Chapter 2
First Law
“The first law of thermodynamics is nothing more
than the principle of the conservation of energy applied
to phenomena involving the production or absorbtion of
heat.” - Max Planck.
In this discussion Q is heat and W is work and the
bold d is used to indicate that the differential is inexact. (In class I will “bar” the “d” to distinguish it
from an exact differential.) By modern convention,
dQ is heat added “to” and
2.1
dW is work done “on”.
The Internal Energy
2.1.1
2. Heat is the energy transfer in response to T differences. A better description would be that Q is
the energy transfer which is not work. Begging the
question? Not if we have a good definition of work.
Components
1. Translational kinetic energy
2. Rotational kinetic energy
3. Vibrational kinetic and potential energy
4. Association energies
(a) atom-atom (chemical energy)
(b) nucleus - electron and electron-electron
(c) nucleon(n/p)-nucleon (nuclear energy)
Any associations that do not change (∆ =  −) can
be literally “swept under the rug” as whatever the association energy is for this sector, it does not change. For
example, chemical reactions do not change the nuclei [all
the nuclei on reactants (R) side are on the products (P)
side], therefore we can neglect the “nucleon association”
sector.
2.1.2
History
Many contributed (Julius Robert Mayer, James Prescott
Joule and Hermann von Helmholtz) [∼1842].
2.1.3
Statement
 = d + d 
(2.1)
1. d = [] = d
22
3. Work is the means of energy transfer which changes
the Hamiltonian (or boundary conditions) and
whereby eigen solutions of the system. For example,
consider a particle in a box. Push on the box and
the energy levels move up and separate. If there is
an Auf-bau and particles occupy levels up to some
point, the energy clearly goes up due to the application of this “work”. The addition of heat is then a
process by which the eigen solutions stay the same
but the “particles” (on average) occupy higher energy levels.
2.4
4. Cases:
(a) General:
 = d + d
P
= d −   + (  F · x ) 
H
P
(b) Always:   =  = 0
As the internal energy is a state function, and
has an exact differential, it has a unique value
at any point in thermospace (specified values
of the thermodynamic variables.)
Heat Capacities

1.  ≡ lim∆ →0 ( ∆
) ≡ ( 
 ) 
Definition 1 The symbol “” is used to indicate an
experimental infinitesimal.
2. Heat capacities are defined for each phase. The input or output of heat during a phase change is not
associated with a heat capacity.
(c) No heat flow (adiabatic): ∆ = 
(d) No Work done:  = d + d =  or
∆ =   Again, you should think of “no work
done” as the same as fixing the  levels.
The work can take many forms, but for d to be
non-zero there must be a change (motion) in the conjugate variable to the applied generalized force.
X
d =
F · x
(2.2)
3. Heat capacities are (within a phase) smooth functions of T [i.e. C(T)]. The functions are often expressed as polynomials as these functions are easy
in integrate. (Now you know what is coming don’t
you, see item below.)
4. Example of C for H2 O.
Fig. Cp (H2 O)

2.2
Work
P
d = (“applied forces”)·(“displacement”)
Examples
Generalized “Force”
Displacement
dW
Mechanical F [N= 
]
x
[m]
Fdx
2
Linear tension 1 [J/m]
L
-1 dL
surface ten. 2 [J/m2 ]
A
-2 dA
pressure p3 [Pa=J/m3 ]
V [m3 ]
-PdV
gravitational mg [N]
h [m]
mgdh
electromotive pot.  [V]
q-charge [Coul.]


M-magnitization HdM
magnetic H [Tesla= 
 ]

chemical  [J/mole]
n-quantity
 dn
In trying to remember the signs, just appreciate that
+ve work increases the energy of the system.
5. The heat required to transform ice into steam, from
T= -10 C to T= 110 C at 1 bar is
R0

=
  ( )
+ +
−10 
R 100 
R 110 

( ) + + 100 
( ) .
0
6. The heat capacity under constant volume is:
 ≡ (


) = (
) 


(2.3)
7. The heat capacity under constant pressure is:
2.3
Heat
1. Leads to altered population of QL (most often translational levels)
2. Energy transfer in response to a Temp. imbalance
 ≡ (


) = (
) 


(2.4)
The enthalpy is defined to have this quality.
8.  =  + the work required to heat and expand
the system against an external P [per mole·K].
3. Energy transfer that is not work
Exercise 2.3.1 Book example 2.2: Heat H2 O in an adiabatic enclosure → 24 Most of the energy goes into
the vaporization of the water (association sector). Only
1.7 kJ into expansion. FOLLOW THE ENERGY!
23


(a) W [mole·K] = R = 8.13 [ 
].
(b) W [mole·K] ∼ 0 for liquids and solids as dV
∼ 0 (“No movey, no worky!”)
2.5
State Functions
Fig. Indicator diagram (See 2.12, pg 29.)
For any state function
R
1. ∆ =   =  − 
H
2.  = 0
3. On the other hand, work and heat are not state
functions and depend on path.
R
 = −   
R
 =
= −   = − ∆
2.8
Designer fxn Enthalpy H

1. Define  ≡  +  
But this work (between these end points) is only one
of an infinite number of possible values.
R
 =  d = value depends on the path
R
 =  d = value depends on the path
(  ) is the result of a Legendre transformation
on  to swap out  as an independent variable and
replace it with 
 = [ ] + (  )
= [  −   ] +   +  
4. Consider loading a remote pallet onto a pickup truck
with a hoist on the end of a swingable boom.
=   +  
Note the sign change of the “PV” term. Recall that

 = −( 
)
(a) drag the pallet to near the truck and then use
the hoist to pick it almost straight up and onto
truck.
2. 
(b) Pick the pallet up where it is, at the end of a
boom, and rotate it into the truck.
1 
=
 
=
d −   +   +   = d +  
 = 
2.6
Reversibility
[d + d ] +   +  
= the heat transfer at constant P.
This is “bench-top” chemistry.
3. Integrate to get: ∆ =  
Consider the two following changes in volume:
4. This justifies the previously stated result
1. Quasi static and reversible. The piston moves up
slowly as the water evaporates.

 ≡ ( 
 ) = (  ) 
2. Sudden and irreversible - jank off the cup of water.
2.9
Fig. reversible vs irreversible
2.7

W
dU vs dH
Consider the internal energy and enthalpy differences between two stationary thermodynamic states. Labeling
these states  and  we are asking for the difference in
the heat required to convert a fixed amount of material
from one equilibrium state at  to another at   The
heat required is more in the constant pressure case as one
must commit enough energy to push back the retaining
walls. If one did not do this the pressure would (almost
always) increase.
W vs W
1. As  = ( 
 ) ⇒ ∆ =


–—+–––+–––—+––—+–––-+––
W
W
0
W
W




down the slide
up the slide
below
above
R
 ( )
 =
=
2. As  = ( 
 ) ⇒ ∆ =
 =
=
24
 ∆
R
 ( )
 ∆

3.  =  + (  ) =  +   Therefore the
magnitudes of the heat transfer at constant  and
the heat transfer at constant  required (to convert
 into ) are related via,
ln



= (  )  −1 = (  )−1 = (  )


1∴0
5. As   1 and   0  and  change in opposite
directions, i.e.
4. Cases
(a) Condensed Phases:  ≈  as these phases
have small coefficients of thermal expansion

[ ≡ 1 ( 
) ∼ 0] thus  is very small.
(b) When the # of moles of gas changes:
 ≈  +(∆)
= − ln
 


Remark 2 What is ? It is the energy required to heat
one mole of an ideal gas by one degree. Later we will
have fun with  which is 0  little brother  =  
Clearly between the three (    ) we only need two!
History has saddled us with all three. (The truth is worse
than that. If we decided to measure  in energy units we
could dump both  and  )
Remark 3 What are the units of the IGL? ENERGY.
The ideal gas law says if you measure the internal energy
content via the   product or via the average kinetic
energy per mole (times 2/3rds - see SM) you get the
same answer.
Adiabatic Expansion
Compression of an IG
and
Under adiabatic (“no through passage”) conditions
1.  = 0 ∴  = 


  = −  = −  = − 

R
R

3.  
 = −

4. If C 6=  ( )


= − ln


(b) adiabatic expansions cool.
( )
( )



= (  )  −1 

= (  )  = (  ) 
  =  
0
Remark 1 I will use small letters (v, u, h, g, and s) to
denote molar quantities. Heat capacities will always be
molar so  refers to the molar heat capacity with units
of  While I am on the subject, note that the entropy 

also has the units of  = 8314[ 
]
2.
(a) adiabatic compressions heat and
6. Again assuming IG,
(c) For isothermal processes (rxns) when the # of
moles of gas is fixed and the gases are close to
ideal,
R
R
R


 = −   = −  = − 

 ln
= ln(  ) = ln(  )( − )
with  ≡
 =  +  
2.10



=  ln 
25
See Fig. 2.13, pg 35.
isothermal
compressions squeeze out heat
expansions
suck in heat
adiabatic
heat
cool