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Internal energy and state variables

Clarify what internal energy includes, why it is a state variable, and how it behaves for ideal gases versus condensed matter.

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  • University Physics Year 1
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Internal energy records microscopic energy stored in a system. Heat and work record how energy crosses its boundary. For a fixed amount of ideal gas, temperature fixes the internal energy, but it does not tell you how much heat was transferred: work can supply or remove some of the energy too.

At a glance

  • Prerequisites: the first law; the ideal-gas equation
  • You will learn: separate microscopic energy from bulk motion; use an ideal gas’s temperature endpoints to calculate internal-energy change; distinguish that change from heat
  • Key result: for a fixed amount and composition of ideal gas, Δ U = n∫_Tᵢ^TfCV(T) dT.
  • Common trap: using Q = nCVΔ T in a process that also does work.

Identify the stored energy

Here U denotes internal energy; other course lessons also use E for this same quantity. It includes microscopic translational, rotational and vibrational energies, interaction energies, and electronic or nuclear contributions relevant to the model. It excludes the bulk centre-of-mass kinetic energy and external-field potential energy of the whole sample.

The amount and composition of matter are fixed in the gas calculations below, which use the thermodynamic ideal-gas model PV = nRT. Saying “U depends only on T” refers to that fixed sample, not to comparing different amounts of gas. Doubling the amount doubles its internal-energy change for the same temperature interval.

For an ideal gas, intermolecular potential energy is negligible. Active internal molecular modes can still contribute: vibration includes intramolecular potential energy as well as kinetic energy. Thus the ideal-gas model does not remove every possible potential-energy contribution.

Use temperature endpoints without assuming the heat

For an ideal gas with fixed n and composition,

dU = nCV(T) dT, Δ U = n∫_Tᵢ^TfCV(T) dT.

Here CV is the molar heat capacity at constant volume, in J mol⁻¹K⁻¹. The gas’s total constant-volume heat capacity is nCV, in J/K. If CV is nearly constant over the temperature interval, Δ U ≈ nCV(Tf-Tᵢ).

The subscript V describes how the heat capacity is measured. In a rigid vessel with no other work, heat input equals internal-energy change. Once the ideal-gas relation U(T) is known, however, the same temperature integral gives Δ U for any process between the stated temperature endpoints. This does not make heat path independent: Q = Δ U-Wₒₙ = Δ U + Wₒᵤₜ. The work must still be known. The OpenStax ideal-gas heat-capacity treatment derives the temperature-only internal-energy relation and distinguishes constant-volume from constant-pressure heating.

For a liquid, solid or real gas, internal energy need not depend on temperature alone. Changes in arrangement and interaction energy can matter. During melting, for example, energy can increase while temperature remains constant.

Internal-energy change from temperature

Worked example 1

Heating a monatomic gas

Problem

Find the internal-energy change of 2.0 mol of classical monatomic ideal gas whose temperature changes from 300 K to 360 K. Use R = 8.314 J mol⁻¹K⁻¹. No process path is specified. Does this information also fix the heat transfer?

Show full solution

  1. Select the temperature-only relation

    Method

    Use CV = 3R/2 and Δ T = 60 K.

    Reason

    The classical monatomic model has three translational energy terms. For this fixed sample, the temperature endpoints determine internal-energy change.

    Working

    Answer

    Δ U = nCVΔ T = (2.0)((3/2)R)(60) = 1496.52 J ≈ +1.5 kJ.

  2. Separate the state change from the transfer

    Working

    Answer

    Internal energy increases by about 1.5 kJ. The heat is not yet determined, because Q = Δ U-Wₒₙ also requires the work. If there is no work, Q = Δ U; if the gas delivers work, more heat is needed for the same internal-energy increase.

Guided practice 2

The same temperature rise at constant pressure

About 7 min

Problem

The same 2.0 mol monatomic ideal gas now warms from 300 K to 360 K at constant pressure through a massless frictionless piston. Expansion is slow, the only work is pressure–volume work, and bulk kinetic and potential energies do not change. Find its internal-energy change, work output and heat input. Compare with heating the gas through the same temperature interval in a rigid vessel with no other work.

Distinguish stored energy, work and heat

Unit: J
Unit: J
Unit: J

Hints

Hint 1

Use each relation for its own job

Temperature fixes ΔU. The change in PV fixes expansion work at constant pressure. The first law then fixes heat.
Show solution step by step

  1. Reuse the internal-energy change

    Method

    Use the same fixed sample and temperature endpoints.

    Reason

    For an ideal gas, changing the process path does not change the temperature-only internal-energy relation.

    Working

    Answer

    Δ U = n(3R/2)Δ T = 1496.52 J.

  2. Calculate work and heat

    Method

    Use PV = nRT at the two constant-pressure endpoints.

    Reason

    The increase in volume also transfers energy out through piston work.

    Working

    Answer

    Wₒᵤₜ = P(Vf-Vᵢ) = nR(Tf-Tᵢ) = 997.68 J, Q = Δ U + Wₒᵤₜ = 1496.52 + 997.68 = 2494.20 J. In the rigid-vessel comparison, Wₒᵤₜ = 0, so Q = 1496.52 J. The final temperatures agree, but the final volumes differ; equal temperature does not make these complete final states identical.

Practice set

Check your understanding 1

Why is internal energy a state variable while heat is a path variable?

Show hint
Compare stored energy with energy transferred during a process.
Show answer
A given equilibrium state has a definite internal energy, so its change depends on the endpoints. Heat records a transfer during a process; different paths between the same complete states can exchange different heat amounts.

Check your understanding 2

An ideal gas has n = 1.5 mol, constant CV = 20.8 J mol⁻¹K⁻¹ and Δ T = -15 K. Find its internal-energy change. Does this show that 468 J left as heat?

Show hint
Keep the negative temperature change, then distinguish state change from transfer.
Show answer
Δ U = 1.5(20.8)(-15) = -468 J ≈ -4.7 × 10² J. This fixes the internal-energy decrease, not the heat. Some or all energy could have left through work; use the first law and the work information before assigning Q.

Check your understanding 3

Can internal energy change during melting at constant temperature? Explain.

Show hint
Temperature does not track every microscopic potential-energy change.
Show answer
Yes. Melting changes molecular arrangement and the interaction-energy contribution, so internal energy can change while temperature stays constant. The fixed-sample ideal-gas rule U = U(T) does not describe a melting solid.

Apply the model to expansion into vacuum

Check your understanding 4

A fixed amount of classical monatomic ideal gas initially at 300 K occupies one compartment of a sealed insulated rigid vessel. The other compartment is empty. A partition is removed, and the gas eventually equilibrates at twice its original volume. Neglect wall and partition heat capacities and energy transfer during partition removal. No shaft or electrical work occurs, and bulk kinetic energy is unchanged between the endpoints. Find the internal-energy change and final temperature. Explain why the volume increase does not imply work output in this process.

Unit: J
Unit: K
Show answer

Insulation gives Q = 0. The outer boundary is rigid and the expansion into the evacuated compartment opposes no external pressure, so no work crosses it: Wₒₙ = 0. The first law gives Δ U = 0. For a fixed amount of classical monatomic ideal gas, 0 = Δ U = n(3R/2)(Tf-Tᵢ), Tf = Tᵢ = 300 K.

The gas has nonzero pressure initially, but multiplying that pressure by the volume increase does not calculate work against vacuum. Nor is there a quasistatic equilibrium pressure curve during the expansion. A reversible adiabatic piston expansion would deliver work and lower temperature; it is a different process. The unchanged-temperature result here relies on the ideal-gas model, not on every gas having temperature-only internal energy. Compare your explanation against these boundary, process and model distinctions.

Before you move on

  • Internal energy is microscopic stored energy, separate from bulk motion.
  • For a fixed amount and composition of ideal gas, temperature endpoints determine Δ U through the molar CV.
  • Knowing Δ U does not determine heat until work is known.
  • A volume increase need not deliver work: check the external pressure and boundary.
  • During a phase change, temperature can stay constant while internal energy changes.

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Syllabus and review details

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