Cool an expanding gas without removing heat

Insulation prevents heat exchange, not energy transfer

Predict what will happen when the same starting gas expands against a load with its walls thermally insulated. With Q = 0, the first law becomes ΔU = −W. If the gas does positive work, its internal energy falls. For this ideal gas, the temperature falls too. Energy leaves through work even though no heat leaves.

Select initial 300 K and final 40 L. The solid isothermal path ends at 300 K. The dashed reversible adiabatic path ends at approximately 188.988 K. The paths share a starting state and final volume, not a final state. Their different internal-energy changes are therefore consistent with internal energy being a state property.

Specify the gas and the process

The lab models one mole of a classical monatomic ideal gas with molar heat capacity Cv = 3R/2, measured in J/(mol K). It assumes constant heat capacities and only pressure-volume work. For this model Cp = 5R/2, so their ratio gamma = Cp/Cv = 5/3. These constants do not describe every gas or every temperature range.

For the reversible adiabatic path, Tf = Ti(Vi/Vf)^(gamma − 1). Subscripts i and f mean initial and final states. Volumes use the same units, temperatures use kelvin, and the exponent here is 2/3. For doubling volume, Tf = 300 × (20/40)^(2/3) ≈ 188.988 K. The final pressure from nRTf/Vf is approximately 39.281 kPa.

Then ΔU = nCv(Tf − Ti) ≈ −1384.429 J. With Q = 0, W ≈ 1384.429 J. Compare the ledger with the isothermal work, about 1728.848 J: the isothermal pressure stays higher during expansion because the bath replenishes energy. The extra work is not evidence of a more efficient cyclic engine; neither displayed path is a complete cycle.

Reverse the volume change

Select final 10 L. The adiabatic temperature rises to about 476.220 K, and the gas requires approximately 2197.644 J of work input. Work by the gas is negative; ΔU is positive. The isothermal compression instead stays at 300 K and rejects approximately 1728.848 J as heat.

Capture the comparison. Explain why the dashed pressure curve is above the solid one during compression but below it during expansion. Set the final volume to 20 L and check that both processes collapse to the unchanged initial state.

Do not apply the curve to every insulated process

Adiabatic only specifies zero heat transfer. The plotted relation additionally requires a reversible ideal path with quasistatic mechanical balance and constant heat capacity. An equivalent relation is Pf Vf^gamma = Pi Vi^gamma. Insulation alone does not justify that equation.

Consider gas released into an evacuated compartment of a rigid, insulated container. There is no opposing external pressure for the expansion work and no heat crossing the outer boundary. Between equilibrium states, Q = W = ΔU = 0. For an ideal gas the final temperature is unchanged. The rapid intermediate states are not the equilibrium points on our dashed curve. This is a counterexample to “every adiabatic expansion cools a gas.” The OpenStax adiabatic-process reference distinguishes free expansion from quasistatic adiabatic work.

Transfer using the energy account

An insulated gas does 600 J of work against a load. Its amount is 2 mol and its constant Cv is 12 J/(mol K). Find its temperature change. Then replace that process with free expansion into vacuum under the stated ideal-gas assumptions.

Check the adiabatic transfer

ΔU = −600 J. The total heat capacity is nCv = 24 J/K, so ΔT = −600/24 = −25 K. Free expansion instead has W = 0 and hence ΔT = 0. The first calculation uses the energy balance and specified heat capacity; it does not need the reversible pressure-volume relation.