Adiabatic Processes for an Ideal Gas

Push a bicycle pump handle in quickly and it warms up in your hand. No heat crossed the wall of the pump. The gas inside simply converted the work you did into internal energy, and its temperature rose. This is an adiabatic process, one in which $q = 0$ throughout. Adiabatic processes matter because they isolate the mechanical side of the first law. With no heat exchange, all of the internal energy change must appear as work, so $\Delta U = w$ exactly. This single constraint lets us predict temperature changes from work alone, and it underlies the operation of diesel engines, sound propagation, and atmospheric lapse rates. This article derives the temperature and work relations for adiabatic expansion and compression of an ideal gas. You will see the reversible case worked out in full. Then you will see the irreversible, constant-external-pressure case. Finally you will compare both against isothermal expansion for the same volume change.

Key takeaways

In any adiabatic process $q = 0$ , so the first law reduces to $\Delta U = w$ .
For an ideal gas, $\Delta U = nC_{V,m}\Delta T$ regardless of path, since $U$ depends only on $T$ .
Adiabatic expansion always cools an ideal gas; adiabatic compression always heats it.
Reversible adiabats are steeper than isotherms on a $p$ – $V$ diagram because $\gamma = C_{p,m}/C_{V,m} > 1$ .
Irreversible adiabatic expansion against constant external pressure gives less cooling than the reversible case for the same final volume.
Key idea: with $q=0$ , work alone drives the temperature change, and how much the gas cools depends entirely on the path taken to get there.

The core insight

Because heat is forbidden to enter or leave, an adiabatic process forces the gas to pay for any expansion work out of its own internal energy account, so temperature and work become two faces of the same first-law statement, $\Delta U = w$ , with the path (reversible vs. irreversible) deciding exactly how much the temperature falls.

The physical picture

Consider an ideal gas sealed in a thermally insulated cylinder with a movable piston. No conducting wall connects the gas to surroundings, so $q = 0$ at every instant of the process, not just at the start and end. The first law for a closed system,

\Delta U = q + w,

collapses to $\Delta U = w$ . Because the internal energy of an ideal gas depends only on temperature (Joule's law, a consequence of zero intermolecular forces), any change in $U$ shows up entirely as a change in $T$ . If the gas does positive work on the surroundings during expansion, $w$ is negative in the convention $w = -\int p_{\text{ext}}\,dV$ , so $U$ falls and the gas cools. If the surroundings do work on the gas during compression, $U$ rises and the gas heats. Two distinct adiabatic paths deserve separate pictures. The reversible adiabat proceeds through a continuous sequence of equilibrium states, with the internal pressure always equal to the external pressure at each infinitesimal step; this path is used later to derive the $pV^\gamma = \text{constant}$ relation. The irreversible adiabat against a constant external pressure — for instance, a piston released against fixed atmospheric pressure — is far simpler to integrate but gives a different, generally smaller, temperature drop for the same volume change, because the gas does less work when it expands against a smaller effective resisting pressure through most of the process.

Figure

Adiabatic vs isothermal paths on a p–V diagram, same starting point

The adiabat starts at the same point as the isotherm but falls away more steeply, because along the adiabat the gas is also cooling as it expands — pressure drops from both the volume increase and the temperature decrease.

Figure

Energy bookkeeping for adiabatic expansion, q=0

The core relations

q = 0 \quad \Rightarrow \quad \Delta U = w

where: $\Delta U$ is the internal energy change and $w$ the work done on the gas, for any adiabatic process.

\Delta U = n C_{V,m}\,\Delta T

where: $n$ is moles of gas, $C_{V,m}$ the molar heat capacity at constant volume, valid for an ideal gas regardless of path (prerequisite: Heat Capacity at Constant Volume).

w = n C_{V,m}(T_2 - T_1) \quad \text{(any adiabatic path)}

where: $T_1, T_2$ are initial and final temperatures; this equals $\Delta U$ because $q=0$ .

T_2 = T_1 - \frac{p_{\text{ext}}(V_2 - V_1)}{n C_{V,m}} \quad \text{(irreversible, constant } p_{\text{ext}}\text{)}

where: $p_{\text{ext}}$ is the fixed opposing pressure during the whole expansion or compression.

T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \quad \text{(reversible adiabat)}

where: $\gamma = C_{p,m}/C_{V,m}$ , the heat capacity ratio (prerequisite: Relation between $C_p$ and $C_V$ ).

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

where: $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ ; this foreshadows the full $pV^\gamma$ derivation covered under Adiabatic Reversible $pVT$ Relations.

Derivation ATemperature change for reversible adiabatic expansion of an ideal gas

Goal. Find how $T$ changes with $V$ when an ideal gas expands or compresses reversibly with $q=0$ at every step.

Step 1. Write the first law for an infinitesimal reversible step. For a reversible change, $dw = -p\,dV$ , and since $q=0$ throughout,

dU = -p\,dV.

Step 2. Express $dU$ using the constant-volume heat capacity. Because $U$ depends only on $T$ for an ideal gas,

dU = nC_{V,m}\,dT,

so combining with Step 1,

nC_{V,m}\,dT = -p\,dV.

Step 3. Substitute the ideal-gas pressure. Using $p = nRT/V$ ,

nC_{V,m}\,dT = -\frac{nRT}{V}\,dV.

Step 4. Separate variables. Divide both sides by $nT$ and rearrange:

C_{V,m}\,\frac{dT}{T} = -R\,\frac{dV}{V}.

Step 5. Integrate from the initial state $(T_1, V_1)$ to the final state $(T_2, V_2)$ . Treating $C_{V,m}$ as constant over the temperature range,

C_{V,m}\ln\frac{T_2}{T_1} = -R\ln\frac{V_2}{V_1}.

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

\ln\frac{T_2}{T_1} = -\frac{R}{C_{V,m}}\ln\frac{V_2}{V_1} \quad\Rightarrow\quad \frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{R/C_{V,m}}.

Step 7. Introduce $\gamma$ . Since $C_{p,m} - C_{V,m} = R$ , we have

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

Step 8. State the boxed result.

\boxed{\,T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}\,}

This says an ideal gas expanding reversibly and adiabatically ( $V_2 > V_1$ ) always cools ( $T_2 < T_1$ ), since $\gamma - 1 > 0$ for any real gas. The companion relation $p V^{\gamma}=\text{constant}$ follows by substituting $T = pV/nR$ , a result developed fully under Adiabatic Reversible $pVT$ Relations.

Worked examples

Example 1 — Reversible adiabatic expansion of a monatomic ideal gas. Suppose 2.00 mol of argon (treated as an ideal monatomic gas, $C_{V,m} = \tfrac{3}{2}R$ ) at initial temperature $T_1 = 500\ \text{K}$ and volume $V_1 = 10.0\ \text{L}$ expands reversibly and adiabatically to $V_2 = 20.0\ \text{L}$ . Find $T_2$ , $\Delta U$ , and $w$ . Heat capacities.

C_{V,m} = \frac{3}{2}(8.314\ \text{J K}^{-1}\text{mol}^{-1}) = 12.47\ \text{J K}^{-1}\text{mol}^{-1}

C_{p,m} = C_{V,m} + R = 12.47 + 8.314 = 20.79\ \text{J K}^{-1}\text{mol}^{-1}

\gamma = \frac{C_{p,m}}{C_{V,m}} = \frac{20.79}{12.47} = 1.667

Temperature from the reversible adiabat relation.

T_2 = T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1} = 500\ \text{K}\times\left(\frac{10.0}{20.0}\right)^{0.667}

Compute $(0.500)^{0.667} = e^{0.667\ln 0.500} = e^{0.667\times(-0.6931)} = e^{-0.4623} = 0.6298$ .

T_2 = 500\ \text{K}\times 0.6298 = 315\ \text{K}

Internal energy change.

\Delta U = nC_{V,m}(T_2 - T_1) = (2.00\ \text{mol})(12.47\ \text{J K}^{-1}\text{mol}^{-1})(315\ \text{K} - 500\ \text{K})

\Delta U = (2.00)(12.47)(-185) = -4614\ \text{J} \approx -4.61\ \text{kJ}

Work. Since $q=0$ , $w = \Delta U = -4.61\ \text{kJ}$ .

Sanity check: the gas expanded and cooled, so $\Delta U < 0$ as expected; $w<0$ correctly shows the gas did work on the surroundings, consistent with the $-p\,dV$ convention.

Example 2 — Comparing reversible vs. irreversible adiabatic expansion to the same final volume. Take 1.00 mol of an ideal diatomic gas ( $C_{V,m} = \tfrac{5}{2}R = 20.79\ \text{J K}^{-1}\text{mol}^{-1}$ , $\gamma = 7/5 = 1.400$ ) initially at $T_1 = 400\ \text{K}$ , $p_1 = 5.00\ \text{atm}$ . It expands adiabatically until $p_2 = 1.00\ \text{atm}$ , once reversibly and once irreversibly against a constant external pressure of $1.00\ \text{atm}$ . Compare final temperatures and work. Initial volume.

V_1 = \frac{nRT_1}{p_1} = \frac{(1.00)(0.08206\ \text{L atm K}^{-1}\text{mol}^{-1})(400\ \text{K})}{5.00\ \text{atm}} = 6.565\ \text{L}

Reversible case. For a reversible adiabat, $p V^\gamma$ is constant, so combined with the ideal gas law one gets $T p^{(1-\gamma)/\gamma} = \text{constant}$ (a relation developed fully in the next module, but usable here as a shortcut):

T_2 = T_1\left(\frac{p_2}{p_1}\right)^{(\gamma-1)/\gamma} = 400\ \text{K}\times\left(\frac{1.00}{5.00}\right)^{0.2857}

$(0.200)^{0.2857} = e^{0.2857\times\ln 0.200} = e^{0.2857\times(-1.609)} = e^{-0.4599} = 0.6313$

T_2^{\text{rev}} = 400\times 0.6313 = 252.5\ \text{K}

w^{\text{rev}} = \Delta U = nC_{V,m}(T_2-T_1) = (1.00)(20.79)(252.5-400) = (20.79)(-147.5) = -3067\ \text{J}

Irreversible case (constant $p_{\text{ext}} = 1.00\ \text{atm}$ ). First find $V_2$ using the ideal-gas law with the unknown $T_2$ ; but here it is simpler to use the energy balance directly. The final volume for the irreversible path is fixed by mechanical equilibrium at the end, $p_2 V_2 = nRT_2$ , combined with

nC_{V,m}(T_2 - T_1) = -p_{\text{ext}}(V_2 - V_1).

Substitute $V_2 = nRT_2/p_2$ :

nC_{V,m}(T_2-T_1) = -p_{\text{ext}}\left(\frac{nRT_2}{p_2} - V_1\right)

20.79(T_2 - 400) = -1.00\times101325\ \text{Pa}\times\left(\frac{(1.00)(8.314)T_2}{101325} - 6.565\times10^{-3}\ \text{m}^3\right)

Using consistent SI units ( $p_{\text{ext}} = 101325\ \text{Pa}$ , $V_1 = 6.565\times10^{-3}\ \text{m}^3$ ):

20.79(T_2 - 400) = -8.314\,T_2 + 665.4

20.79\,T_2 - 8316 = -8.314\,T_2 + 665.4

29.10\,T_2 = 8981.4 \quad\Rightarrow\quad T_2^{\text{irr}} = 308.6\ \text{K}

w^{\text{irr}} = nC_{V,m}(T_2-T_1) = (20.79)(308.6-400) = (20.79)(-91.4) = -1900\ \text{J}

Comparison. The reversible path cools the gas to 252.5 K and releases 3067 J of work; the irreversible path against the same final pressure only cools it to 308.6 K, releasing 1900 J. The irreversible expansion does less work and produces less cooling for reaching the same final pressure, because the gas pushes against the full 1 atm from the very first instant rather than against a continuously falling internal pressure.

Limiting cases & common traps

Situation	Behavior
$\gamma \to 1$ (very many internal degrees of freedom)	Adiabat approaches the isotherm; less cooling per unit expansion
Monatomic gas, $\gamma = 5/3$	Steepest adiabats; largest temperature drop for given $\Delta V$
Free expansion into vacuum, $p_{\text{ext}}=0$	Still adiabatic ( $q=0$ ) but $w=0$ , so $\Delta T = 0$ for an ideal gas
Reversible vs. irreversible, same $V_2$	Reversible path always extracts more work and cools more

The chemist behind it

The formal machinery here — heat capacities, the ratio $\gamma$ , and its role in adiabatic processes — was consolidated in the 19th century as calorimetry and thermodynamics matured together. Among the key contributors was the French engineer and physicist Sadi Carnot, whose analysis of idealized heat engines implicitly relied on reversible adiabatic and isothermal strokes, laying groundwork later formalized by Clausius and Kelvin into the full apparatus used here.

Origins · 1 minSadi Carnot and how this was discovered

Sadi Carnot

1796–1832 · pioneered the reversible cycle whose adiabatic branches define the temperature-volume relation used throughout this article

Ask the tutor: How does the reversible adiabat relation $T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}$ combine with the ideal gas law to give the full $pV^\gamma = \text{constant}$ expression?

Why is q exactly zero in an adiabatic process, not just approximately small

Adiabatic is a defining condition, not an approximation: the system is either thermally insulated by a real barrier, or the process happens so fast that negligible heat can cross the boundary in the available time, as in sound waves or rapid compression strokes. In both cases $q=0$ is imposed as a constraint on the process, exactly, by design or by timescale, rather than emerging as a small residual to be neglected.

Does the formula deltaU equals nCv times deltaT apply to any adiabatic process, or only reversible ones

It applies to any process for an ideal gas, reversible or not, because $U$ depends only on $T$ for an ideal gas, independent of path. What changes between reversible and irreversible paths is not this relation but how much work is actually done, since $w$ depends on the path through $p_{\text{ext}}$ , and that in turn fixes the final $T_2$ differently for each path.

How does adiabatic expansion relate to isothermal expansion for the same initial state and volume change

Both start from the same $p_1, V_1, T_1$ , but an isotherm keeps $T$ fixed by allowing heat to flow in as the gas expands, while the adiabat has no such heat input and so cools. On a $p$ – $V$ diagram the adiabat is always steeper than the isotherm through the same point, and it lies below the isotherm at any larger volume, doing less total work for the same expansion.

Can an ideal gas be compressed adiabatically without heating up

No, for a compression with $p_{\text{ext}} > 0$ , work is done on the gas, so $w>0$ and thus $\Delta U>0$ , meaning $T$ must rise. The only way to avoid heating during compression of an ideal gas is to do zero work on it, which is not a compression at all. This heating is why diesel engines ignite fuel purely from compression, without a spark.

Why does gamma determine how steep the adiabat is compared to the isotherm

The exponent $\gamma - 1$ in $T_1V_1^{\gamma-1} = T_2V_2^{\gamma-1}$ controls how strongly temperature responds to a volume change. A larger $\gamma$ , as for monatomic gases with fewer degrees of freedom to store energy, means a given volume increase produces a larger temperature drop, and hence a faster-falling pressure, which is exactly what makes the $p$ – $V$ curve steeper than the constant- $T$ isotherm.