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---
title: "Adiabatic process"
chunk: 5/7
source: "https://en.wikipedia.org/wiki/Adiabatic_process"
category: "reference"
tags: "science, encyclopedia"
date_saved: "2026-05-05T10:56:55.166101+00:00"
instance: "kb-cron"
---
P
=
P
1
(
V
1
V
)
γ
.
{\displaystyle P=P_{1}\ \left({\frac {V_{1}}{V}}\right)^{\gamma }~.}
Substituting this into (b2) gives
W
=
∫
V
1
V
2
P
1
(
V
1
V
)
γ
d
V
.
{\displaystyle W=\int _{V_{1}}^{V_{2}}P_{1}\ \left({\frac {V_{1}}{V}}\right)^{\gamma }\ \mathrm {d} V~.}
Integrating, we obtain the expression for work,
W
=
P
1
V
1
γ
V
2
1
−
γ
−
V
1
1
−
γ
1
−
γ
=
P
2
V
2
−
P
1
V
1
1
−
γ
.
{\displaystyle {\begin{aligned}W&=P_{1}\ V_{1}^{\gamma }\ {\frac {V_{2}^{1-\gamma }-V_{1}^{1-\gamma }}{1-\gamma }}\\[1ex]&={\frac {P_{2}\ V_{2}-P_{1}\ V_{1}}{1-\gamma }}.\end{aligned}}}
Substituting γ = ⁠α + 1/α⁠ in the second term,
W
=
−
α
P
1
V
1
γ
(
V
2
1
−
γ
−
V
1
1
−
γ
)
.
{\displaystyle W=-\alpha \ P_{1}\ V_{1}^{\gamma }\ \left(V_{2}^{1-\gamma }-V_{1}^{1-\gamma }\right)~.}
Rearranging,
W
=
−
α
P
1
V
1
(
(
V
2
V
1
)
1
−
γ
−
1
)
.
{\displaystyle W=-\alpha \ P_{1}\ V_{1}\ \left(\left({\frac {V_{2}}{V_{1}}}\right)^{1-\gamma }-1\right)~.}
Using the ideal gas law and assuming a constant molar quantity (as often happens in practical cases),
W
=
−
α
n
R
T
1
(
(
V
2
V
1
)
1
−
γ
−
1
)
.
{\displaystyle W=-\alpha \ n\ R\ T_{1}\left(\left({\frac {V_{2}}{V_{1}}}\right)^{1-\gamma }-1\right)~.}
By the continuous formula,
P
2
P
1
=
(
V
2
V
1
)
−
γ
,
{\displaystyle {\frac {P_{2}}{P_{1}}}=\left({\frac {V_{2}}{V_{1}}}\right)^{-\gamma }\ ,}
or
(
P
2
P
1
)
−
1
γ
=
V
2
V
1
.
{\displaystyle \left({\frac {P_{2}}{P_{1}}}\right)^{-{\frac {1}{\gamma }}}={\frac {V_{2}}{V_{1}}}~.}
Substituting into the previous expression for W,
W
=
−
α
n
R
T
1
(
(
P
2
P
1
)
γ
−
1
γ
−
1
)
.
{\displaystyle W=-\alpha \ n\ R\ T_{1}\left(\left({\frac {P_{2}}{P_{1}}}\right)^{\frac {\gamma -1}{\gamma }}-1\right)~.}
Substituting this expression and (b1) in (b3) gives
α
n
R
(
T
2
−
T
1
)
=
α
n
R
T
1
(
(
P
2
P
1
)
γ
−
1
γ
−
1
)
.
{\displaystyle \alpha \ n\ R\ (T_{2}-T_{1})=\alpha \ n\ R\ T_{1}\ \left(\left({\frac {P_{2}}{P_{1}}}\right)^{\frac {\gamma -1}{\gamma }}-1\right)~.}
Simplifying,
T
2
−
T
1
=
T
1
(
(
P
2
P
1
)
γ
−
1
γ
−
1
)
,
T
2
T
1
−
1
=
(
P
2
P
1
)
γ
−
1
γ
−
1
,
T
2
=
T
1
(
P
2
P
1
)
γ
−
1
γ
.
{\displaystyle {\begin{aligned}T_{2}-T_{1}&=T_{1}\ \left(\left({\frac {P_{2}}{P_{1}}}\right)^{\frac {\gamma -1}{\gamma }}-1\right)\ ,\\{\frac {T_{2}}{T_{1}}}-1&=\left({\frac {P_{2}}{P_{1}}}\right)^{\frac {\gamma -1}{\gamma }}-1\ ,\\T_{2}&=T_{1}\ \left({\frac {P_{2}}{P_{1}}}\right)^{\frac {\gamma -1}{\gamma }}~.\end{aligned}}}
=== Derivation of discrete formula and work expression ===
The change in internal energy of a system, measured from state 1 to state 2, is equal to
At the same time, the work done by the pressure–volume changes as a result from this process, is equal to
Since we require the process to be adiabatic, the following equation needs to be true
By the previous derivation,
Rearranging (c4) gives
P
=
P
1
(
V
1
V
)
γ
.
{\displaystyle P=P_{1}\ \left({\frac {V_{1}}{V}}\right)^{\gamma }~.}
Substituting this into (c2) gives
W
=
∫
V
1
V
2
P
1
(
V
1
V
)
γ
d
V
.
{\displaystyle W=\int _{V_{1}}^{V_{2}}P_{1}\ \left({\frac {V_{1}}{V}}\right)^{\gamma }\ \mathrm {d} V~.}
Integrating we obtain the expression for work,
W
=
P
1
V
1
γ
V
2
1
−
γ
−
V
1
1
−
γ
1
−
γ
=
P
2
V
2
−
P
1
V
1
1
−
γ
.
{\displaystyle W=P_{1}\ V_{1}^{\gamma }{\frac {V_{2}^{1-\gamma }-V_{1}^{1-\gamma }}{1-\gamma }}={\frac {P_{2}\ V_{2}-P_{1}\ V_{1}}{1-\gamma }}~.}
Substituting γ = ⁠α + 1/α⁠ in second term,
W
=
−
α
P
1
V
1
γ
(
V
2
1
−
γ
−
V
1
1
−
γ
)
.
{\displaystyle W=-\alpha \ P_{1}V_{1}^{\gamma }\ \left(V_{2}^{1-\gamma }-V_{1}^{1-\gamma }\right)~.}
Rearranging,
W
=
−
α
P
1
V
1
(
(
V
2
V
1
)
1
−
γ
−
1
)
.
{\displaystyle W=-\alpha \ P_{1}\ V_{1}\ \left(\left({\frac {V_{2}}{V_{1}}}\right)^{1-\gamma }-1\right)~.}
Using the ideal gas law and assuming a constant molar quantity (as often happens in practical cases),
W
=
−
α
n
R
T
1
(
(
V
2
V
1
)
1
−
γ
−
1
)
.
{\displaystyle W=-\alpha \ n\ R\ T_{1}\ \left(\left({\frac {V_{2}}{V_{1}}}\right)^{1-\gamma }-1\right)~.}
By the continuous formula,
P
2
P
1
=
(
V
2
V
1
)
−
γ
,
{\displaystyle {\frac {P_{2}}{P_{1}}}=\left({\frac {V_{2}}{V_{1}}}\right)^{-\gamma }\ ,}
or
(
P
2
P
1
)
−
1
γ
=
V
2
V
1
.
{\displaystyle \left({\frac {P_{2}}{P_{1}}}\right)^{-{\frac {1}{\gamma }}}={\frac {V_{2}}{V_{1}}}~.}
Substituting into the previous expression for W,
W
=
−
α
n
R
T
1
(
(
P
2
P
1
)
γ
−
1
γ
−
1
)
.
{\displaystyle W=-\alpha \ n\ R\ T_{1}\ \left(\left({\frac {P_{2}}{P_{1}}}\right)^{\frac {\gamma -1}{\gamma }}-1\right)~.}
Substituting this expression and (c1) in (c3) gives
α
n
R
(
T
2
−
T
1
)
=
α
n
R
T
1
(
(
P
2
P
1
)
γ
−
1
γ
−
1
)
.
{\displaystyle \alpha \ n\ R\ (T_{2}-T_{1})=\alpha \ n\ R\ T_{1}\ \left(\left({\frac {P_{2}}{P_{1}}}\right)^{\frac {\gamma -1}{\gamma }}-1\right)~.}
Simplifying,
T
2
−
T
1
=
T
1
(
(
P
2
P
1
)
γ
−
1
γ
−
1
)
,
T
2
T
1
−
1
=
(
P
2
P
1
)
γ
−
1
γ
−
1
,
T
2
=
T
1
(
P
2
P
1
)
γ
−
1
γ
.
{\displaystyle {\begin{aligned}T_{2}-T_{1}&=T_{1}\ \left(\left({\frac {P_{2}}{P_{1}}}\right)^{\frac {\gamma -1}{\gamma }}-1\right)\ ,\\{\frac {T_{2}}{T_{1}}}-1&=\left({\frac {P_{2}}{P_{1}}}\right)^{\frac {\gamma -1}{\gamma }}-1\ ,\\T_{2}&=T_{1}\ \left({\frac {P_{2}}{P_{1}}}\right)^{\frac {\gamma -1}{\gamma }}~.\end{aligned}}}
== Graphing adiabats ==
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