2.4, 2.10
Applications of the First Law to ideal gases
What we know
Equation of state
The fist law
pv = RT
du = dq + dw
Internal energy is function
only of temperature
Specific heats are related
"expansion work"
dh = dq + vdp
du = cv dT
c p − cv = R
dw = − pdv
dh = c p dT
du = cv dT
dh = c p dT
c p − cv = R
dw = − pdv
The first law of thermodynamics for an ideal gas
du = dq + dw
dh = dq + vdp
cv dT = dq − pdv
c p dT = dq + vdp
Isothermal vs. Adiabatic
An isothermal process in one in which
the initial and final temperatures are
the same.
dT = 0
An adiabatic process in one in which no
heat is exchanged between the system and
its surroundings.
dq = 0
Isothermal processes are not
necessarily adiabatic.
Reversible Changes
• A reversible change is one that can be
reversed by an infinitesimal modification
of a variable.
• In a reversible expansion or compression
penv = p gas
Lets consider an isothermal expansion
du = cv dT
cv dT = dq − pdv
dT = 0
Because Internal energy is function
only of temperature, the internal
energy of the gas is unchanged
dq = pdv
The first law of thermodynamics for an isothermal expansion
dq = pdv
isothermal expansion
If we want to integrate
only expansion
work"
dq = pdv
dw = − pdv
What do we need????
∫ dw ≠ 0
Path!!!
Lets consider the path of an
isothermal reversible expansion
A
B
Equation of state is satisfied
during all the stages of the
expansion
pv = RT
v1
v2
isothermal expansion
If we want to integrate
dq = pdv
dv
dq = RT
v
⎛ v2 ⎞
Δq = RT ln⎜⎜ ⎟⎟
⎝ v1 ⎠
RT
p=
v
pv = RT
Δq = RT ∫
v2
v1
RT
v=
p
dv
v
⎛ p1 ⎞
Δq = RT ln⎜⎜ ⎟⎟
⎝ p2 ⎠
⎛ v2 ⎞
⎛ p1 ⎞
Δq = RT ln⎜⎜ ⎟⎟ = RT ln⎜⎜ ⎟⎟
⎝ v1 ⎠
⎝ p2 ⎠
isothermal expansion
du = cv dT
dh = c p dT
c p − cv = R
Constant volume process
du = dq + dw
dw = − pdv
dv = 0
du = dq − pdv
du = dq
du = cv dT
dq = cv dT
The amount of heat required to
raise the temperature of the
gas from T1 to T2 at constant
volume
Δq = cv (T2 − T1 )
Constant volume
du = cv dT
dh = c p dT
Constant pressure process
dh = dq + vdp
dh = c p dT
The amount of heat required to
raise the temperature of the
gas from T1 to T2 at constant
pressure
dw = − pdv
c p − cv = R
dp = 0
dh = dq
dq = c p dT
Δq = c p (T2 − T1 )
Constant pressure
Adiabatic process
dq = 0
Process in which NO HEAT is
exchange between the system and
its environment
du = dq + dw
du = dw
w
Fist law for a
reversible adiabatic
process
du = − pdv
An adiabatic compression
increases the internal
energy of the system
q
q
Adiabatic process
The first law of thermodynamics for an ideal gas
cv dT = dq − pdv
Adiabatic process
cv dT = − pdv
c p dT = dq + vdp
dq = 0
c p dT = vdp
Fist law for adiabatic expansion
Adiabatic process
dq = 0
Poisson’s equations
Lets consider a reversible adiabatic EXPANSION for an ideal gas
cv dT = − pdv
pv = RT
RT
cv dT = −
dv
v
Lets assume
constant
cv
RT
p=
v
dT
dv
cv
= −R
T
v
There is a final and initial state
cv ∫
T2
T1
v2 dv
dT
= −R∫
v1 v
T
Poisson’s equations
cv ∫
T2
T1
v2 dv
dT
= −R∫
v1 v
T
⎛ T2 ⎞ ⎛ v1 ⎞
⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟
⎝ T1 ⎠ ⎝ v2 ⎠
R
cv
Reverse process
⎛ v2 ⎞
⎛ T2 ⎞
cv ln⎜⎜ ⎟⎟ = − R ln⎜⎜ ⎟⎟
⎝ v1 ⎠
⎝ T1 ⎠
During an adiabatic
expansion of a gas, the
temperature decreases
adiabatic Compression
Work is done on the gas and the temperature increases
⎛ T2 ⎞ ⎛ v1 ⎞
⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟
⎝ T1 ⎠ ⎝ v2 ⎠
R
⎛ p2 ⎞ ⎛ v1 ⎞
⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟
⎝ p1 ⎠ ⎝ v2 ⎠
⎛ T2 ⎞ ⎛ p2 ⎞
⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟
⎝ T1 ⎠ ⎝ p1 ⎠
cv
cp
cv
pv = RT
R
cp
isothermal expansion Vs reversible adiabatic expansion
pv = RT
A given pressure
decrease produces a
smaller volume increase
in the adiabatic case
relative to the
isothermal case
Temperature also decreases
during the adiabatic expansion
P, v diagram
Dry Adiabatic Processes in the Atmosphere
T0 ⎛ p 0 ⎞
= ⎜⎜ ⎟⎟
T ⎝ p ⎠
R
cp
dq = 0
For reversible adiabatic
processes for an ideal gas
Lifting Processes
Orographic Lifting
Frontal Lifting
p
T
Low-level convergence
Vertical Mixing
Lifting:
A dry
adiabatic
process as
long
CONDENSATION
DOES NOT
OCCUR!!
If for
p 0 = 1000 mb
T0 ⎛ p 0 ⎞
= ⎜⎜ ⎟⎟
T ⎝ p ⎠
Where
θ
θ
R
cp
R
cp
the temperature is θ ,
θ
⎛ p0 ⎞
= ⎜⎜ ⎟⎟
T ⎝ p⎠
for dry air is
R
cp
⎛ p0 ⎞
θ = T ⎜⎜ ⎟⎟
⎝ p ⎠
R
cp
R
R
R
2
=
= 5
= = 0.286
c p cv + R 2 R + R 7
Potential Temperature
Temperature a parcel of gas (e.g. dry air)
would have if compressed (or expanded) in
an reversible adiabatic process from a
state,
and T , to a a pressure of p 0 = 1000 mb
p
Also a state variable, invariant during a
reversible adiabatic process:
Conservative quantity!!!!
Consider a
Temperature profile
with
Lapse rate
Γ = 6 Ο C km -1
dT
Γ=−
dz
T2
p = 1000 mb
For p < 1000 mb
p > 1000 mb
⇒
⇒
θ >T
θ <T
⇒
T1
θ =T
Adiabatic compression
to lower the parcel
Adiabatic expansion to
rise the parcel
Considering Water vapor
⎛ p0 ⎞
θ = T ⎜⎜ ⎟⎟
⎝ p ⎠
R
R
??
cp
cp
c p = (1 − q v )c pd + q v c pv ≈ c pd (1 + 0.87 q v )
Specific heat
for moist air
d
v
For dry air
And knowing
(from Ch. 1)
R = Rd (1+ 0.608qv )
⎛ p0 ⎞
θ = T ⎜⎜ ⎟⎟
⎝ p ⎠
Considering water vapor
For water vapor
R Rd
=
c p c pd
Rd (1− 0.26 qv )
qv
Specific humidity
⎛ 1 + 0.608qv ⎞ R d
⎜⎜
⎟⎟ ≈
(1 − 0.26qv )
⎝ 1 + 0.87qv ⎠ c pd
c pd
Potential Temperature for Moist Air
The difference between dry-air and moist
air potential temperature is generally
less than 0.1 degree.
Adiabatic expansion or compression
of moisture air can be treated as if
it were dry air
Important: θ is not
conserved if a change of
phase occurs
Liquid to vapor
solid to Liquid
the temperature a parcel which
contains no moisture would have
to equal the density of a parcel
at a specific temperature and
humidity.
Virtual Temperature
Tv
temperature a parcel at a specific
pressure level and virtual
temperature would have if it were
lowered or raised to 1000 mb.
Virtual Potential
Temperature
θv
Virtual Potential Temperature
Neglect the dependence on water vapor from the exponent, and
replace temperature by virtual temperature
⎛ p0 ⎞
θ = T ⎜⎜ ⎟⎟
⎝ p ⎠
Rd (1− 0.26 qv )
c pd
⎛ p0 ⎞
θ v = Tv ⎜⎜ ⎟⎟
⎝ p⎠
Rd
c pd
Adiabatic Ascent of a Parcel
T
Rate of decrease of
temperature with height
(first law, enthalpy form)
θ
constant
c p dT = vdp
First law for
adiabatic
process
If there are no large vertical accelerations
(hydrostatic relation applies)
− gdz = vdp
dT
Γ=−
dz
c p dT = − gdz
g
Γd =
≈ 9.8 Ο C km -1
c pd
For moist air
Γ=
g
c pd (1+ 0.87qv )
Lapse rate
for dry air