Adiabatic Reversible pVT Relations

Push a bicycle pump quickly and the barrel warms your hand. No heat entered the gas. Compression alone raised its temperature. This is the fingerprint of an adiabatic process. Turning that qualitative fact into precise arithmetic is the job of this article. We already know that adiabatic means $q = 0$ . So all internal energy change comes from work. What we have not yet pinned down is different. It is how pressure, volume, and temperature move together. This happens during a reversible adiabatic change of an ideal gas. That relationship, $pV^\gamma = \text{constant}$ , is one of the most quoted formulas in thermodynamics. It is used everywhere from engine cycles to the speed of sound. This article derives $pV^\gamma = \text{constant}$ and its twin $TV^{\gamma-1} = \text{constant}$ from the First Law. It works two fully numeric examples. It also flags the traps that catch students who confuse adiabats with isotherms. It sets up the later discussion of the Joule expansion. There, a gas expands adiabatically but irreversibly, and the temperature story changes completely.

Key takeaways

For a reversible adiabatic ideal-gas process, $pV^\gamma = \text{constant}$ , with $\gamma = C_{p,m}/C_{V,m}$ .
The companion relation $TV^{\gamma-1} = \text{constant}$ follows directly by substituting the ideal gas law.
Both relations come from combining $dU = C_V\,dT$ with $dw = -p\,dV$ under $q = 0$ .
Adiabats are steeper than isotherms on a $p$ – $V$ diagram because $\gamma > 1$ .
Reversible adiabatic compression always heats an ideal gas; reversible adiabatic expansion always cools it.
Key idea: setting $q=0$ in the First Law turns $dU = -p\,dV$ into a separable differential equation whose integral is $pV^\gamma = \text{constant}$ .

The core insight

Because a reversible adiabatic process has no heat exchange, every joule of work done on or by the ideal gas must show up as internal energy change; combining $dU = C_{V,m}\,dT$ with $dw = -p\,dV$ and eliminating $p$ via $pV = nRT$ produces a single separable equation that integrates exactly to $pV^\gamma = \text{constant}$ .

The physical picture

Consider an ideal gas sealed in a thermally insulated cylinder with a frictionless, reversible piston. No heat can cross the walls, so $q = 0$ throughout. If the piston moves inward, it does work on the gas; if the piston moves outward, the gas does work on the piston. In either direction, since $q = 0$ , the First Law reduces to

dU = dw

For an ideal gas, $U$ depends only on $T$ , so $dU = C_{V,m}\,dT$ (per mole). The reversible work is $dw = -p\,dV$ . These two facts, held together only by the ideal gas law $pV = nRT$ , are the entire content of the derivation below. The key physical difference from an isothermal expansion is that here temperature is not held fixed by a heat bath. As the gas expands adiabatically, it does work at the expense of its own internal energy, so $T$ falls as $V$ rises. Compression does the reverse: work done on the gas raises $T$ as $V$ shrinks. This coupling between $T$ and $V$ , absent in an isothermal process, is what produces the extra factor of $\gamma$ in the pressure–volume relation compared to Boyle's law. Plotted on a $p$ – $V$ diagram, an adiabat and an isotherm can cross at one point, but the adiabat is always steeper there. Physically: along the adiabat, compressing the gas raises $T$ , and that extra heating pushes $p$ up faster than the isotherm's temperature-fixed $p \propto 1/V$ curve would predict.

Figure

Adiabat versus isotherm through the same starting point

Figure

Temperature drop during reversible adiabatic expansion

The core relations

pV^\gamma = \text{constant}

where $\gamma = C_{p,m}/C_{V,m}$ is the heat capacity ratio for the ideal gas undergoing the reversible adiabatic change.

TV^{\gamma-1} = \text{constant}

where $T$ and $V$ are related through the same reversible adiabatic path, with $\gamma - 1 = R/C_{V,m}$ .

Tp^{(1-\gamma)/\gamma} = \text{constant}

where this equivalent form is useful when pressure, rather than volume, is the natural control variable.

Derivation AFrom the First Law to pV^γ = constant

Goal. Starting only from $q = 0$ , $dU = C_{V,m}\,dT$ , $dw = -p\,dV$ , and $pV = nRT$ , derive $pV^\gamma = \text{constant}$ and $TV^{\gamma-1} = \text{constant}$ for one mole of ideal gas undergoing a reversible adiabatic change.

Step 1: Apply the First Law with q = 0. For any infinitesimal step, $dU = dq + dw$ . Adiabatic means $dq = 0$ , so

dU = dw

For a reversible change, the work is $dw = -p\,dV$ (expansion work against the instantaneous gas pressure, since the process is quasistatic).

Step 2: Express dU using the ideal-gas heat capacity. For an ideal gas $U$ is a function of $T$ alone (Joule's law), so per mole,

dU = C_{V,m}\,dT

Combining with Step 1,

C_{V,m}\,dT = -p\,dV

Step 3: Eliminate p using the ideal gas law. Since $pV_m = RT$ for one mole, $p = RT/V_m$ . Substituting,

C_{V,m}\,dT = -\frac{RT}{V_m}\,dV

Step 4: Separate variables. Divide both sides by $T$ and rearrange so that all temperature terms sit on the left and all volume terms on the right:

C_{V,m}\,\frac{dT}{T} = -R\,\frac{dV_m}{V_m}

This step is valid because $C_{V,m}$ is treated as constant over the temperature range considered (true for a monatomic ideal gas exactly, and a good approximation for diatomics away from vibrational excitation).

Step 5: Integrate between the initial and final states.

C_{V,m}\ln\left(\frac{T_2}{T_1}\right) = -R\ln\left(\frac{V_{m,2}}{V_{m,1}}\right)

Step 6: Convert to a single power-law relation. Divide through by $C_{V,m}$ and exponentiate:

\left(\frac{T_2}{T_1}\right) = \left(\frac{V_{m,1}}{V_{m,2}}\right)^{R/C_{V,m}}

Define $\gamma \equiv C_{p,m}/C_{V,m}$ . For an ideal gas, $C_{p,m} - C_{V,m} = R$ (Mayer's relation), so

\frac{R}{C_{V,m}} = \gamma - 1

Substituting gives

T_1 V_{m,1}^{\gamma - 1} = T_2 V_{m,2}^{\gamma - 1}

which is the boxed relation

\boxed{TV^{\gamma-1} = \text{constant}}

Step 7: Convert to the pressure–volume form. Use $T = pV_m/R$ at each state and substitute into $TV_m^{\gamma-1} = \text{constant}$ :

\frac{pV_m}{R}\,V_m^{\gamma-1} = \text{constant}

pV_m^{\gamma} = \text{constant}\times R

Since $R$ is a true constant, absorb it into the constant on the right to obtain the final result

\boxed{pV^{\gamma} = \text{constant}}

Consistency check. For $\gamma \to 1$ (as if $C_{p,m} \approx C_{V,m}$ , unphysical for an ideal gas but useful as a limit), the relation collapses toward $pV = \text{constant}$ , the isothermal Boyle's law form — confirming that the extra temperature coupling built into $\gamma > 1$ is exactly what distinguishes an adiabat from an isotherm.

Worked examples

Example 1 — Foundational: adiabatic compression of argon. A sample of argon gas (monatomic, treat as ideal) occupies $V_1 = 5.00\ \text{L}$ at $p_1 = 1.00\ \text{atm}$ and $T_1 = 298\ \text{K}$ . It is compressed reversibly and adiabatically to $V_2 = 1.00\ \text{L}$ . Find $p_2$ and $T_2$ . For a monatomic ideal gas, $C_{V,m} = \tfrac{3}{2}R$ and $C_{p,m} = \tfrac{5}{2}R$ , so

\gamma = \frac{C_{p,m}}{C_{V,m}} = \frac{5/2\,R}{3/2\,R} = \frac{5}{3} = 1.667

Apply $p_1V_1^\gamma = p_2V_2^\gamma$ :

p_2 = p_1\left(\frac{V_1}{V_2}\right)^{\gamma} = (1.00\ \text{atm})\left(\frac{5.00}{1.00}\right)^{1.667}

\left(5.00\right)^{1.667} = e^{1.667\ln 5.00} = e^{1.667 \times 1.609} = e^{2.683} = 14.63

p_2 = 14.6\ \text{atm}

Now find $T_2$ from $T_1V_1^{\gamma-1} = T_2V_2^{\gamma-1}$ , with $\gamma - 1 = 0.667$ :

T_2 = T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1} = (298\ \text{K})\left(5.00\right)^{0.667}

(5.00)^{0.667} = e^{0.667\times 1.609} = e^{1.073} = 2.925

T_2 = 298\ \text{K}\times 2.925 = 872\ \text{K}

Sanity check. Verify with the ideal gas law directly: $p_2V_2/T_2$ should equal $p_1V_1/T_1$ .

\frac{p_1V_1}{T_1} = \frac{(1.00)(5.00)}{298} = 0.01678\ \text{atm L K}^{-1}

\frac{p_2V_2}{T_2} = \frac{(14.6)(1.00)}{872} = 0.01674\ \text{atm L K}^{-1}

These agree to within rounding, confirming internal consistency. Compressing argon to one-fifth its volume adiabatically raises the temperature by nearly 600 K — this is the physical basis of diesel ignition.

Example 2 — Applied: work done and final state for a diatomic gas expansion. Two moles of nitrogen gas ( $N_2$ , diatomic, $C_{V,m} = \tfrac{5}{2}R$ over the temperature range of interest, vibrations frozen out) start at $T_1 = 500\ \text{K}$ , $p_1 = 10.0\ \text{atm}$ . The gas expands reversibly and adiabatically until $p_2 = 2.00\ \text{atm}$ . Find $T_2$ , $V_1$ , $V_2$ , and the work done by the gas. First, $\gamma$ for a diatomic ideal gas:

\gamma = \frac{C_{p,m}}{C_{V,m}} = \frac{7/2\,R}{5/2\,R} = \frac{7}{5} = 1.400

Use the $T$ – $p$ form of the adiabat, $Tp^{(1-\gamma)/\gamma} = \text{constant}$ , so

T_2 = T_1\left(\frac{p_2}{p_1}\right)^{(\gamma-1)/\gamma}

with exponent $(\gamma-1)/\gamma = 0.400/1.400 = 0.2857$ .

T_2 = 500\ \text{K}\times\left(\frac{2.00}{10.0}\right)^{0.2857} = 500\ \text{K}\times(0.200)^{0.2857}

(0.200)^{0.2857} = e^{0.2857\times\ln(0.200)} = e^{0.2857\times(-1.609)} = e^{-0.460} = 0.631

T_2 = 500\ \text{K}\times 0.631 = 316\ \text{K}

Get the volumes from $pV = nRT$ , using $R = 0.08206\ \text{L atm mol}^{-1}\text{K}^{-1}$ :

V_1 = \frac{nRT_1}{p_1} = \frac{(2.00)(0.08206)(500)}{10.0} = 8.206\ \text{L}

V_2 = \frac{nRT_2}{p_2} = \frac{(2.00)(0.08206)(316)}{2.00} = 25.93\ \text{L}

Check via the $pV^\gamma$ relation (using SI-consistent units is not required here since both sides use L·atm, but let's confirm the ratio matches):

\frac{V_2}{V_1} = \left(\frac{p_1}{p_2}\right)^{1/\gamma} = (5.00)^{1/1.4} = (5.00)^{0.7143} = e^{0.7143\times1.609} = e^{1.149} = 3.155

Predicted $V_2 = 8.206\ \text{L}\times 3.155 = 25.90\ \text{L}$ , matching the 25.93 L found above (small rounding difference). Good. Now the work done by the gas. Since $q = 0$ , $w_{by\,gas} = -\Delta U = -nC_{V,m}(T_2-T_1)$ :

w_{by\,gas} = -n\left(\frac{5}{2}R\right)(T_2 - T_1) = -(2.00)\left(\frac{5}{2}\times 8.314\ \text{J mol}^{-1}\text{K}^{-1}\right)(316-500)\ \text{K}

= -(2.00)(20.79\ \text{J mol}^{-1}\text{K}^{-1})(-184\ \text{K}) = 7648\ \text{J} \approx 7.65\ \text{kJ}

The gas expands, cools from 500 K to 316 K, and delivers about 7.65 kJ of work to the surroundings — all of it drawn from its own internal energy, since no heat entered the system.

Limiting cases & common traps

Situation	What happens	Common error
$\gamma \to 1$	Adiabat approaches isotherm's $pV=\text{const}$ shape	Forgetting real gases never reach $\gamma=1$ ; monatomic ideal gas gives the largest $\gamma = 5/3$
Reversible adiabatic expansion	$T$ always falls	Assuming $T$ is constant because "no heat is added" — heat and temperature are not the same thing
Reversible adiabatic compression	$T$ always rises	Confusing this with isothermal compression, where $w$ and $q$ both flow but $T$ stays fixed
Using $C_{V,m}$ vs $C_{p,m}$ in $\Delta U$	Adiabatic $w$ always uses $C_{V,m}$ , since $\Delta U = nC_{V,m}\Delta T$ regardless of the path	Mistakenly plugging in $C_{p,m}$ because pressure is changing
Free (irreversible) adiabatic expansion into vacuum	$q=0$ , $w=0$ , so $\Delta U = 0$ and $T$ is unchanged for an ideal gas	Applying $pV^\gamma=\text{constant}$ to an irreversible expansion — the formula requires a reversible path

The chemist behind it

The exponent $\gamma$ and the study of adiabatic sound propagation trace back to Pierre-Simon Laplace, who corrected Newton's formula for the speed of sound in air by recognizing that sound compressions and rarefactions happen too fast for heat to equilibrate — they are adiabatic, not isothermal. Laplace's insight that $c_{sound} \propto \sqrt{\gamma RT/M}$ rather than $\sqrt{RT/M}$ brought theory into agreement with measurement, and the ratio $\gamma = C_{p,m}/C_{V,m}$ is still sometimes called the Laplace constant in older texts.

Origins · 1 minPierre-Simon Laplace and how this was discovered

Pierre-Simon Laplace

1749–1827 · French mathematician and physicist who corrected Newton's sound-speed formula by introducing the adiabatic heat-capacity ratio, foreshadowing the pV^γ relation used throughout thermodynamics and acoustics

Ask the tutor: What changes in the math and the physics when we move from this reversible adiabatic expansion to the irreversible Joule expansion into a vacuum?

Why does the reversible path matter for pV^γ = constant

The derivation uses $dw = -p\,dV$ , which is only valid when the external pressure equals the gas pressure at every instant — that is the definition of a reversible, quasistatic process. In an irreversible adiabatic expansion, such as expansion against a constant external pressure lower than the gas pressure, $q$ is still zero, but the work is $-p_{ext}\Delta V$ , not $-p\,dV$ integrated along the gas's own pressure. The relation $pV^\gamma=\text{constant}$ simply does not hold for the irreversible path, even though the First Law itself still applies.

Does pV^γ = constant apply to real gases

Only approximately. The derivation assumes an ideal gas, both for $pV=nRT$ and for $U$ depending on $T$ alone. Real gases have intermolecular forces, so $U$ depends weakly on $V$ as well as $T$ , and $\gamma$ itself varies with temperature and pressure. For many practical engineering calculations — compressors, engines, atmospheric adiabats — the ideal-gas adiabat is still a good working approximation, but precise work requires a real-gas equation of state and measured heat capacities.

How is this different from the isothermal reversible expansion formula

Isothermal reversible expansion holds $T$ fixed, giving $w = -nRT\ln(V_2/V_1)$ and $pV=\text{constant}$ (Boyle's law). Adiabatic reversible expansion holds $q=0$ instead, allowing $T$ to change, giving the steeper relation $pV^\gamma=\text{constant}$ . Both are reversible ideal-gas processes, but they respond to different constraints — constant temperature versus zero heat flow — and therefore trace different paths on a $p$ – $V$ diagram, always crossing at exactly one point for a given starting state.

Why is γ always greater than 1

Because $C_{p,m} = C_{V,m} + R$ for an ideal gas (Mayer's relation), and $R > 0$ , we always have $C_{p,m} > C_{V,m}$ , so $\gamma = C_{p,m}/C_{V,m} > 1$ . Physically, at constant pressure the gas must also do expansion work as it absorbs heat, so more heat is needed per kelvin of temperature rise than at constant volume — hence $C_{p,m}$ exceeds $C_{V,m}$ , and the ratio always exceeds unity.

What comes next after this relation

The next natural question is what happens when an ideal gas expands adiabatically but without any opposing pressure at all — the Joule expansion into a vacuum. There $q=0$ and $w=0$ simultaneously, so $\Delta U=0$ , and for an ideal gas that forces $\Delta T = 0$ as well. Comparing that irreversible case with the reversible adiabat here sharpens the distinction between "no heat" and "no temperature change," and sets up the idea of internal pressure for real gases.