Chemistry 360
Spring 2015
Dr. Jean M. Standard
February 4, 2015
Review of Expansion/Compression of Gases
A review of various types of compression and expansion processes is given below. The examples focus on ideal
gases, but most of the equations can also be developed using the various equations of state for real gases. The
examples concentrate on determining the values of the First Law quantities q, w, and ΔU , and in addition the
enthalpy change ΔH is also given. In most cases, the processes considered are reversible processes.
I. CONSTANT TEMPERATURE (ISOTHERMAL) PROCESSES
€
For a reversible isothermal process,
w = −
∫
€
P dV
Using the ideal gas equation,
€
w = − nRT ℓn(V2 /V1)
or
€
w = − nRT ℓn( P1 / P2 ) .
#∂ U &
Since %
( = 0 for an ideal gas, and dT = 0 for an isothermal process, we have that ΔU = 0 . Then, from the
$ ∂ V 'T
€
First Law, q = −w .
€
H
=
U
+ PV ,
Using
the
definition
€
€
€
ΔH = ΔU + Δ( PV ) .
€
Substituting the ideal gas expression,
€
ΔH = ΔU + Δ( nRT ) ,
or
ΔH = ΔU + nRΔT .
€
So, for an isothermal reversible expansion or compression of an ideal gas, since ΔU = 0 and ΔT = 0 , we also
have that ΔH = 0 .
€
Summary – Isothermal Reversible Processes
€
q=
w=
ΔU =
€
€
€
€
ΔH =
€
–w
− nRT ℓn(V2 /V1 )
0
0
€
€
2
II. CONSTANT PRESSURE (ISOBARIC) PROCESSES
Since P is constant, w = − Pext ΔV , and if the process is reversible, then w = − P ΔV .
We showed in class that if C p is independent of temperature for an ideal gas, then
€
€
q p = ΔH = C p ΔT .
€
From the First Law,
€
ΔU = q + w = C p ΔT − P ΔV .
Summary – Isobaric Reversible Processes
q€= ΔH = C p ΔT
−P ΔV
w=
€
€
€
C p ΔT − P ΔV (For an isobaric process, this is actually the same as C v ΔT .)
ΔU =
€
ΔH =
€
C p ΔT
€
€
€ VOLUME
III. CONSTANT
€ (ISOCHORIC) PROCESSES
Since the process is constant volume, dV = 0 . The work is therefore
w = −
€
∫
Pext dV
w = 0.
We showed in class that if C v is independent of temperature for an ideal gas, then
€
qv = ΔU = C v ΔT .
€
From the definition of enthalpy,
€
ΔH = ΔU + Δ( PV )
= C v ΔT + PΔV + VΔP .
Since the process is constant volume, ΔV = 0 and therefore
€
ΔH = C v ΔT + VΔP .
€
Summary – Isochoric Processes
q = ΔU€ = C v ΔT
w=
€
€
ΔU =
€
ΔH =
€
€
€
€
0
C v ΔT
C v ΔT + VΔP (For an isochoric process, this is actually the same as C p ΔT .)
€
3
IV. ADIABATIC PROCESSES
For an adiabatic process, q = 0 by definition. Then, from the First Law,
w = ΔU .
€
#∂ U &
For an ideal gas, since %
( = 0 , we have as long as C v is independent of temperature that
€
$ ∂ V 'T
€
ΔU = C v ΔT .
€
In order to calculate the temperature difference in the equation above, an expression for the initial and final
temperatures must be developed. If the€process is reversible, then
dw = − P dV .
We also know for an ideal gas that dU = C v dT , and for an adiabatic process dU = dw . We can substitute
these two relations to obtain
€
C v dT = − P dV .
€
€
Using the ideal gas equation for P and integrating yields
€
C v ℓn(T 2 /T1) = − nR ℓn(V2 /V1) .
Using properties of logarithms, we have
€
" T %C v
" V %−nR
ln$ 2 '
= ln$ 2 ' .
# T1 &
# V1 &
Taking the exponential of both sides yields
€
" T %C v
" V %−nR
= $ 2'
$ 2'
# T1 &
# V1 &
€
Cv
" V %nR
or " T 2 %
= $ 1' .
$ '
# T1 &
# V2 &
Isolating the temperature ratio yields
€
" V %nR /C v
T2
= $ 1'
.
T1
# V2 &
Using the relation C p = C v + nR , we can solve to get nR = C p − C v the temperature ratio becomes
€
€
€
" V %(C p −C v ) /C v
T2
= $ 1'
.
€
T1
# V2 &
4
We can define the exponent as
γ =
C p − Cv
Cv
.
Note that this is not the same γ as used in the CHE 361 Heat Capacity of Gases experiment. Substituting, the
temperature ratio in a reversible adiabatic
€ process is
" V %γ
T2
= $ 1' .
T1
# V2 &
If the initial and final pressures are known rather than the initial and final volumes, the temperature ratio can be
rewritten as
€
" P %nR /C p
T2
= $ 2'
.
T1
# P1 &
Summary – Adiabatic Reversible Processes
q=
0€
w=
ΔU =
ΔU
C v ΔT
C p ΔT
€
ΔH =
€
€
€
€
€
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