A derivation of the rst law of thermodynamics
A derivation of the rst law of thermodynamics for a uid is presented here.The basic
elements of the derivation presented here follow the arguments given in Holton,An In
troduction to Dynamic Meteorology,2004.Consistent with our earlier description of uid
mechanics,1d uid ow is assumed.(The corresponding derivation in 3d is more narural,
but requires a good knowledge of vector calculus.)
We begin with a restatement of the equations of motion of uid mechanics in 1d.Mass
conservation is expressed by the equation
d
dt
+
@v
@x
= 0:(1)
For convenience the variable A used earlier to denote the crosssectional area of the pipe or
duct is assumed to be constant,and therefore has been factored out.This assumption puts
the focus on the compressibility of the uid,which is of central importance in thermody
namics.Newton's second law for an inviscid (frictionless) uid in 1d is
dv
dt
=
@p
@x
:
After multiplication by v,this equation can be written
d
dt
1
2
v
2
= v
@p
@x
:(2)
Consider the product pv,whose dimensions are N=m
2
m=s = (Nm=s)=m
2
= (J=s)=m
2
.
This quantity is related to the time rate of change of work done per unit area.Specically,
the local change in this quantity is the local time rate of change of work done by the uid
per unit area.Thus
dWork
dt
= [(pv)
x+x
(pv)
x
]yz =
"
@(pv)
@x
x
#
yz =
@(pv)
@x
V:(3)
Here xyz = V is a small element of volume.
The general form of the equation describing the time rate of change of the total energy
of a small element of mass V,including an unspecied source/sink term,is
dE
dt
+
dWork
dt
=
dq
dt
:(4)
Here E is the sum of all contributions to the energy of the uid that do not contribute to
the work done,Work is the work done by the uid and q is the heat added to or taken
from the the uid.(Heat exchange mechanisms are radiation,conduction and latent heat
exchange,i.e.,heat exchange associated with phase changes.) Let e
0
denote the internal (or
microscopic) energy per unit mass of the uid.The external (or macroscopic) energy per
unit mass of the uid is v
2
=2.Thus E = V (e
0
+v
2
=2).Let q
0
denote the heat gained or
lost by the uid per unit mass,so q = V q
0
.Then,combining Eqs.(4) and (3) gives
V
"
de
0
dt
+
d
dt
v
2
2
!#
+V
@
@x
(pv) = V
dq
0
dt
:(5)
Note that @(pv)=@x = p @v=@x+v @p=@x.Dividing Eq.(5) by V and subtracting Eq.(2)
gives
de
0
dt
+
p
@v
@x
=
dq
0
dt
:(6)
Substituting @v=@x =
1
d=dt [from Eq.(1)] gives
de
0
dt
p
2
d
dt
=
dq
0
dt
:(7)
This equation is a statement of the rst law of thermodynamics.Had we included gravity in
our derivation,the nal result,Eq.(7),would have been unchanged;gravitational potential
energy would have been included in Eqs.(2) and (5),but those terms would have cancelled
in Eq.(6).
The rst law,Eq.(7),is written in a slightly dierent the form in your textbook.In that
book { and most introductory textbooks { the time dependence of thermodynamic processes
is not considered.Multiplication of Eq.(7) by dt puts the emphasis on changes in the
thermodynamic variables.Also,to simplify the treatment of enclosed systems whose volume
may change,we multiply Eq.(7) by mass m,introduce e = me
0
,q = mq
0
,and note that
because = m=V,
2
d=dt = m
1
dV=dt.With these trivial changes,Eq.(7) becomes
de +pdV = dq:(8)
Note that +pdV is the work done by the system,while pdV is the work done on the system.
The preceding derivation should make clear that e is the internal energy of the uid asso
ciated with random motions of the molecules that comprise the uid,while the source/sink
term dq is the heat added to or lost from the uid.Consistent with kinetic theory,the
internal energy e depends only on the temperature of the uid.For example,if the uid is
at rest,it has zero macroscopic kinetic energy because the mean velocity of the molecules in
the uid is zero,but an increase in temperature of the uid increases rms molecular speeds,
thereby increasing the internal { or microscopic { energy of the uid.
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