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Heat capacities of gases

Derive gas heat capacities at constant volume and pressure using molecular degrees of freedom and Cp − Cv = R.

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  • University Physics Year 1
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The same ideal gas can require different heat inputs for the same temperature rise. At fixed volume, no boundary expansion work occurs. At fixed pressure, the gas expands and transfers energy to its surroundings as work. The internal-energy rise is the same for the same ideal-gas amount and temperature interval; the heat input need not be.

Use the ideal gas equation, heat-capacity definitions and the first law. Here Q > 0 is heat into the gas, Wₒₙ > 0 is work into it, and Δ U = Q + Wₒₙ. Work output is Wₒᵤₜ = -Wₒₙ.

Distinguish molar capacities from sample capacities

In this lesson CV and CP are molar capacities, in J/(mol K). The sample has capacities nCV and nCP. Assume a fixed amount and composition, only pressure–volume work and well-defined thermodynamic states, with negligible bulk kinetic and potential-energy changes. An ideal gas has U = U(T) at fixed amount, so dU = nCV(T) dT. Although constant-volume heating defines CV, this ideal-gas internal-energy relation is valid between temperatures reached by other paths too. It is the heat relation δ Q = nCV dT that needs constant volume and no other work.

Derive the difference using the expansion energy

At fixed volume, boundary work vanishes and δ QV = dU = nCV dT. For a quasistatic constant-pressure path, δ QP = dU + P dV. Differentiating PV = nRT at fixed P and n gives P dV = nR dT. Therefore δ QP = n(CV + R) dT, CP-CV = R. The extra heat is the expansion work. This molar relation remains true when ideal-gas capacities vary with temperature; it is not a universal difference for a real gas, liquid or solid. The MIT discussion of heat capacities distinguishes the defining paths from the broader ideal-gas energy relation.

Connect capacities to active molecular motion

In a classical regime with f active quadratic energy terms and constant f, equipartition gives U-U₀ = (f/2)nRT, so CV = fR/2 and CP = (f/2 + 1)R. A monatomic ideal gas has three translational terms, giving CV = 3R/2 and γ = CP/CV = 5/3.

For a diatomic molecule with translation and rotation active but vibration frozen, f = 5, giving CV = 5R/2 and γ = 7/5. A thermally active vibrational mode has both kinetic and potential quadratic terms. Which modes are active depends on temperature relative to their quantum energy spacings; the room-temperature diatomic values are an approximation for an appropriate gas and range, not a universal count.

If modes activate gradually, use a measured or modelled CV(T) rather than differentiating f(T)RT/2 as though equipartition had supplied a continuous mode count. The ratio γ(T) = 1 + R/CV(T) may then vary, and a constant-γ adiabatic formula needs separate justification.

Comparing heating at fixed volume and pressure

Worked example 1

Same temperature rise, different heat input

Problem

Find the heat required to warm 1.5 mol of monatomic ideal gas by 40 K at constant volume and at constant pressure. Assume only pressure–volume work and constant heat capacities. Use R = 8.314 J mol⁻¹K⁻¹.
Show full solution

  1. Calculate constant-volume heating

    Method

    Use CV = 3R/2 in QV = nCVΔ T.

    Reason

    There is no expansion work at fixed volume, so all heat raises internal energy.

    Working

    Answer

    QV = nCVΔ T = 1.5((3/2)R)(40) ≈ 748 J.

  2. Include constant-pressure expansion

    Method

    Use CP = CV + R = 5R/2.

    Reason

    Heating at fixed pressure also supplies expansion work, requiring nRΔ T of extra heat.

    Working

    Answer

    QP = nCPΔ T = 1.5((5/2)R)(40) ≈ 1247 J.

  3. Compare the heat inputs

    Working

    Answer

    QV ≈ 748 J and QP ≈ 1247 J (about 0.75 and 1.2 kJ). The unrounded difference is nRΔ T ≈ 499 J, the constant-pressure work output.

Guided practice 2

Keep Mayer’s relation when capacity varies

About 8 min

Problem

A supplied ideal-gas model has molar CV(T) = R[3/2 + T/(300 K)] over 300–360 K. Heat 0.50 mol through that interval in separate constant-volume and quasistatic constant-pressure experiments, with only pressure–volume work. Use R = 8.314 J/(mol K). Find the internal-energy increase, constant-pressure work output and constant-pressure heat input. Explain why the final volumes need not agree even though the internal-energy changes do.

Work through the model, then compare your reasoning

Unit: J
Unit: J
Unit: J

Hints

Hint 1

Check the model

Integrate C_V(T); then use C_P(T) − C_V(T) = R instead of replacing the capacity by a fixed value.
Show solution step by step

  1. Integrate the state change

    Method

    Use Δ U = n∫ CV(T) dT.

    Reason

    For this ideal gas, internal energy depends only on temperature and amount.

    Working

    Answer

    Δ U = 0.50R[(3/2)(60) + (360²-300²)/(2 × 300)] = 0.50R(156) = 648.492 J. The kelvin factors in the supplied model make the bracket an interval with units K.

  2. Account for boundary work

    Method

    Use W_(out,P) = nRΔ T and QP = Δ U + W_(out,P).

    Reason

    The extra constant-pressure heat goes into expansion work.

    Working

    Answer

    W_(out,P) = 0.50(8.314)(60) = 249.42 J, so QP = 897.912 J. At fixed volume QV = 648.492 J and pressure rises. At fixed pressure volume rises by the temperature ratio. These final states have different P,V, but the same ideal-gas U(T) because their amounts and final temperatures agree.

Check the missing work channel

Check your understanding 1

A sealed rigid vessel contains 1.0 mol of a classical monatomic ideal gas that warms by 20 K. The gas receives 200 J of heat and is stirred through an externally driven paddle. Neglect wall and paddle heat storage. Take R = 8.314 J/(mol K), neglect macroscopic energy changes and find the work input. A colleague uses Q = nCVΔ T to reject the data. Explain the missing assumption.

Unit: J
Show answer

Δ U = n(3R/2)Δ T = 249.42 J. The first law gives Wₒₙ = Δ U-Q = 49.42 J. Rigid walls remove pressure–volume work, not paddle work. The relation for Δ U still applies, but equating it with heat requires no other work channel. The data are consistent with the stated stirring input.

Practice set

Check your understanding 2

A sample contains 2 mol of diatomic ideal gas with translation and rotation active but vibration frozen. Find CV and CP per mole.

Show hint
Use five active quadratic terms and CP-CV = R.
Show answer
CV = 5R/2 ≈ 20.8 J mol⁻¹K⁻¹ and CP = 7R/2 ≈ 29.1 J mol⁻¹K⁻¹. Do not multiply by two: the question asks for molar values, not the whole sample’s capacities.

Check your understanding 3

For the same amount of ideal gas heated over the same temperature interval with only pressure–volume work, show that QP-QV = nRΔ T.

Show hint

Subtract the two heat integrals and use CP(T)-CV(T) = R.

Show answer

QP-QV = n∫_Tᵢ^Tf(CP-CV) dT = n∫_Tᵢ^TfR dT = nRΔ T. This remains true even if the heat capacities vary with temperature; the extra heat becomes expansion work.

Check your understanding 4

Find γ = CP/CV for a classical monatomic ideal gas.

Show hint
Use CV = 3R/2 and add R to get CP.

Show answer

γ = (5R/2)/(3R/2) = 5/3 ≈ 1.67. The common factor R/2 cancels, so γ is dimensionless.

What to keep

For an ideal gas, molar CP-CV = R, while Δ U = n∫ CV(T) dT depends on its temperature interval. Heat depends on the path and work channels. Use equipartition values only where the corresponding modes are thermally active and their capacity model fits the range.

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Syllabus and review details

No official syllabus alignment is listed for this lesson.

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