15 KiB
| title | chunk | source | category | tags | date_saved | instance |
|---|---|---|---|---|---|---|
| Adiabatic process | 4/7 | https://en.wikipedia.org/wiki/Adiabatic_process | reference | science, encyclopedia | 2026-05-05T10:56:55.166101+00:00 | kb-cron |
or 25.1 bar. This pressure increase is more than a simple 10:1 compression ratio would indicate; this is because the gas is not only compressed, but the work done to compress the gas also increases its internal energy, which manifests itself by a rise in the gas temperature and an additional rise in pressure above what would result from a simplistic calculation of 10 times the original pressure. We can solve for the temperature of the compressed gas in the engine cylinder as well, using the ideal gas law, P V = n R T (n is amount of gas in moles and R the gas constant for that gas). Our initial conditions being 100 kPa of pressure, 1 L volume, and 300 K of temperature, our experimental constant (n R ) is:
P
V
T
=
c
o
n
s
t
a
n
t
2
=
10
5
P
a
×
10
−
3
m
3
300
K
=
0.333
P
a
m
3
K
−
1
.
{\displaystyle {\begin{aligned}{\frac {P\ V}{T}}&={\mathsf {constant}}_{2}\\&={\frac {10^{5}~{\mathsf {Pa}}\times 10^{-3}~{\mathsf {m}}^{3}}{300~{\mathsf {K}}}}\\&=0.333~{\mathsf {Pa}}\ {\mathsf {m}}^{3}{\mathsf {K}}^{-1}~.\end{aligned}}}
We know the compressed gas has V = 0.1 L and P = 2.51×106 Pa, so we can solve for temperature:
T
=
P
V
c
o
n
s
t
a
n
t
2
=
2.51
×
10
6
P
a
×
10
−
4
m
3
0.333
P
a
m
3
K
−
1
=
753
K
.
{\displaystyle {\begin{aligned}T&={\frac {P\ V}{{\mathsf {constant}}_{2}}}\\&={\frac {2.51\times 10^{6}~{\mathsf {Pa}}\times 10^{-4}~{\text{m}}^{3}}{0.333~{\mathsf {Pa}}\ {\mathsf {m}}^{3}{\mathsf {K}}^{-1}}}\\&=753~{\mathsf {K}}.\end{aligned}}}
That is a final temperature of 753 K, or 479 °C, or 896 °F, well above the ignition point of many fuels. This is why a high-compression engine requires fuels specially formulated to not self-ignite (which would cause engine knocking when operated under these conditions of temperature and pressure), or that a supercharger with an intercooler to provide a pressure boost but with a lower temperature rise would be advantageous. A diesel engine operates under even more extreme conditions, with compression ratios of 16:1 or more being typical, in order to provide a very high gas pressure, which ensures immediate ignition of the injected fuel.
=== Adiabatic free expansion of a gas ===
For an adiabatic free expansion of an ideal gas, the gas is contained in an insulated container and then allowed to expand in a vacuum. Because there is no external pressure for the gas to expand against, the work done by or on the system is zero. Since this process does not involve any heat transfer or work, the first law of thermodynamics then implies that the net internal energy change of the system is zero. For an ideal gas, the temperature remains constant because the internal energy only depends on temperature in that case. Since at constant temperature, the entropy is proportional to the volume, the entropy increases in this case, therefore this process is irreversible.
=== Derivation of P–V relation for adiabatic compression and expansion === The definition of an adiabatic process is that heat transfer to the system is zero, δQ = 0. Then, according to the first law of thermodynamics,
where dU is the change in the internal energy of the system and δW is work done by the system. Any work (δW) done must be done at the expense of internal energy U, since no heat δQ is being supplied from the surroundings. Pressure–volume work δW done by the system is defined as
However, P does not remain constant during an adiabatic process but instead changes along with V. It is desired to know how the values of dP and dV relate to each other as the adiabatic process proceeds. For an ideal gas (recall ideal gas law PV = nRT) the internal energy is given by
where α is the number of degrees of freedom divided by 2, R is the universal gas constant and n is the number of moles in the system (a constant). Differentiating equation (a3) yields
Equation (a4) is often expressed as dU = n CV dT because CV = α R . Now substitute equations (a2) and (a4) into equation (a1) to obtain
−
P
d
V
=
α
P
d
V
+
α
V
d
P
,
{\displaystyle -P\ \mathrm {d} V=\alpha \ P\ \mathrm {d} V+\alpha \ V\ \mathrm {d} P\ ,}
factor −P dV:
−
(
α
+
1
)
P
d
V
=
α
V
d
P
,
{\displaystyle -(\alpha +1)\ P\ \mathrm {d} V=\alpha V\ \mathrm {d} P\ ,}
and divide both sides by P V :
−
(
α
+
1
)
d
V
V
=
α
d
P
P
.
{\displaystyle -(\alpha +1)\ {\frac {\mathrm {d} V}{V}}=\alpha \ {\frac {\mathrm {d} P}{P}}~.}
After integrating the left and right sides from V0 to V and from P0 to P and changing the sides respectively,
ln
(
P
P
0
)
=
−
α
+
1
α
ln
(
V
V
0
)
.
{\displaystyle \ln \left({\frac {P}{P_{0}}}\right)=-{\frac {\alpha +1}{\alpha }}\ \ln \left({\frac {V}{V_{0}}}\right)~.}
Exponentiate both sides, substitute α + 1/α with γ, the heat capacity ratio
(
P
P
0
)
=
(
V
V
0
)
−
γ
,
{\displaystyle \left({\frac {P}{P_{0}}}\right)=\left({\frac {V}{V_{0}}}\right)^{-\gamma }\ ,}
and eliminate the negative sign to obtain
(
P
P
0
)
=
(
V
0
V
)
γ
.
{\displaystyle \left({\frac {P}{P_{0}}}\right)=\left({\frac {V_{0}}{V}}\right)^{\gamma }~.}
Therefore,
(
P
P
0
)
(
V
V
0
)
γ
=
1
,
{\displaystyle \left({\frac {P}{P_{0}}}\right)\left({\frac {V}{V_{0}}}\right)^{\gamma }=1\ ,}
and
P
0
V
0
γ
=
P
V
γ
=
c
o
n
s
t
a
n
t
.
{\displaystyle P_{0}\ V_{0}^{\gamma }=P\ V^{\gamma }={\mathsf {constant}}~.}
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 (b4) gives