Adiabatic compression

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

γ = 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} = n R / (γ - 1)

. The proof outline is

C_{V} d T = - p d V

from the first law of thermodynamics, followed by

p = n RT / V

from the ideal gas law and integration. The result is

p V^{γ} = constant, T V^{γ - 1} = constant .

(1)

These finite power laws belong to this reversible, constant-

γ

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 predicts

\frac{T _{2}}{T _{1}} = r^{γ - 1}, \frac{p _{2}}{p _{1}} = r^{γ}, W_{in} = - W_{out} = C_{V} (T_{2} - T_{1}) = \frac{p _{1} V _{1}}{γ - 1} (r^{γ - 1} - 1) .

(2)

All 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

γ = 7/5

p_{1} V_{1} = n R T_{1} = 1 kJ

and

r = 2

. Then

T_{2} / T_{1} = 2^{0.4} ≃ 1.31951

and

p_{2} / p_{1} = 2^{1.4} ≃ 2.63902

. The input work and internal energy increase are both

0.79877 kJ

, while

Q_{in} = 0

. No initial temperature or apparatus performance is implied by the chosen energy scale. Independent integration gives the signed outward work

W_{out} = \int_{V_{1}}^{V_{2}} p d V = \frac{p _{2} V _{2} - p _{1} V _{1}}{1 - γ},

(3)

which is negative during compression and agrees with

- Δ U

. This integral uses equilibrium gas pressure only because the path is reversible.

Figure 1.
Reversible adiabatic compression and adiabatic expansion: temperature changes balance signed work
. Original analytical ideal gas model with constant $γ = 1.4$ and common initial state $p_{1} V_{1} = 1 kJ$ . The two runs go from $V_{1}$ separately to $V_{1} /2$ and $2 V_{1}$ ; they are not a compression followed by its reversal. Shaded pressure-volume areas give work magnitudes, with signs supplied by the direction of motion. The dashed isothermal process lines are comparisons, not additional adiabatic processes. The right panel displays $Δ U = - W_{out}$ for each run. No curves are measured data. **Details**.

In the sensor experiment of adiabatic process, compare the pressure-volume integral with

C_{V} Δ 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