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Adiabatic compression decreases the volume of a sample without heat transfer across its boundary. For the fixed-composition ideal gas and endpoint assumptions in adiabatic process, compression work increases internal energy; positive heat capacity at constant volume then means a higher final temperature. The effect comes from supplied work, even though no heat enters. The amount of temperature increase depends on the work actually delivered to the gas, not on the adiabatic process condition alone.

The reversible ideal-gas adiabat supplies a particularly useful quantitative benchmark. Assume a fixed amount of substance $n$, a quasistatic process, negligible friction, only pressure-volume work, and constant heat capacity ratio $\gamma =C_p/C_V>1$. Here $C_p$ and $C_V$ are the whole sample's heat capacity at constant pressure and heat capacity at constant volume. With $R$ the molar gas constant, Mayer's relation gives $C_V=nR/(\gamma -1)$. The proof outline is $C_{V}dT=-pdV$ from the first law of thermodynamics, followed by $p=nRT/V$ from the ideal gas law and integration. The result isThese finite power laws belong to this reversible, constant-$\gamma$ ideal gas model. If heat capacity ratio varies, use the integral relation in heat capacity ratio; if the compression is irreversible, use the actual work balance, as in irreversible adiabatic piston compression. A short stroke by itself justifies neither power law. ^[ref]

Writing $r=V_1/V_2>1$ for the compression ratio, the reversible model predictsAll temperatures in these ratios are absolute. The larger pressure rise than in an isothermal compression, where $p_2/p_1=r$, explains the greater local stiffness described by isentropic compressibility.

For an original stipulated example, set $\gamma =7/5$, $p_1{V}_1=nR{T}_1=1\mathrm{kJ}$ and $r=2$. Then $T_2/T_1=2^{0.4}\simeq 1.31951$ and $p_2/p_1=2^{1.4}\simeq 2.63902$. The input work and internal energy increase are both $0.79877\mathrm{kJ}$, while $Q_{\mathrm{in}}=0$. No initial temperature or apparatus performance is implied by the chosen energy scale. Independent integration gives the signed outward workwhich is negative during compression and agrees with $-\Delta U$. This integral uses equilibrium gas pressure only because the path is reversible.

In the sensor experiment of adiabatic process, compare the pressure-volume integral with $C_V\Delta T$ as an energy check only when internal pressure is sufficiently uniform and piston dynamics permit that boundary-work estimate. Account for friction separately before comparing it with hand input work. PASCO's manual supplies a compression procedure and an illustrative logarithmic-slope result $1.39$, compared with the model value $1.40$ for a diatomic gas; that is its sample result, not new measurements here. ^[ref]

Adiabatic expansion increases the accessible volume without heat transfer. In a loaded piston expansion of the fixed-composition ideal gas considered in adiabatic process, positive outward work comes from internal energy, so the final temperature decreases. This provides a cooling mechanism without removing heat during the stroke. In the Carnot cycle, reversible adiabatic expansion connects two different temperatures between its isothermal processes.^[ref]

For the same constant-$\gamma$ reversible ideal-gas adiabat, put $b=V_2/V_1>1$. The scaling and first law of thermodynamics giveWith the figure's stipulated $\gamma =1.4$, $p_1{V}_1=1\mathrm{kJ}$ and $b=2$, the final ratios are $T_2/T_1\simeq 0.75786$ and $p_2/p_1\simeq 0.37893$. Outward work is $0.60535\mathrm{kJ}$ and $\Delta U=-0.60535\mathrm{kJ}$. Reversing this particular path from its cooled endpoint would require the same work input; the figure's separate compression starts at the original state, so its input is different. Endpoint entropy supplies another independent check:The two logarithm terms cancel using $C_V(\gamma -1)=nR$, confirming the reversible path is an isentropic process. ^[ref]

PASCO's expansion demonstration first compresses trapped air and waits for thermal equilibrium before the rapid expansion. Its manual warns that friction-warmed cylinder regions contact the gas during expansion, compromising quantitative comparison. ^[ref] The hold separates cooling caused by expansion from simply undoing the earlier hot compression. As proposed controls, observe the temperature dip and subsequent warming, compare rapid and slow strokes from matched initial states, and assess wall heat transfer and probe lag before fitting a reversible model.

Cooling is not a universal consequence of adiabatic expansion: free expansion of an ideal gas has no outward work and returns to the same equilibrium temperature. Its entropy increases, so the reversible power laws do not describe it. Reuse free expansion, Joule's free-expansion experiment and temperature sensitivity of a free-expansion experiment for that apparatus, its energy budget and the real-gas qualifications. The useful distinction is the load: expansion cools this ideal gas when it must spend internal energy doing work.

For constant-pressure heat addition $A\to B$ and two reversible ideal-gas adiabats, set $r=(p_h/p_l{)}^{(\gamma -1)/\gamma }$. The Brayton cycle has ${\eta }_B=1-1/r$, maximum temperature $T_B$ and minimum $T_D=T_A/r$. A Carnot cycle using those extreme temperatures has the strictly greater efficiency displayed whenever $T_B>T_A$. Reversibility of the isobaric heating requires continuously matched reservoir temperatures, rather than heat transfer solely at the maximum temperature.

The ideal Brayton cycle consists of adiabatic compression, constant-pressure heat addition, adiabatic expansion, and constant-pressure heat rejection. Its ideal working substance is an ideal gas with constant heat capacities, and the insulated stages are reversible thermodynamic processes.

For pressure ratio $r_p=p_h/p_l$ and heat capacity ratio $\gamma =C_p/C_V$, a reversible ideal-gas adiabat has $T\propto p^{(\gamma -1)/\gamma }$. Thus $T_2/T_1=T_3/T_4=k=r_p^{(\gamma -1)/\gamma }$, and the constant-pressure heat transfers have $Q_{\mathrm{out}}/Q_{\mathrm{in}}=(T_4-T_1)/(T_3-T_2)=1/k$. The thermal efficiency isThis formula requires constant heat capacities and internally reversible insulated stages. An arbitrary dissipative adiabatic process does not satisfy the same pressure-temperature relation.

Compressibility measures how readily a sample's volume decreases when its pressure increases. For a simple compressible system with fixed particle number $N$ and composition, specify the constraint $X$:Here $P$ denotes pressure, also written $p$ elsewhere. The minus sign makes compressibility positive on a mechanically stable equilibrium branch. Its units are inverse pressure, $\mathrm{P}{\mathrm{a}}^{-1}$; its reciprocal $K_X=1/{\kappa }_X$ is the constrained bulk modulus, measured in pascal units. Holding temperature fixed gives isothermal compressibility; holding entropy fixed gives isentropic compressibility ${\kappa }_S$ along an isentropic process. The latter describes a reversible adiabatic compression, since an irreversible adiabatic process can increase entropy. ^[ref]

For an ideal gas, the ideal gas law $PV=N{k}_{B}T$ and the reversible ideal-gas adiabat give, at the same equilibrium state,Here $C_P$ and $C_V$ are the whole sample's heat capacity at constant pressure and heat capacity at constant volume, and $\gamma$ is the heat capacity ratio. Constant $\gamma$ is required for the finite-path formula $P{V}^{\gamma }=\mathrm{constant}$. Reversible adiabatic compression raises the temperature, making the pressure rise more steeply than in an isothermal compression. Thus keeping entropy fixed makes the gas harder to compress locally. For a classical monatomic gas at $100\mathrm{kPa}$, $\gamma =5/3$, so ${\kappa }_T=10^{-5}\mathrm{P}{\mathrm{a}}^{-1}$ and ${\kappa }_S=6\times 10^{-6}\mathrm{P}{\mathrm{a}}^{-1}$; the corresponding values of the bulk modulus are $100\mathrm{kPa}$ and $166.67\mathrm{kPa}$.

For a smooth, stable single thermodynamic phase at positive temperature, Maxwell relations also implywhere $\alpha$ is the coefficient of thermal expansion. The ideal gas result is one example of this broader link between mechanical and thermal responses. The equality ${\kappa }_T={\kappa }_S$ is possible when $\alpha =0$ and the other responses remain finite. These formulas compare equilibrium partial derivatives, rather than arbitrary fast or dissipative compression histories. ^[ref]

The first law applies conservation of energy to a thermodynamic system. For a closed system with unchanged bulk kinetic energy and potential energy, writeHere $U$ is internal energy, $Q>0$ is heat entering the system and $W>0$ is work done by the system on its surroundings. Compression therefore gives negative $W$. Some references define work positive when done on the system; their plus sign represents the same balance with a different convention. If bulk kinetic energy or potential energy changes, include it in the total energy balance. ^[ref]

Supplying $1000\mathrm{J}$ of heat while a gas does $300\mathrm{J}$ of work increases its internal energy by $700\mathrm{J}$. Compressing it with $200\mathrm{J}$ of input work in an adiabatic process instead gives $Q=0$, $W=-200\mathrm{J}$ and $\Delta U=200\mathrm{J}$.

The distinction between $d$ and $\delta$ records state versus path: internal energy is a state function, whereas heat and work depend on the process. For an ideal gas expanding reversibly at constant temperature, $W=N{k}_{B}T\log (V_2/V_1)$ and $Q=W$, since internal energy of an ideal gas depends only on temperature. Choose $N{k}_{B}T=1000\mathrm{J}$ and $V_2=2V_1$: then $Q=W=693.15\mathrm{J}$. An insulated free expansion into vacuum between the same equilibrium endpoints has $Q=W=0$. Both have $\Delta U=0$, yet their transfers differ. Their common entropy increase is $N{k}_B\log 2$. ^[ref]

For a simple compressible equilibrium system of fixed composition, combining the laws givesHere $p$ is pressure and $V$ is volume. This identity relates neighboring equilibrium states. Identifying $TdS$ with actual heat and $pdV$ with actual work requires a suitable reversible thermodynamic process; an irreversible piston calculation uses the external resisting pressure for boundary work. This distinction prevents applying a reversible ideal-gas adiabat to an irreversible adiabatic piston compression.

For one molecule whose translational motion separates from its internal degrees of freedom, the classical one-particle partition function iswhere $q_{\mathrm{int}}$ contains rotational, vibrational, and other internal states. For $N$ indistinguishable noninteracting molecules, the classical ideal-gas partition function is $Z_N=z_1^N/N!$. The thermodynamic pressure therefore obeyssoThe internal structure changes the energy, entropy, and heat capacity through $q_{\mathrm{int}}(T)$, but it does not change the ideal gas law while the internal partition function is independent of volume and the molecules have negligible interactions.

For a dilute monatomic gas $q_{\mathrm{int}}=1$. Using the Stirling formula in $Z_N=(V/{\lambda }^3{)}^N/N!$ and differentiating the Helmholtz free energy $F=-k_{B}T\log Z_N$ gives the Sackur-Tetrode equationWith $N$ fixed, an isentropic process therefore satisfies $V{T}^{3/2}=\text{constant}$, or $T{V}^{2/3}=\text{constant}$. Eliminating $T$ with the ideal-gas law yields the monatomic reversible ideal-gas adiabat

An adiabatic process exchanges no heat with its surroundings, so $\delta Q=0$. For a reversible process it is also isentropic. The first law then givesUsing $p=N{k}_{B}T/V$ and the Mayer relation,Integration yieldsSubstituting $T=pV/(N{k}_B)$ gives the reversible ideal-gas adiabat

An adiabatic process has no heat transfer, $\delta Q=0$. With work done by the gas taken as positive, the first law of thermodynamics givesUsing $dE=C_{V}dT$ and $p=N{k}_{B}T/V$,Integration, followed by the Mayer relation $N{k}_B/C_V=\gamma -1$, yieldsEliminating $T$ with the equation of state gives the reversible ideal-gas adiabat

In the Carnot cycle, let $1\to 2$ be reversible isothermal expansion at $T_h$, $2\to 3$ reversible adiabatic expansion, $3\to 4$ isothermal compression at $T_c$, and $4\to 1$ adiabatic compression. The pressure–volume curve is clockwise. The entropy–temperature figure below uses temperature on the horizontal axis and entropy on the vertical axis, literally an $S$–$T$ plot: isotherms are vertical and adiabats horizontal. The hot isotherm increases entropy by $\Delta S$ and absorbs $Q_h=T_h\Delta S$; the cold one rejects $Q_c=T_c\Delta S$. Thus $W=(T_h-T_c)\Delta S$ and ${\eta }_C=1-T_c/T_h$.

For the second engine, choose the working direction with heat addition during $A\to B$ at $p_1$ and rejection during $C\to D$ at $p_2$. A reversible ideal-gas adiabat obeys $T\propto p^{(\gamma -1)/\gamma }$. DefineThe two constant-pressure heat transfers are $Q_{\mathrm{in}}=c_p(T_B-T_A)$ and $Q_{\mathrm{out}}=c_p(T_C-T_D)=Q_{\mathrm{in}}/r$, where rejected heat is quoted positive. Therefore the Brayton cycle efficiency isfor $\gamma =5/3$. All heat-capacity amount conventions cancel in the ratio.

Since $T_B>T_A$ and $T_C=T_B/r$, $T_D=T_A/r$, the maximum is at B and the minimum at D. The Brayton-Carnot efficiency comparison isThe inequality is strict for a nondegenerate cycle with positive heat input. Although each process can be reversible, the isobaric heat transfer spans a range of temperatures; making the whole cycle reversible requires a succession of matched-temperature reservoirs, rather than just the hottest and coldest reservoirs.

For an ideal diatomic gas with active rotation and frozen vibration, $\gamma =7/5$ andHere $0<p_2/p_1<1$. This comparison holds at the same pressure ratio and assumes temperature-independent heat capacities in the stated mode regime. Frozen rotation or excited vibration would require their actual heat capacities.

Use an ideal gas to illustrate the Carnot cycle, with hot and cold heat baths at $T_h=T_2$ and $T_c=T_1$. The four processes are all reversible thermodynamic processes. Lettered states below belong only to this illustration; the engine later in the question uses numbered states.

Starting at $A$, perform the following four stages:

The clockwise enclosed area in the pressure-volume diagram is the positive net work per cycle, $W=\oint pdV=Q_h-Q_c$.

Because the working substance returns to its original thermodynamic state, its net entropy change is zero. There is no entropy change on either reversible ideal-gas adiabat; on each isothermal process the entropy transfer is the exchanged heat divided by the bath's absolute temperature. HenceThe first law of thermodynamics over a complete cycle gives $W=Q_h-Q_c$. The thermal efficiency is thereforeThe use of kelvins is essential. This result does not depend on the working substance. For the ideal gas illustration, the two reversible ideal-gas adiabats imply $V_C/V_B=V_D/V_A=(T_h/T_c{)}^{1/(\gamma -1)}$, so $\ln (V_B/V_A)=\ln (V_C/V_D)$; the explicit heat expressions in (a) give the same ratio.

For the numbered engine, denote the low and high pressures by $p_1=p_4=p_l$ and $p_2=p_3=p_h$. Let $R_g$ denote the molar gas constant. The ideal gas law makes the constant-pressure branches straight rays in a temperature-volume diagram:The high-pressure ray is steeper. For a constant heat capacity ratio $\gamma$, each reversible ideal-gas adiabat obeysConsequently $1\to 2$ curves upward and left during compression; $2\to 3$ goes upward and right along the steeper ray; $3\to 4$ curves downward and right during expansion; and $4\to 1$ goes downward and left along the shallower ray.

The plot is an original calculation for $\gamma =7/5$, $p_h/p_l=10$ and $V_3/V_2=2$; its state directions follow the original paper. The dashed extensions identify the two constant-pressure rays, not additional stages of the cycle.

This is the Brayton cycle. Assume fixed gas amount, constant heat capacities, and internally reversible adiabatic processes, as intended by the ideal engine model. Put $r_p=p_h/p_l=p_2/p_1>1$ andCombining $p{V}^{\gamma }=\text{constant}$ with the ideal gas law gives $T\propto p^{(\gamma -1)/\gamma }$ along either reversible ideal-gas adiabat. ThereforeDuring a constant-pressure process the heat supplied equals the enthalpy increase. If $C_p$ is the molar heat capacity at constant pressure, the positive heat input and output areThe insulated branches exchange no heat, so the first law of thermodynamics gives $W=Q_{\mathrm{in}}-Q_{\mathrm{out}}$. HenceAn adiabatic process alone need not be an isentropic process; the relations used here also require internal reversibility. Dissipative compression or expansion would require extra information and would not, in general, obey this efficiency formula.