freedom contributes half a thermal energy unit on average. Count the degrees of freedom explicitly specified here.

Learning goal: Check the degrees-of-freedom assumption behind the monatomic energy formula.

Show solution and calculation

Approach: add three translational and two rotational degrees of freedom.
Five degrees give coefficient five halves and $\Delta U=166.28\,\mathrm J$ .
The monatomic expression would give only 99.768 J.

Check: the corrected value is five thirds of the original value. Ideality neglects intermolecular interactions here but does not automatically remove molecular rotation. This conclusion applies to the specified temperature and degrees-of-freedom model.

\begin{aligned}f&=3+2=5\\\Delta U&=\frac f2nR\Delta T=\frac52(0.20)(8.314)(40)\,\mathrm J=166.28\,\mathrm J\\\frac{\Delta U}{\Delta U_{\mathrm{mono}}}&=\frac53\end{aligned}

[2][4]

When three degrees of freedom are insufficient

The premises are a classical dilute monatomic ideal gas and thermal equilibrium states. Interaction energy, quantum effects and thermal electronic excitation must be negligible. Polyatomic gases, chemical reactions, ionisation and condensed matter need a different description of internal energy.[4]

If amount changes, ΔU is not simply $\frac32nR\Delta T$ with an arbitrary n. Separate states give $U_2-U_1=\frac32R(n_2T_2-n_1T_1)$ in the same model; an open system additionally needs balances for energy and material transport. The rigid-vessel example explicitly concerns a closed system.[5]

What to carry into your next calculation

For an appropriate monatomic ideal gas, thermal energy is $U=\frac32nRT$ . At fixed amount, temperature difference alone determines its change. Heat additionally depends on process and work. Extra molecular degrees of freedom or quantum effects require a different caloric model.[2][5]

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

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This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

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An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Highlighted model premises and the difference between heat and state change in question headings; added relevant paragraph connections to particle energy and the first law.
October 1, 2026Expanded explanations and learning transitions; added a further fully worked example and an original quantitative diagram; clarified assumptions and calculation steps.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.

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Quanta. Internal energy of a monatomic ideal gas. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/innere-energie-einatomiges-idealgas

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Frequently asked questions about Internal energy of a monatomic ideal gas

What does Internal energy of a monatomic ideal gas describe physically?+

$U=\frac32nRT$ describes internal energy of a classical monatomic ideal gas sample at equilibrium. It sums random translational energy of all atoms; motion of the whole sample is excluded. A constant atomic ground-state contribution is omitted through the chosen energy zero. At fixed n, U in this model depends only on absolute temperature. For example, 0.50 mol at 300 K gives about 1870.754 J. Heat and work instead describe transfers during a process, not separate contributions stored in a state. For temperature changes, $\Delta U=\frac32nR\Delta T$ at fixed n is particularly useful.

How can I check a calculation involving Internal energy of a monatomic ideal gas?+

For 0.80 mol between 290 and 410 K, $U_1\approx2893.433\,\mathrm J$ , $U_2\approx4090.716\,\mathrm J$ and $\Delta U\approx1197.283\,\mathrm J$ . Independently check the difference with $nC_{V,m}\Delta T$ and $C_{V,m}=\frac32R$ . Both methods must agree. In a rigid closed vessel without other work, $Q=\Delta U$ ; do not transfer that equality unchecked to a moving piston. An isothermal process at fixed n instead must give $\Delta U=0$ . Despite constant U, heat and work can be nonzero. Therefore also check $\Delta U=Q-W_{\mathrm{by}}$ , with work done outward by the gas defined as positive. Initial and final states must also use the same energy zero. These process balances assume a closed system with unchanged or negligible bulk kinetic and external potential energy.

What assumptions and limits apply to Internal energy of a monatomic ideal gas?+

The model assumes a dilute classical monatomic ideal gas with negligible interaction energy and no relevant electronic excitation. Initial and final states must admit thermodynamic description. Polyatomic molecules can store rotational and vibrational energy; ionisation and chemical conversion also require additional contributions. Classical equipartition can fail at low temperature too. $\Delta U=\frac32nR\Delta T$ requires unchanged amount of substance. If amount changes, $U_2-U_1=\frac32R(n_2T_2-n_1T_1)$ is the appropriate state difference within this model. An open installation additionally needs material and energy transport balances and cannot be treated as a closed vessel.

How do I check units and scale for Internal energy of a monatomic ideal gas?+

With n in mol, R in J/(mol·K) and T in K, nRT has units joule. Absolute temperature matters: 20 °C corresponds to 293.15 K. For ΔT, however, kelvin and Celsius differences have the same numerical value. A useful scale is $\frac Un\approx3741.508\,\mathrm{J\,mol^{-1}}$ at 300 K. Doubling n at equal T doubles U; doubling T at fixed n does the same. For an independent unit check, use $U=\frac32pV$ when the ideal gas law applies and p is absolute pressure. Pa·m³ also gives joules, not force.

When should I combine Internal energy of a monatomic ideal gas with another relation?+

Combine the state formula with $\Delta U=Q-W_{\mathrm{by}}$ when a process specifies heat or work. For a rigid closed vessel without additional work, $Q=\frac32nR\Delta T$ . In reversible isothermal expansion, $p=\frac{nRT}{V}$ gives $W_{\mathrm{by}}=nRT\ln\frac{V_2}{V_1}$ ; because $\Delta U=0$ , $Q=W_{\mathrm{by}}$ . For example, 0.50 mol at 300 K and doubled volume gives about 864.472 J positive work and equal heat input. Another process needs its own relation or a specified work value. The U formula alone does not establish the path or Q, although it determines the state change. These process balances assume a closed system with unchanged or negligible bulk kinetic and external potential energy.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Vol. 2, §2.2: Pressure, Temperature, and RMS Speed

More formulas: Heat and thermodynamics

Q = m \cdot c \cdot \Delta T

Heat Capacity: Q = m·c·ΔT

P = \sigma \cdot A \cdot T^4

Stefan-Boltzmann Law

\Delta L = \alpha L_0 \Delta T

Linear thermal expansion

Q = m L

Latent heat during a phase change

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Symbol

Meaning

Unit

U

Internal energy

n

Amount of substance

mol

R

Universal gas constant

J/(mol·K)

T

Absolute temperature

Derivation & background

Exam blueprint

Validity range

Derivation steps

Sum mean translational atomic energy and convert to amount using NA.

1Consider a classical monatomic ideal gas sample at equilibrium; neglect electronic excitation.
2Three translational degrees each contribute $\frac12k_BT$ , giving $\frac32k_BT$ per atom.
3Sum over all N atoms: $U=\frac32Nk_BT$ . The constant atomic ground-state contribution is omitted through the energy reference.
4Using $N=nN_A$ and $R=N_Ak_B$ gives $U=\frac32nRT$ . Shared gas motion remains separate.
5At fixed n, subtract: $\Delta U=\frac32nR(T_2-T_1)$ .
6Then add the appropriate process balance: $\Delta U=Q-W_{\mathrm{by}}$ for a closed system without external-energy change; $Q=\Delta U$ only at zero net work.

How do you calculate with Internal energy of a monatomic ideal gas?

Here is how to work through a typical Internal energy of a monatomic ideal gas (U = 3/2 nRT) task step by step:

Task

Calculate the quantity from the worked example for “Internal energy of a monatomic ideal gas”.

Solution path

For $n=1.0\,\mathrm{mol}$ of a classical monatomic ideal gas at $T=300\,\mathrm K$ , $U=\frac32nRT\approx3.74\,\mathrm{kJ}$ , using the stated thermal-energy reference.

Energy for the random motion of atoms

When does the formula describe internal energy of an entire gas sample?

U=\frac32Nk_{\mathrm B}T=\frac32nRT

N counts atoms; n measures amount of substance. The chosen energy zero omits constant atomic ground-state contributions.[1]

From an atom to amount of substance

U=N\langle E_{\mathrm{tr}}\rangle=\frac32nN_{\mathrm A}k_{\mathrm B}T=\frac32nRT

$R=N_Ak_B$ is the molar gas constant. The sum assumes negligible interaction energy.[7]

Changes in U and heat capacity

\begin{aligned}&\Delta U=\frac32nR(T_2-T_1)\\&C_{V,m}=\frac32R\end{aligned}

The first equation applies at fixed amount even when volume changes.[2]

\begin{aligned}&T=\frac{2U}{3nR}\\&n=\frac{2U}{3RT}\\&U=\frac32pV\end{aligned}

The first rearrangements use the same energy zero. $U=\frac32pV$ additionally uses $pV=nRT$ ; p is absolute pressure.[1]

Temperature and amount set thermal energy

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Is internal-energy change always equal to supplied heat?

\Delta U=Q-W_{\mathrm{by}}

Three examples: energy, heat and process

Heating in a rigid vessel

Given: $n=0.80\,\mathrm{mol}$ , $T_1=290\,\mathrm K$ , $T_2=410\,\mathrm K$ . Classical monatomic ideal gas in a rigid closed vessel, with no other work or external energy changes.
Find: Find U1, U2, ΔU and Q.

Approach: For the classical monatomic ideal gas use $U=\frac32nRT$ . In a rigid vessel with no other work, $Q=\Delta U$ . Determine initial state energy.
$U_1=\tfrac32(0.80)(8.31446261815324)(290)=2893.43299\,\mathrm J$
Calculate final state at unchanged n.
$U_2=4090.71561\,\mathrm J$
Connect difference and energy balance.
$\begin{aligned}\Delta U&=1197.28262\,\mathrm J\\{} W_{\mathrm{by}}&=0\\{} Q&=\Delta U\end{aligned}$

Result: U rises from about 2.893 to 4.091 kJ; required heat is about 1.197 kJ.

Check: Direct calculation: $\Delta U=nC_{V,m}\Delta T=0.80\cdot12.47169393\cdot120\,\mathrm J=1197.28262\,\mathrm J$ . $\frac{U_2}{U_1}=\frac{410}{290}$ confirms temperature proportionality.

Isothermal expansion supplied by heat

Given: $n=0.50\,\mathrm{mol}$ , $T=300\,\mathrm K$ , $V_2=2V_1$ . Closed classical monatomic ideal gas in a reversible isothermal piston process; only volume work, external energies unchanged.
Find: Find thermal internal energy, its change, work by the gas and heat.

Approach: Isothermally $\Delta U=0$ ; reversible work is $W_{\mathrm{by}}=nRT\ln\!\left(\frac{V_2}{V_1}\right)$ , and $Q=W_{\mathrm{by}}$ . Temperature remains constant, so U does too.
$\begin{aligned}U_1&=U_2=\tfrac32nRT=1870.75409\,\mathrm J\\{} \Delta U&=0\end{aligned}$
Work integration uses $p=\frac{nRT}{V}$ .
$W_{\mathrm{by}}=\int_{V_1}^{2V_1}\frac{nRT}{V}\,dV=nRT\ln2\approx864.47195\,\mathrm J$
The first law requires equal heat input.
$Q=\Delta U+W_{\mathrm{by}}=864.47195\,\mathrm J$

Result: About 864.47 J of heat enters and leaves the sample as work; stored energy U remains unchanged.

Check: Final volume is doubled and absolute pressure halved. Thus pV and $\frac32pV$ remain unchanged. Positive $W_{\mathrm{by}}$ during expansion matches the sign of Q.

The same temperature change determines the same internal energy change, but not equal heat. At constant pressure, supplied heat must additionally cover volume work.[2][5][6]

Equal warming, different heat demand

Given: $n=0.300\,\mathrm{mol}$ of a classical monatomic ideal gas, $T_1=300\,\mathrm K$ , $T_2=340\,\mathrm K$ . A: rigid and closed, no other work. B: closed, reversible constant-pressure heating, only volume work. External energies unchanged.
Find: Find the common internal energy change and heat along each path.

Approach: internal energy change depends only on the temperatures. At constant pressure, the gas equation gives the additional work.
$\begin{aligned}\Delta U&=\frac32nR\Delta T\\W_A&=0\\W_B&=p\Delta V=nR\Delta T\\Q_A&=\Delta U\\Q_B&=\Delta U+W_B\end{aligned}$
Substitute: temperature change is 40 K.
$\begin{aligned}\Delta U&=\frac32(0.300)(8.314462618)(40)\,\mathrm J\\&\approx149.660\,\mathrm J\\W_B&=(0.300)(8.314462618)(40)\,\mathrm J\approx99.774\,\mathrm J\\Q_A&\approx149.660\,\mathrm J\\Q_B&\approx249.434\,\mathrm J\end{aligned}$

Result: Both paths raise internal energy by about 150 J. Path B requires about 249 J of heat because about 99.8 J additionally leaves as work.

What temperature alone reveals

Five exercises with solutions

Calculate · Foundation

Find U for 0.25 mol of an appropriate monatomic ideal gas at 400 K.

Show hint

Insert n and T.

Learning goal: Calculate thermal energy of an appropriate monatomic ideal gas.

Show solution and calculation

Approach: multiply energy per mole of an appropriate classical monatomic ideal gas by amount.
Retain the selected thermal-energy reference.

Check: twice the amount at the same temperature would have twice this energy; units are joules.

\begin{aligned}U&=\frac32nRT\\&=\frac32(0.25)(8.314462618)(400)\,\mathrm J\\&\approx1247.16939\,\mathrm J\end{aligned}

Calculate · Application

1.20 mol cools by 50 K. Find ΔU without assuming a process for Q.

Show hint

ΔT is negative; ΔU is path independent.

Learning goal: Find energy change from temperature difference without assuming heat transfer.

Show solution and calculation

Approach: apply the same caloric equation to both states at fixed amount, then subtract.
Temperature change is −50 K, and internal energy decreases by about 748.3 J.

Check: cooling agrees with the negative sign. Heat is unspecified without work information; only in a rigid vessel with no other work would $Q=\Delta U$ .

\begin{aligned}\Delta U&=\frac32nR\Delta T\\&=\frac32(1.20)(8.314462618)(-50)\,\mathrm J\\&\approx-748.301636\,\mathrm J\end{aligned}

Interpret · Application

Volume is halved at constant n and T. What happens to p and U?

Show hint

Combine the ideal gas law and U(T).

Learning goal: Distinguish pressure change from unchanged internal energy in isothermal volume change.

Show solution and calculation

Approach: use the equation of state for pressure and the monatomic caloric model for internal energy.
Halving volume doubles pressure.
Internal energy stays unchanged at equal n and T.

Check: in reversible compression, work by the gas is negative; $Q=W_{\mathrm{by}}<0$ maintains constant temperature. Increased pressure alone does not establish increased internal energy.

\begin{aligned}\frac{p_2}{p_1}&=\frac{V_1}{V_2}=2\\U_2&=\frac32nRT=U_1\\\Delta U&=0\end{aligned}

Interpret · Application

Two identical vessels each contain 0.50 mol of the same classical monatomic ideal gas at 320 K. One rests; the other moves uniformly as a whole at 20 m/s. Compare thermal internal energies of the gas fillings. Must their total energies in the laboratory frame also agree? Use $R=8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}$ .

Show hint

Temperature describes random motion relative to mean flow. A common drift is not a temperature increase.

Learning goal: Separate thermal internal energy from bulk motion.

Show solution and calculation

Approach: use the same thermal expression for U for both fillings.
Both have 1995.36 J of thermal internal energy.
In the laboratory, the moving gas additionally has bulk kinetic energy of its total mass; gas mass would be needed to calculate it.

Check: a mere inertial-frame change must not alter thermal energy at unchanged temperature. Vessel energy is outside this gas balance.

\begin{aligned}U_1=U_2&=\frac32nRT=\frac32(0.50)(8.314)(320)\,\mathrm J\\&=1995.36\,\mathrm J\\E_{\mathrm{bulk}}&=\frac12M_{\mathrm{gas}}u^2\end{aligned}

[1][2]

Error analysis · Application

A model gas contains rigid diatomic molecules. Translation and two rotational degrees of freedom are classically active; vibration is frozen and the gas is ideal. For 0.20 mol heated by 40 K, someone uses $\Delta U=\frac32nR\Delta T$ . Correct the coefficient and ΔU using $R=8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}$ . Why is “ideal” insufficient justification for the factor three halves?

Show hint

Each classically active quadratic degree of freedom contributes half a thermal energy unit on average. Count the degrees of freedom explicitly specified here.

Learning goal: Check the degrees-of-freedom assumption behind the monatomic energy formula.

Show solution and calculation

Approach: add three translational and two rotational degrees of freedom.
Five degrees give coefficient five halves and $\Delta U=166.28\,\mathrm J$ .
The monatomic expression would give only 99.768 J.

\begin{aligned}f&=3+2=5\\\Delta U&=\frac f2nR\Delta T=\frac52(0.20)(8.314)(40)\,\mathrm J=166.28\,\mathrm J\\\frac{\Delta U}{\Delta U_{\mathrm{mono}}}&=\frac53\end{aligned}

[2][4]

When three degrees of freedom are insufficient

What to carry into your next calculation

Sources for this explanation

Quanta · Updated October 2, 2026

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An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Highlighted model premises and the difference between heat and state change in question headings; added relevant paragraph connections to particle energy and the first law.
October 1, 2026Expanded explanations and learning transitions; added a further fully worked example and an original quantitative diagram; clarified assumptions and calculation steps.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.

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Quanta. Internal energy of a monatomic ideal gas. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/innere-energie-einatomiges-idealgas

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