Isothermal, adiabatic, isobaric, and isochoric processes

One gas, four different controls

A sealed cylinder contains n = 0.250 mol of an assigned ideal gas at T_i = 320 K in V_i = 4.00 L. Its piston initially rests at p_i ≈ 166.3 kPa absolute. Suppose we let the gas occupy more volume. Will it cool, take in heat, or stay at the same pressure? The answer depends on what the apparatus holds fixed and how it moves.

Choose the gas as the system. It exchanges no matter with its surroundings. Heat Q is energy transferred into the gas because of a temperature difference; work W_by is energy transferred outward by mechanical action. For the closed-system changes here, endpoint motion and height changes are negligible, and the energy balance is ΔU = Q − W_by. Thus expansion work is positive in our convention. Internal energy U belongs to a state; Q and W_by belong to a process.

The opening gas obeys the ideal gas equation of state (EOS), pV = nRT, with absolute pressure p, kelvin temperature T and R = 8.314462618… J/(mol K). To calculate energy changes, separately assign the constant molar heat capacity c_v = 2.50R. Then ΔU = n c_v(T_f − T_i). This caloric assumption is extra information, not a consequence of the pressure equation alone. The gas is hypothetical; the numbers are not a measured material dataset.

First follow the four controls, the work sign, the four ideal-gas examples and the two paths with common endpoints. Return later to the finite irreversible changes and sections marked “Further”. Published preparation is “Physical systems, states, boundaries, and environments”, “Conservation of mass, momentum, and energy”, and “Pressure, buoyancy, and fluid statics”. Essential gas and first-law terms are introduced here, so the planned neighboring thermodynamics lessons are not required live prerequisites.

Name the constraint before choosing a formula

Isothermal means the system temperature stays constant during a process for which a system temperature is meaningful. Adiabatic means there is no heat transfer across the selected boundary during the process. Isobaric means system pressure stays constant; isochoric means system volume stays constant. These names constrain different quantities and do not by themselves specify an ideal gas, a speed, a work value or an energy change.

A gas in good thermal contact with a large reservoir can approach an isothermal process if heat transfer keeps pace with the controlled motion. Insulation can approach an adiabatic process; a sufficiently short event can also have negligible heat transfer compared with the relevant energy changes. “Fast” alone does not prove this approximation, and “slow” alone does not guarantee constant temperature.

A slowly moving piston against a balanced constant load can approximate an isobaric process. A rigid vessel gives an isochoric process. A constant external load during a sudden release does not make the gas pressure uniform or constant throughout the motion. Likewise, a rigid wall prevents volume work but does not prevent heat transfer or work through a stirrer or electrical connection.

The constraints can overlap in special cases. An insulated rigid vessel with no other work can have Q = W_by = ΔU = 0. Isothermal and adiabatic are nevertheless different statements. A gas can absorb heat while expanding at constant temperature, and an insulated gas can change temperature while doing work.

A state path and the work at its boundary

A quasistatic change passes through states close enough to internal equilibrium that one gas pressure and temperature describe each stage. A reversible limit requires more: no dissipative friction, unrestrained expansion or finite thermal driving difference that leaves an uncompensated change when system and surroundings are restored. A slow process can still be irreversible. For example, slow motion with sliding friction dissipates energy.

For reversible pressure–volume work, W_by = ∫ from V_i to V_f p dV. The differential dV is positive in expansion and negative in compression. On a p–V diagram, the signed area under the appropriate curve represents this integral. An expansion followed backward along the same reversible curve has the opposite work. A vertical, constant-volume leg has zero pressure–volume work.

The general mechanical accounting uses the force actually transmitted at the moving boundary. During rapid piston motion, gas pressure can vary across the cylinder and the piston can store kinetic energy. One cannot replace that boundary-force history by an arbitrary bulk equilibrium p(V) curve. For a horizontal frictionless piston under a constant remote resistance p_ext, rest endpoints give ∫(p_b − p_ext)dV = ΔK_piston = 0. Here p_b is the area-averaged normal traction exerted by the gas on the piston. Thus the total gas work ∫p_b dV equals p_ext(V_f − V_i), provided no other apparatus energy or transfer is omitted. The instantaneous pressures need not be equal; the finite-release example uses internal gas dissipation to settle the motion.

We will draw continuous gas-state curves only for the explicitly quasistatic paths. For a finite release or free expansion, equilibrium endpoints can be marked without inventing intermediate equilibrium states. A line representing an external resistance belongs to the work accounting, not automatically to the gas equation of state.

Isothermal: heat can replace energy carried away as work

Let the opening ideal gas expand reversibly from 4.00 L to 8.00 L while a reservoir maintains T = 320 K. At fixed n and T, p = nRT/V, so pressure decreases inversely with volume. The final pressure is about 83.145 kPa. A reservoir alone is not a piston controller: the external resisting pressure must also be lowered in small enough steps to approach mechanical reversibility.

Integrating the ideal pressure gives W_by = nRT ln(V_f/V_i). The logarithm takes the dimensionless volume ratio. Here W_by ≈ 461.052 J. The ideal caloric model has ΔU = 0 when T is unchanged. The first law therefore requires Q = W_by ≈ 461.052 J into the gas. Energy enters by heat and leaves by work while the gas internal energy remains unchanged.

This is why “constant temperature means no heat” fails. In reversible isothermal compression along the same curve, work is done on the gas, W_by is negative, and the gas must reject heat to maintain its temperature. Without adequate heat exchange the same mechanical operation generally changes temperature.

The relation ΔU = 0 here uses the consistent ideal-gas energy law U = U(T). For a nonideal fluid, internal energy can change with volume even at fixed temperature. A later return section gives an explicit counterexample. The word isothermal by itself supplies neither that energy law nor reversibility.

Adiabatic: expansion work can lower the temperature

Now insulate the cylinder and reduce the opposing load reversibly as the same opening gas expands to 8.00 L. No heat enters or leaves. The gas does work using its internal energy, so ΔU = −W_by. For the assigned positive c_v, a decrease of U means a decrease of T.

Along this reversible ideal-gas path, n c_v dT = −p dV = −nRT dV/V. Dividing by nT and integrating gives c_v ln(T_f/T_i) = −R ln(V_f/V_i) when c_v is constant. Define c_p = c_v + R and γ = c_p/c_v = 1.40 for this model. Then TV^(γ−1) is constant and pV^γ is constant.

Doubling volume gives T_f = 320 × 2^(−0.40) ≈ 242.515 K and p_f ≈ 63.0118 kPa. The gas delivers W_by = n c_v(T_i − T_f) ≈ 402.656 J and Q = 0. Compared with the isothermal expansion to the same volume, the reversible adiabatic pressure falls faster because the gas both dilutes and cools.

These power laws require fixed amount and composition, an ideal EOS, constant heat capacities, a reversible quasistatic path and pressure–volume work only. Insulation alone supplies only the no-heat condition. Finite irreversible expansion, variable heat capacity and a nonideal EOS need their own equations. During reversible compression the signs reverse and the ideal gas warms.

Isobaric: a balanced load and a heat supply

Keep the gas pressure at p_i while slowly supplying heat and allowing the piston to rise. A constant balanced load sets the pressure in the idealized quasistatic limit. Since pV = nRT with p and n fixed, V/T is constant. Expanding from 4.00 L to 8.00 L therefore raises the assigned gas temperature from 320 K to 640 K.

Pressure–volume work is the rectangle W_by = p_i(V_f − V_i) = nR(T_f − T_i), about 665.157 J. The internal energy increases by n c_v(T_f − T_i) ≈ 1662.893 J. The heat input is their sum, Q ≈ 2328.050 J. Some input raises the gas energy and some transfers energy to the load.

For this ideal gas, Q = n(c_v + R)ΔT = n c_pΔT at constant pressure. Molar c_p describes this constant-pressure heating response; it is not a replacement for c_v in the ideal state relation ΔU = n c_vΔT. The full heat-capacity lesson develops measurement, temperature dependence and other materials.

Constant pressure does not mean constant temperature or zero work. In a general simple fluid under the same constant-pressure, pressure–volume-only conditions, Q = Δ(U + pV), the enthalpy change. The simple linear relation V ∝ T still needs an ideal EOS. Heating through a phase change or changing composition requires additional state information.

Isochoric: heat can change pressure without volume work

Lock the piston or use a rigid sealed vessel, then heat the opening gas from 320 K to 640 K. Its volume remains 4.00 L. For this ideal gas, p/T is constant, so absolute pressure rises from about 166.289 kPa to 332.579 kPa.

Because dV = 0, pressure–volume work is zero even while the pressure rises. Under our no-other-work conditions, Q = ΔU = n c_vΔT ≈ 1662.893 J. Less heat is required than for the isobaric change through the same temperature interval because the isobaric gas also transfers work outward.

The two experiments end at different volumes and pressures. Their equal ideal ΔU follows the equal temperature change, not an assertion that their final states coincide. In a nonideal fluid, U can depend on volume too; the ideal comparison needs its stated model.

A rigid vessel with an electrical heater or moving stirrer can receive work through a non-volume channel. Constant volume would still make pressure–volume work zero, but Q would then not equal ΔU unless that extra work were included separately. Also distinguish a uniform controlled heating path from rapid heating with temperature gradients; equilibrium endpoints alone do not establish every intermediate state.

Compare the four paths without equating their endpoints

All four illustrative paths start at the same equilibrium point. Three expand to 8.00 L; the isochoric path stays at 4.00 L and is heated to 640 K. Their endpoints differ. The purpose of the common-start diagram is to show which constraint shapes each curve, not to claim that four different heats reach one common final state.

For ideal reversible expansion through the same initial point, an isothermal has slope dp/dV = −p/V while a constant-γ adiabat has slope −γp/V. Since γ > 1, the adiabat is steeper at that common point. A constant-pressure path is horizontal and a constant-volume path is vertical. The latter is not a function p(V) over an interval, but its volume work is still unambiguously zero.

Work comparisons must identify what is held the same. In the displayed expansions to the same volume, isobaric work exceeds isothermal work, which exceeds reversible adiabatic work. Those paths require different heat exchanges and end at different temperatures. This ordering is a comparison of these specified curves, not a theorem that any arbitrary process with one of these names has that work.

Two routes to exactly the same final state

Now fix both endpoints: start at 4.00 L and 320 K, and finish at 8.00 L and 640 K. For the fixed amount of ideal gas, both endpoint pressures are p_i. Path A first expands reversibly at 320 K to 8.00 L, then heats at fixed volume to 640 K. Path B first heats at fixed volume to 640 K, then expands reversibly at 640 K to 8.00 L. Each vertical leg can be carried out with controlled thermal contact; reversibility of heat transfer requires a sequence of closely matched reservoir temperatures.

The volume work on A is nR(320 K)ln 2 ≈ 461.052 J. On B it is nR(640 K)ln 2 ≈ 922.103 J. There is no volume work on either vertical leg. Path B expands while hotter, so its pressure is higher at every intermediate volume and its area is greater.

Both paths have ΔU = n c_v(640 − 320) ≈ 1662.893 J because U is a state function and the endpoints coincide. Therefore Q_A ≈ 2123.944 J and Q_B ≈ 2584.996 J. The extra heat on B equals its extra outward work; neither heat nor work is a state property.

A p–V plot locates equilibrium states, while a process specification supplies the path and exchanges. Knowing initial and final p,V,T does not generally identify the intervening area. The two-stage examples make that missing information visible without replacing the full first-law lesson.

A finite adiabatic release obeys a different equation

Consider an insulated cylinder with a horizontal frictionless piston, initially held at the opening state. Release the piston against a constant remote resistance p_ext = p_i/2 ≈ 83.1446 kPa. Internal viscous and acoustic dissipation in the gas allows the motion to settle. The piston starts and ends at rest, and the final equilibrium gas pressure equals p_ext. The gas need not have one uniform pressure or temperature during motion. Assume no heat or other work, no external damper, and no net change in other apparatus energy. Energy delivered against the specified resistance is the outward work, not an omitted energy change.

For the gas, W_by = ∫p_b dV, where p_b is the area-averaged normal boundary traction. The horizontal frictionless piston gives ∫(p_b − p_ext)dV = ΔK_piston = 0 between rest endpoints, so the complete gas-work integral equals p_ext(V_f − V_i). This does not require instantaneous equality of p_b and p_ext. The endpoint energy balance is therefore n c_v(T_f − T_i) = −p_ext(V_f − V_i). The final ideal EOS is p_extV_f = nRT_f. These two equations, rather than pV^γ = constant, determine the endpoint. With c = c_v/R and r = p_ext/p_i, elimination gives T_f/T_i = (c + r)/(c + 1) and V_f/V_i = (c + r)/[(c + 1)r].

For c = 2.50 and r = 0.500, T_f ≈ 274.286 K and V_f ≈ 6.85714 L. The outward work is about 237.556 J, Q = 0 and ΔU ≈ −237.556 J. A reversible adiabatic expansion to the same final pressure ends instead at about 262.507 K and 6.56268 L, delivering about 298.763 J. The final pressure is shared; the other endpoint properties are not.

The external-resistance rectangle represents the complete work for this apparatus, not a horizontal gas-state trajectory. Temporary piston motion and gas flow are allowed; their endpoint energy changes vanish. Piston friction or an external damper can take energy in addition to the specified remote-load work, even if none of that dissipated energy returns to the gas as heat. A spring retaining energy or an apparatus still moving at the final endpoint also adds terms. Those changes require a new balance; calling the gas adiabatic does not remove them.

Free expansion: equal endpoint temperatures without a reversible isotherm

Place the opening gas in one half of a rigid insulated vessel, with the other half evacuated. Remove the partition and wait until the gas reaches equilibrium throughout the doubled accessible volume. The outer boundary does not move, and the vacuum offers no resisting load. With no other transfers, W_by = 0 and Q = 0, so ΔU = 0.

For the assigned ideal gas with positive c_v, the final temperature is again 320 K. Its accessible volume is now 8.00 L and its final pressure is about 83.1446 kPa. These are the same gas endpoints as the reversible isothermal expansion, yet that earlier process transferred about 461.052 J both as heat into the gas and as work outward. Equal endpoints do not make the exchanges equal.

During the free expansion, the gas rushes into the empty space and is not described by a succession of uniform equilibrium states. Equal initial and final temperatures do not prove a spatially uniform constant-temperature trajectory during the transient. In particular, integrating nRT/V along an imagined reversible isotherm would calculate the work of that imagined comparison, not the actual zero work.

The ideal temperature result is conditional. An interacting fluid can have volume-dependent internal energy, so a no-heat, no-work expansion need not preserve temperature. Nor is this a throttling calculation: steady flow through a valve has matter crossing the chosen system boundary and requires a different balance.

Further: reversible adiabats with variable heat capacity

The reversible ideal-gas differential c_v(T)dT/T = −R dV/V remains useful when c_v varies with temperature. Integrate the actual function: ∫ from T_i to T_f c_v(T)dT/T + R ln(V_f/V_i) = 0. A single constant γ power law cannot generally replace this integral.

As a separate assigned caloric model, let c_v(T) = R[2.00 + 0.500T/(320 K)], positive throughout the interval considered. It equals 2.50R initially but decreases as the gas cools. Reversible adiabatic expansion from the opening T_i,V_i to twice the volume requires 2 ln[T_f/(320 K)] + 0.500(T_f − 320 K)/(320 K) + ln 2 = 0.

Solving gives T_f ≈ 240.730 K, rather than the constant-c_v model’s 242.515 K. Integrating n c_v(T)dT gives ΔU ≈ −401.724 J and W_by ≈ 401.724 J, with Q = 0. The altered final temperature comes from an altered caloric law; it is not evidence about a measured gas or a correction to the exact constant-c_v model.

Real molecular modes, temperature ranges and composition can require temperature-dependent property data. The heat-capacity subject develops those choices. This example isolates why an apparently familiar pressure EOS does not justify silently holding γ fixed.

Further: the same process name with a nonideal EOS

The four process names do not depend on ideal-gas behavior. For a simple fluid, reversible work along an isotherm is still ∫p(T,V)dV, but the EOS changes the curve. In a thermodynamically consistent one-component fluid, the molar identity (∂U_m/∂v_m)_T = T(∂p/∂T)_v − p shows why isothermal energy change need not vanish. Here v_m = V/n and U_m = U/n.

For the assigned van der Waals model p = RT/(v_m − b) − a/v_m², with v_m > b, that identity gives U_m = u₀(T) − a/v_m. A reversible isothermal expansion therefore has ΔU = na(1/v_i − 1/v_f) and W_by = n[RT ln((v_f − b)/(v_i − b)) + a(1/v_f − 1/v_i)]. Their sum is Q = nRT ln((v_f − b)/(v_i − b)). The unknown T-only function cancels from this fixed-temperature difference.

For a separate hypothetical fluid with n = 0.100 mol, T = 350 K, a = 0.200 Pa m⁶/mol², b = 3.00 × 10⁻⁵ m³/mol and molar volume increasing from 0.000600 to 0.001200 m³/mol, the model gives ΔU ≈ 16.6667 J, W_by ≈ 192.602 J and Q ≈ 209.269 J. The supplied path is above this model’s critical temperature and has negative isothermal slope throughout; it stays on a single homogeneous branch. These are model calculations, not real-fluid validation.

For a general consistent simple fluid, a reversible adiabat with volume work only satisfies c_v(T,v_m)dT + T(∂p/∂T)_v dv_m = 0. This follows by inserting dU_m = c_vdT + [T(∂p/∂T)_v − p]dv_m into dU_m = −p dv_m. It requires both a thermal EOS and suitable caloric information. Even the direction of temperature change depends on these coefficients; Q = 0 alone is not a universal cooling rule.

Further: polytropic curves organize more idealized paths

An explicitly quasistatic ideal-gas trajectory is sometimes represented as pV^m = constant, where m is a dimensionless exponent and n remains the gas amount. Combining with the ideal EOS gives T_f/T_i = (V_f/V_i)^(1−m). For m ≠ 1, integrating p dV yields W_by = nR(T_f − T_i)/(1 − m). With the assigned constant c_v, Q = n c_v(T_f − T_i) + W_by.

For m = 0 the curve is isobaric. For m = γ, this reversible constant-c_v ideal-gas trajectory is adiabatic. The case m = 1 must use the separate logarithmic isothermal formula; inserting m = 1 into the displayed fraction produces an indeterminate expression, not zero work. Constant volume is best stated directly as dV = 0.

At m = 1.20 and twice the opening volume, the assigned model gives T_f ≈ 278.576 K, W_by ≈ 430.521 J, ΔU ≈ −215.261 J and Q ≈ 215.261 J. The gas cools while receiving heat because it transfers more energy outward as work than it receives as heat. Temperature change alone does not identify the sign of Q.

A fitted pV^m relation is a description of a curve over some range. It does not by itself establish reversibility, a microscopic heat-transfer mechanism or a constant heat-capacity model. Compressor and engine approximations need their own boundary choices, property ranges and losses; a closed-system gas path is not automatically a steady-flow machine model.

Further: a bounded entropy diagnostic

Entropy S is a state property developed with the second law. For the assigned constant-c_v ideal gas, a relation between equilibrium endpoints is ΔS = n c_v ln(T_f/T_i) + nR ln(V_f/V_i). It can be evaluated using an imagined reversible comparison even when the actual change is irreversible. It is not a rule to divide the actual transient heat by an assumed gas temperature.

The reversible adiabatic power law makes these two logarithmic terms cancel, giving ΔS = 0. Insulated free expansion instead gives ΔS = nR ln 2 ≈ 1.44079 J/K, even though Q = 0. The finite constant-resistance release with the stated frictionless piston and internal gas dissipation gives ΔS ≈ 0.319317 J/K. These examples separate no heat transfer from constant entropy.

For a closed adiabatic system the second law permits entropy production, so adiabatic irreversible changes can increase S. A reversible adiabat is isentropic, meaning constant S. Constant system entropy alone does not generally prove that the whole process is both adiabatic and reversible: heat exchange and internal production require their own accounting.

This is a diagnostic return path, not a derivation of entropy or the second law. General entropy balances, heat-transfer boundary temperatures, engines and limits on conversion belong to the later entropy and engines subjects. The gas and apparatus conditions remain essential even when an endpoint formula is compact.

Choose the experiment, then check the calculation

Start by naming the system and its exchanges: fixed gas amount or flowing material, moving or rigid boundary, thermal contact or insulation, imposed load and any additional work. State whether the gas remains close to internal equilibrium. A temperature, pressure or volume held by a controller is a physical condition to justify, not merely a letter to set constant in a formula.

Next choose the EOS and caloric model over the intended range. Use the constraint to relate states, evaluate the appropriate boundary work, and obtain the remaining energy transfer from ΔU = Q − W_by. Check signs, absolute temperature, absolute pressure, units, endpoints and limiting cases. For a finite irreversible event, solve endpoint balances with actual external conditions rather than drawing an unsupported gas-state path.

The four constraints describe useful limits for cylinders, sealed vessels, thermal reservoirs and rapid insulated events. They also reveal what a measurement has not told us. Equal endpoint temperatures do not prove an isothermal trajectory; no heat does not prove pV^γ; constant external resistance does not prove constant gas pressure; constant volume rules out volume work rather than every kind of work.

Return to the first-law subject for general energy bookkeeping, heat capacities for caloric properties, entropy for irreversibility, and flow or engines for their complete systems. The useful habit here is to connect each process equation to the constraint, realization, model and path that make it valid.

Worked examples

Try it yourself

1. Does an isothermal expansion necessarily have Q = 0?

   Show the answer and reasoning

   No.

   For reversible expansion of this ideal gas at constant T, ΔU = 0 but W_by > 0, so positive heat Q = W_by is required.

2. Does insulation alone justify pV^γ = constant?

   Show the answer and reasoning

   No.

   The displayed power law also needs a reversible quasistatic ideal-gas path, constant heat capacities, fixed composition and pressure–volume work only.

3. A rigid vessel is electrically heated with no heat crossing its outer boundary. Must ΔU be zero?

   Show the answer and reasoning

   No.

   Isochoric means zero pressure–volume work. Electrical work can enter the system; the balance must include that channel instead of treating it as heat.

4. A piston is suddenly released against constant p_ext. Is the gas necessarily isobaric throughout?

   Show the answer and reasoning

   No.

   The gas can have nonuniform, changing pressure and the piston can accelerate. A constant external resistance is not a uniform constant gas pressure.

5. Reverse the reversible isothermal opening expansion. What are the work and heat signs?

   Show the answer and reasoning

   W_by < 0 and Q < 0.

   Compression transfers work into the gas; heat must leave to keep its ideal internal energy unchanged. The magnitudes are about 461.052 J for the exact reverse endpoints.

6. Which coefficient belongs in ΔU=n c_vΔT for the assigned ideal gas during an isobaric change?

   Show the answer and reasoning

   c_v, despite the isobaric path.

   The ideal energy state relation uses c_v for any equilibrium endpoint pair. At constant pressure, Q includes expansion work and equals n c_pΔT under the stated conditions.

7. For paths A and B, why does different p–V area not imply different ΔU?

   Show the answer and reasoning

   The endpoints are the same and U is a state function.

   The area determines their reversible volume work. Different heats balance that difference while ΔU stays about 1662.893 J.

8. Do the four paths in the common-start table reach one common final state?

   Show the answer and reasoning

   No.

   Their constraints produce different final temperatures, pressures or volumes. Work comparisons there refer only to the specified curves and endpoint choices.

9. Should one integrate nRT/V from 4.00 L to 8.00 L to find the actual insulated free-expansion work?

   Show the answer and reasoning

   No; actual work is zero.

   The equilibrium isotherm would describe a different reversible comparison. The rigid outer boundary and vacuum provide no actual resisting work.

10. Does equal T before and after free expansion establish a uniform isothermal transient?

    Show the answer and reasoning

    No.

    The gas is out of equilibrium during the rush into vacuum. Endpoints do not establish a meaningful uniform gas temperature at every intermediate moment.

11. At p_f = p_i/2, do the stated finite constant-resistance release with a horizontal frictionless piston and internal gas dissipation, and the reversible adiabatic expansion, have the same final volume?

    Show the answer and reasoning

    No: approximately 6.85714 L and 6.56268 L respectively.

    Both have no heat transfer, but their mechanical paths and work differ. The finite-release work equality uses the frictionless piston and rest endpoints; internal gas dissipation permits settling. They must be solved with their own balances and conditions.

12. Can initial c_v=2.50R alone justify holding γ=1.40 throughout the separate variable-c_v example?

    Show the answer and reasoning

    No.

    The caloric law changes with temperature. Its reversible adiabat follows the integral of c_v(T)/T, not a constant-exponent substitute.

13. For the separate nonideal isotherm, why is Q larger than W_by by about 16.6667 J?

    Show the answer and reasoning

    The internal energy increases at fixed temperature.

    This model has U_m=u₀(T)−a/v_m. Increasing molar volume raises U_m, so Q=ΔU+W_by exceeds work by that positive energy change.

14. How can insulated ideal free expansion have positive ΔS when Q = 0?

    Show the answer and reasoning

    Irreversibility can produce entropy.

    The endpoint state-function formula gives nR ln2≈1.44079 J/K. One must not replace that calculation by actual Q/T for an irreversible change.

15. If a process absorbs heat in one stage and rejects the same amount in another, is it adiabatic?

    Show the answer and reasoning

    Not from that information.

    Zero net heat over the combined event differs from no heat transfer during every stage. Adiabatic requires the latter boundary condition, or a justified approximation to it.

What to carry forward

• A process name fixes a constraint; the system, apparatus, EOS, caloric law and path determine what follows.
• For the declared convention, ΔU=Q−W_by. Reversible volume work is a signed p–V area; a finite irreversible event needs its actual boundary accounting.
• Isothermal does not mean no heat, and adiabatic does not mean constant temperature, reversible or universally pV^γ=constant.
• Constant pressure and constant volume heating differ because a moving boundary can transfer work; ideal ΔU still uses c_v.
• Heat and work can differ between identical endpoints, while internal energy and entropy changes are state differences.
• Variable caloric functions, nonideal interactions and irreversible relaxation require their own equations and do not inherit an ideal reversible power law.