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Adiabatic processes

Develop adiabatic-process relations for ideal gases and connect temperature change to compression and expansion work.

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  • University Physics Year 1
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Adiabatic means no heat transfer through the chosen boundary. It does not mean constant temperature, no energy transfer or reversibility. An insulated gas can cool by producing work, warm through work input, or undergo a free expansion with no change in its ideal-gas endpoint temperature.

Use the first law, ideal gas law and gas heat capacities. With Q > 0 into the gas and Wₒₙ > 0 into it, Δ U = Q + Wₒₙ. For an adiabatic process, Δ U = Wₒₙ = -Wₒᵤₜ, assuming negligible bulk energy changes.

Justify the boundary approximation

Insulation or a short process time can make heat transfer small compared with the relevant energy change, but “fast” alone does not prove Q ≈ 0. Estimate the possible boundary heat during the actual interval. A small heat term relative to a large stored energy is not necessarily small relative to the work or temperature change being predicted.

For a fixed amount of ideal gas, Δ U = n∫ CV(T) dT. That relation applies between its equilibrium endpoints even when the route is irreversible. Relating work to -∫ P dV using one gas pressure requires a quasistatic mechanically balanced boundary path; a prescribed finite external load is a different work model.

Derive the reversible power law

For reversible adiabatic expansion with only pressure–volume work, nCV(T) dT = -P dV = -nRT/V dV. Thus CV(T)dT/T = -R dV/V. If CV is constant over the interval, integration gives TV^(γ-1) = constant, PV^γ = constant, γ = CP/CV = 1 + R/CV. The constants have the appropriate units for the chosen state variables; do not take logarithms of dimensional quantities without forming ratios. If capacity varies, the correct relation is instead ∫_Tᵢ^TfCV(T)/T dT = -R ln Vf/Vᵢ. Do not replace a varying capacity by a constant γ without checking the resulting approximation. Reversible adiabatic means constant entropy; an irreversible adiabatic process can generate entropy, so “adiabatic” alone does not justify the power law.

Compare the two reversible curves

Two reversible expansions from one initial state

From relative volume one to three, the solid isothermal curve falls from relative pressure one to one third. The dashed reversible monatomic adiabatic curve falls from one to about 0.160. Work output is the area under the selected curve over its volume interval, multiplied by the initial pressure–volume product.

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From relative volume one to three, the solid isothermal curve falls from relative pressure one to one third. The dashed reversible monatomic adiabatic curve falls from one to about 0.160. Work output is the area under the selected curve over its volume interval, multiplied by the initial pressure–volume product.From relative volume one to three, the solid isothermal curve falls from relative pressure one to one third. The dashed reversible monatomic adiabatic curve falls from one to about 0.160. Work output is the area under the selected curve over its volume interval, multiplied by the initial pressure–volume product.
The adiabatic pressure falls faster because the gas cools while producing work. At equal final volume, the curves have different final pressures and temperatures.
Open full-size graph
What to noticeCalculated ideal-gas models, not measurements. The gas amount is fixed; the isotherm remains at the initial temperature. The adiabatic path is reversible, with constant monatomic heat-capacity ratio 5/3 and only pressure–volume work. No equilibrium path is implied for free expansion.
View figure data
Figure data: Two reversible expansions from one initial state
Relative volume, V/Vᵢ (dimensionless)IsothermalReversible adiabatic
1.001.00001.0000
1.050.95240.9219
1.100.90910.8531
1.150.86960.7922
1.200.83330.7380
1.250.80000.6894
1.300.76920.6458
1.350.74070.6064
1.400.71430.5708
1.450.68970.5383
1.500.66670.5088
1.550.64520.4817
1.600.62500.4569
1.650.60610.4340
1.700.58820.4130
1.750.57140.3935
1.800.55560.3754
1.850.54050.3587
1.900.52630.3431
1.950.51280.3286
2.000.50000.3150
2.050.48780.3023
2.100.47620.2904
2.150.46510.2792
2.200.45450.2687
2.250.44440.2588
2.300.43480.2495
2.350.42550.2407
2.400.41670.2324
2.450.40820.2246
2.500.40000.2172
2.550.39220.2101
2.600.38460.2034
2.650.37740.1971
2.700.37040.1910
2.750.36360.1853
2.800.35710.1798
2.850.35090.1746
2.900.34480.1696
2.950.33900.1648
3.000.33330.1602

Both curves start in the same state. During the adiabatic expansion, energy leaves as work without replacement by heat, so temperature falls and pressure drops faster. At the same final volume, this model has a smaller work area than the isotherm at the starting temperature, but it also has a different final state. This comparison is not a comparison of two routes between identical thermodynamic endpoints.

Energy change during adiabatic expansion

Worked example 1

Adiabatic expansion

Problem

One mole of classical monatomic ideal gas expands adiabatically from 360 K to 300 K. Only pressure–volume work occurs, and bulk kinetic and potential energies are unchanged between endpoints. Use R = 8.314 J/(mol K). Find its internal-energy change and work output. Must you know whether the expansion was reversible?

Show full solution

  1. Find the internal-energy change

    Method

    Use Δ U = nCV(Tf-Tᵢ) with CV = (3/2)R.

    Reason

    An ideal gas’s internal energy depends only on temperature; the monatomic heat capacity sets the change per kelvin.

    Working

    Answer

    Δ U = nCV(Tf-Tᵢ) = (1)((3/2)R)(-60 K) ≈ -7.48 × 10² J

  2. State the energy and work

    Working

    Answer

    Since Q = 0, Wₒₙ = Δ U and Wₒᵤₜ = -Wₒₙ: Δ U = -7.48 × 10² J, Wₒᵤₜ = +7.48 × 10² J The gas supplies positive work output by losing internal energy, consistent with its cooling. The endpoint internal-energy relation and first law do not require reversibility; a reversible pressure–volume power law would require additional assumptions.

Guided practice 2

Choose the assumptions before using the exponent

About 8 min

Problem

One mole of classical monatomic ideal gas begins at 300 K in 5.0 L and expands reversibly and adiabatically to 10 L, with only pressure–volume work. Use CV = 3R/2 and R = 8.314 J/(mol K). Find the final temperature, final pressure and work output. State which assumptions justify the volume–temperature power law and check the energy-transfer signs.

Work through the model, then compare your reasoning

Unit: K
Unit: kPa
Unit: J

Hints

Hint 1

Check the model

Adiabatic removes heat. Reversibility and constant capacity supply the power law; they are additional assumptions.
Show solution step by step

  1. Select the reversible relation

    Method

    Use γ = 5/3 and Tf/Tᵢ = (Vᵢ/Vf)^(2/3).

    Reason

    The active monatomic degrees of freedom fix the constant classical capacity.

    Working

    Answer

    Tf = 300(1/2)^(2/3) = 188.988… K. The temperature decreases because the gas supplies work without receiving heat.

  2. Recover pressure and work

    Method

    Use Pf = nRTf/Vf and Wₒᵤₜ = nCV(Tᵢ-Tf).

    Reason

    The state law and the first law provide separate consistency checks.

    Working

    Answer

    Pf = nRTf/Vf = 157124.754… Pa = 157.124754… kPa. Wₒᵤₜ = (3/2)(8.314)(300-188.988…) = 1384.428… J. Hence Wₒₙ = -1384.428… J and Δ U has that same negative value, while Q = 0. Check that Pf/Pᵢ = (Vᵢ/Vf)^(5/3) with Pᵢ = 498.84 kPa.

Solve an adiabatic path without the reversible power law

Check your understanding 1

One mole of classical monatomic ideal gas starts at 300 K in 5.0 L. It expands adiabatically against a constant external pressure of 100 kPa until final mechanical balance and internal thermal equilibrium. There is no other energy transfer; any transient motion has died away, and piston energy storage and friction are negligible. Do not assume a reversible path. Use R = 8.314 J/(mol K). Find Tf, Vf and work output from the first law and final state equation. Does PᵢVᵢ^(5/3) = PfVf^(5/3) hold?

Unit: K
Unit: L
Unit: J
Show answer

The work output is Pₑₓₜ(Vf-Vᵢ), not an integral of a reversible gas-pressure path. At the endpoint Pf = Pₑₓₜ and Vf = nRTf/Pₑₓₜ. The first law gives nCV(Tf-Tᵢ) = -Pₑₓₜ(Vf-Vᵢ), so n(CV + R)Tf = nCVTᵢ + PₑₓₜVᵢ. This yields Tf ≈ 204.1 K, Vf = 16.9652 L and Wₒᵤₜ = 1196.52 J. The ratio PfVf^(5/3)/(PᵢVᵢ^(5/3)) ≈ 1.54, not one. Check your explanation against the actual load, endpoint state equation and energy balance, rather than assuming that zero heat implies reversibility. Heat was zero, but the finite driving pressure difference made the path irreversible. The first-law and endpoint approach remains valid when the reversible exponent law does not.

Practice set

Check your understanding 2

With only pressure–volume work, an ideal gas is compressed adiabatically. Is Wₒₙ positive or negative?

Show hint

Identify which side supplies energy through work.

Show answer

Positive: compression transfers energy to the gas by work. With Q = 0, its internal energy increases by Wₒₙ.

Check your understanding 3

An adiabatic process supplies total work Wₒₙ = +180 J to a closed system, with no bulk kinetic or potential-energy change. Find Δ U.

Show hint

Set the heat-transfer term to zero in the first law.

Show answer

Δ U = Q + Wₒₙ = 0 + 180 = +180 J. The positive sign denotes an energy gain.

Check your understanding 4

An ideal gas expands reversibly and adiabatically with constant γ > 1. Does its pressure increase or decrease?

Show hint
Rearrange PV^γ = constant to make P the subject.
Show answer
Pressure decreases: P ∝ V^(-γ), so increasing volume reduces pressure.

What to keep

Adiabatic fixes the heat-transfer term. The first law and the appropriate work model determine the energy change. Add reversibility, an ideal gas and constant capacity before using PV^γ = constant; otherwise use the endpoint energy balance or the variable-capacity relation.

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Syllabus and review details

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