Download Key concepts from last class • The internal energy, U, is the total

Key concepts from last class
• The internal energy, U, is the total energy of the system: It is the sum of the kinetic
and potential energies contributed by all the atoms, ions and molecules present in
the system. It depends on temperature and in general on pressure.
• When interactions between molecules can be neglected (ideal gas), the only
contributions to the internal energy are the kinetic energy of the molecules.
• Temperature is a measure of the average kinetic energy.
• For a monoatomic ideal gas, the only contribution is the translational kinetic energy,
which gives
• For a polyatomic ideal gas,
is also a function of temperature, but the value
depends on the number of atoms and bonds.
• We can change the internal energy by adding or subtracting energy (heat or work):
U = w + q.
Key equations / conclusions
We know how to calculate work done by/on the system. It depends on the
external pressure and the volume change :
We know how to calculate heat for a process that involves changing the
temperature at either constant pressure or constant volume :
,
.
,
.∆
(Cm,V is constant)
,
.
,
.∆
(Cm,p is constant)
We know ∆
and because w = 0 for a process at constant volume, ∆
We also know that, for an ideal gas, U depends only on temperature. As a
consequence, if Tf = Ti, then U = 0
Internal Energy
145
Because for a monoatomic ideal gas all the energy is translational kinetic energy:
It makes sense that Um does not depend on the volume in the case of an ideal gas.
Changing the volume changes the average distance between molecules. Because
ideal gases are assumed not to experience any repulsive or attractive interactions,
their internal energy is independent of the average distance between molecules.
T = constant
Same internal energy if interactions are negligible (ideal behavior)
Internal Energy
146
In the case of a real gas, attractive forces dominate when molecules are closer
together. (we discussed this when talking about Van der Waals gases)
T = constant
Now the three states have different internal energy. The internal energy is lowered
by attractive forces.
Important : The internal energy depends only on temperature for an ideal gas, but not
for a real gas. In the case of a real gas, the density of the gas matters because it
affects the potential energy of the system.
Internal Energy
147
Because for a monoatomic ideal gas all the energy is translational kinetic energy:
For polyatomic ideal gases (those that obey PV = nRT), we need to take into account
vibrational and rotational modes, but the internal energy still depends on temperature
only.
For a real gas, the internal energy is lowered when attractive forces dominate
(molecules closer together). Therefore, Um depends not only on T, but also on the
density of the gas.
State and Path Variables
148
Previously, we said…
Heat and work cannot be stored: they are transient quantities that only apply to a system that
undergoes a change in its state:
There is no change in heat, because there is no initial and final heats. Instead, we’ll talk about
the heat (and/or work) involved in changing the system from an initial to a final state.
Note: We could talk about a change in temperature (volume, pressure, etc).
Initial
Vinitial
Tinitial
Pinitial
qinitial
Final
Vfinal
Tfinal
Pfinal
q
20°C
>20°C
150°C
<150°C
qfinal
Hum..Some variables are used to describe the system in a particular state while others are
used to describe the change in state. (more to come!)
State and Path Variables
149
Consider a system consisting of a pure substance (e.g. a pure liquid). Specifying the
pressure, volume and temperature is sufficient to specify many other properties of
the liquid (e.g. its density, refractive index, surface tension, etc).
These properties are called state functions: they depend only on the state of the
system, but not on how the system arrived to that state. If we measure the
temperature of a cup of hot water, we don’t care if the water was heated on a hot
plate or in strong sunlight.
Initial
Vinitial
Tinitial
Pinitial
qinitial
Final
Vfinal
Tfinal
Pfinal
q
20°C
>20°C
150°C
<150°C
qfinal
State and Path Variables
150
Internal energy (U) is a state function
U is a state function because it is the sum of the kinetic and potential energies of all
the molecules and atoms that make up the system.
You don’t need to know how the energy was transferred to the system to be able to
define it.
Uinitial 
Vinitial
Tinitial
Pinitial
qinitial
Initial
Final
Ufinal 
Vfinal
Tfinal
Pfinal
q
20°C
>20°C
150°C
<150°C
qfinal
State and Path Variables
151
And we also said…
So w  0 after all…
What if we do the
one in the figure?
Will it be the
same?
I guess it makes sense… you
need to do work to do all this
stuff, even if you end up in the
same place you started
I’ll try when I get home, but it
looks like it not only depends on
the initial and final states, but
also on the particular path you
follow
Initial = final
In this tutorial, we saw that the work
performed by the system depends on the
particular path we follow to go from the
initial and final state. Therefore, specifying
the initial and final states is not enough. We
also need to know the exact path.
State and Path Variables
152
Heat and Work are path variables. We can’t talk about the “initial heat” and “final
heat”, because heat is not a property that describes a system. Instead, it depends on
the path we followed to take the system from an initial state (characterized by a set of
state functions), to a final state (characterized by other values of the state functions).
Uinitial 
Vinitial
Tinitial
Pinitial
Initial
winitial
150°C
w
Ufinal 
Vfinal
Tfinal
Pfinal
q
20°C
qinitial
Final
>20°C
<150°C
qfinal
wfinal
State and Path Variables
The Path shown in the figure is cyclic because
the initial and final states are the same.
This tells us that p, T, V, U, etc are all
equal to zero.
However, q and w are in principle NOT zero.
Note that we use  to specify a change in a
state function.
There is no q because there is no “initial q” and
“final q”
153
Initial = final
Internal Energy
154
∆
initial
final
You can follow different paths to go from 1 to 6. The change in internal energy is
always the same: U = U6 - U1. However, the amount of work and heat will depend on
the path
Internal Energy
155
∆
What about this path?
initial
final
You can use this to your advantage: You can use the path that is easiest to calculate,
or even create another one.
However, if you need to calculate q and/or w you are in trouble…
gas
1 bar
1.1 bar
1 bar from atmosphere +
0.1 bar from weight.
Frictionless piston,
can move up or down
1 bar
Frictionless piston,
can move up or down
156
w
1.1 bar
Stoppers don’t allow
the piston to move
(up or down)
q
1.1 bar
T , V 
1.1 bar
q
P , T 
P.V = n.R.T
Internal Energy
157
An important result
∆
We already saw that w = 0 if there’s no change in volume. Therefore, if you perform
a process that does not change the volume of the system, the change in internal
energy is simply the heat
∆
It means “keeping volume constant ”
We already defined the molar heat capacity at constant volume, Cm,V :
,

,
.
,
.∆
(Cm,V is constant)
Internal Energy
158
Example: Calculate U, q and w for the process depicted in the figure. Assume one
mole of a gas that behaves ideally. CV = 12.5 J/(K.mol)
∆
1
1
2
You may have information you don’t need
3p0
3p0
3p0
V0, 3p0
V0, 3p0
V0, p0
3
How can I reduce the pressure
while keeping the volume
constant?
2
Internal Energy
159
Example: Calculate U, q and w for the process depicted in the figure. Assume one
mole of a gas that behaves ideally. CV = 12.5 J/(K.mol)
∆
1
1
2
3p0
3p0
V0, 3p0
V0, 3p0
3
3p0
q
2
V0, p0
How can I reduce the pressure
while keeping the volume
constant?
We need to cool down the gas.
Internal Energy
160
Example: Calculate U, q and w for the process depicted in the figure. Assume one
mole of a gas that behaves ideally. CV = 12.5 J/(K.mol)
∆
1
3p0
3p0
3p0
V0, 3p0
V0, 3p0
V0, p0
2
0 if volume does not change
1
2
3
How can I reduce the pressure
while keeping the volume
constant?
We need to cool down the gas.
We need to cool down the gas. T2 < T1  we are lowering the internal energy of the
gas, 
< 0, w = 0
Internal Energy
161
Example: Calculate U, q and w for the process depicted in the figure. Assume one
mole of a gas that behaves ideally. CV = 12.5 J/(K.mol)
∆
1
3p0
3p0
3p0
V0, 3p0
V0, 3p0
V0, p0
2
0 if volume does not change
3
1

2
,
.∆
How can I reduce the pressure
while keeping the volume
constant?
We need to cool down the gas.
We need to cool down the gas. T2 < T1  we are lowering the internal energy of the
gas, 
< 0, w = 0
Internal Energy
162
Example: Calculate U, q and w for the process depicted in the figure. Assume one
mole of a gas that behaves ideally. CV = 12.5 J/(K.mol)
3
1
3
2
3
You may have information you don’t need
Internal Energy
163
Example: Calculate U, q and w for the process depicted in the figure. Assume one
mole of a gas that behaves ideally. CV = 12.5 J/(K.mol)
3
1
3
3
2
1
2
3p0
3p0
3p0
p0
V0, 3p0
V0, 3p0
V0, p0
V0, p0
3
p0
3V0, p0
Internal Energy
164
Example: Calculate U, q and w for the process depicted in the figure. Assume one
mole of a gas that behaves ideally. CV = 12.5 J/(K.mol)
3
1
3
3
2
1
3p0
2
3p0
3p0
3
p0
p0 w
w=0
V0, 3p0
V0, 3p0
q
V0, p0
V0, p0
q
3V0, p0
Internal Energy
165
Example: Calculate U, q and w for the process depicted in the figure. Assume one
mole of a gas that behaves ideally. CV = 12.5 J/(K.mol)
No (or minimal) math answers:
3
1
3
w1-2 = 0 (no change in volume), w2-3 = -2p0V0 (constant pext)
•
q1-2 = n.Cv.(T2-T1) < 0 (constant volume, we have the Cv value)
•
T3 = T1 (P1V1 = P3V3)
•
3
2
•
1
U1-3 = 0 (ideal gas, T does not
change)
q2-3 = n.CP.(T3-T2) > 0 (but we do not have the Cp value)
2
3p0
3p0
3p0
V0, 3p0
V0, 3p0
V0, p0
q
3
p0
p0 w
V0, p0
q
3V0, p0
Internal Energy
166
Example: Calculate U, q and w for the process depicted in the figure. Assume one
mole of a gas that behaves ideally. CV = 12.5 J/(K.mol)
No (or minimal) math answers:
3
1
3
2
3
•
w1-2 = 0 (no change in volume), w2-3 = -2p0V0 (constant pext)
•
q1-2 = n.Cv.(T2-T1) < 0 (constant volume, we have the Cv value)
•
T3 = T1 (P1V1 = P3V3)
•
U1-3 = 0 (ideal gas, T does not
change)
q2-3 = n.CP.(T3-T2) > 0 (but we do not have the Cp value)
U1-3 = 0 = q1-3 + w1-3

q1-3 = -w1-3
w1-3 = w1-2 + w2-3 = -2p0V0
(we didn’t need the Cv)
q1-3 = -w1-3 = 2p0V0


Internal Energy
167
A cool result on the side…
q1-2 = n.Cv.(T2-T1)
q2-3 = n.CP.(T3-T2) = n.CP.(T1-T2)
q1-2, w1-2
1
2
q1-3 = 2p0V0
q2-3, w2-3
3
q1-2
q2-3
(T3 = T1)
q1-3
n.Cv.(T2-T1) + n.CP.(T1-T2) = 2p0V0
-n.Cv.(T1-T2) + n.CP.(T1-T2) = 2p0V0
n(CP – Cv).(T1-T2) = 2p0V0
3p0V0 = nRT1
p0V0 = nRT2
n(CP – Cv). 2p0V0 /(nR) = 2p0V0
(T1-T2) = (2p0V0 )/(nR)
n(CP – Cv). 2p0V0 /(nR) = 2p0V0
(CP – Cv) / R = 1
(CP – Cv) = R
(ideal gas)
Internal Energy
168
Example: Calculate U, q and w for the process depicted in the figure. Assume one
mole of a gas that behaves ideally. CV = 12.5 J/(K.mol)
3
1
9
2
Internal Energy
169
Example: Calculate U, q and w for the process depicted in the figure. Assume one
mole of a gas that behaves ideally. CV = 12.5 J/(K.mol)
1
3p0
3p0
3
1
9
2
V0, 3p0
3V0, 3p0
How can I increase the volume
while keeping the external
pressure constant?
2
Internal Energy
170
Example: Calculate U, q and w for the process depicted in the figure. Assume one
mole of a gas that behaves ideally. CV = 12.5 J/(K.mol)
1
3
1
3p0
3p0 w
9
2
V0, 3p0
3V0, 3p0
q
How can I increase the volume
while keeping the external
pressure constant?
U = q + w
∆
0 temperatureincreased ,
0,
0
2
Internal Energy
171
Example: Calculate U, q and w for the process depicted in the figure. Assume one
mole of a gas that behaves ideally. CV = 12.5 J/(K.mol)
1
3
1
9
2
3p0
2
3p0 w
V0, 3p0
3V0, 3p0
q
U = q + w
Easy: -6p0V0
,
.∆
But we have the Cv! What do we do?
Internal Energy
172
∆
1
U1-3 + U3-2
0 (no change in temperature, ideal gas)
2
∆
,
.∆
U3-2
U1-3
3
Careful: this is true for an ideal gas, because U depends only on temperature.
0, and we are in deep trouble…
Otherwise, ∆
U1-3 + U3-2 =
,
. ∆ (even if the process is not done at constant volume)
Internal Energy
173
Example: Calculate U, q and w for the process depicted in the figure. Assume one
mole of a gas that behaves ideally. CV = 12.5 J/(K.mol)
1
3
1
2
3p0
3p0 w
9
2
V0, 3p0
q
3V0, 3p0
U = q + w

Easy: -6p0V0
U - w = q
,
.∆

Describing the state of a system: m, V, n, T, P… Units!
Equations of state: The ideal gas, the van der Waals gas.
Types of Energy: kinetic, potential. Units.
Work: mechanical work against a constant or variable force.
Thermodynamics : definitions
Energy exchanges: work and heat.
pV work
Heat: Heat capacity.
Internal energy: molecular interpretation
Internal energy: First principle
Heat capacity
Enthalpy: Definition, Reversible processes.
174
Internal Energy
175
Monoatomic
ideal gas
U: kinetic (translational, rotational, vibrational) + potential
Ideal gas
Temperature is a measure of the average kinetic energy.
When you increase the temperature of a system, the energy you supply is “stored” in
these motions.
The more ‘places’ to store energy, the more energy you need to supply to increase
the temperature by a given amount (e.g. 1 C)  the higher Cm,v and the Cm,p

,
.∆
(if
,
is constant)
The more ways that a molecule has of taking up energy, the higher the heat capacity.
Constant Volume Heat Capacity
176
R = 8.31 J/(K.mol)
Gas
Cv (J/(K.mol))
CP (J/(K.mol))
CP-CV (J/(K.mol))
Helium (He)
12.46
20.77
8.31
Nitrogen (N2)
20.8
29.1
8.3
Methane (CH4)
27.4
35.8
.
8.4
Ethane (C2H6)
44.4
52.9
.
.
8.5
Gas
Cv (J/(K.mol))
CP (J/(K.mol))
CP-CV (J/(K.mol))
Helium (He)
1.5 R= 3/2 R
2.5 R = 5/2 R
8.31 = R
Nitrogen (N2)
2.5 R = 5/2 R
3.5 R = 7/2 R
8.3 = R
Methane (CH4)
3.3 R
4.3. R
8.4 = 1.01R
Ethane (C2H6)
5.3 R
6.3. R
.
8.5 = 1.02 R
Heat Capacity
177
At constant pressure, as you increase
the temperature, the volume will
increase pushing the piston up. That
means you are performing work, and
part of the heat from the burner is
wasted (you need more energy to
produce the same increase in
temperature).
Changes in volume are negligible for
solids and liquids.
Constant volume: All the heat you
transfer is used to increase the kinetic
energy (i.e. temperature) of the gas
From “The molecules of Life” by Kuriyan et al.
Heat Capacity
178
=
,
,

+R
,
,
For a gas that obeys pV= nRT
Liquids and solids (hard to compress)
If you are interested in a proof check any p.chem. book
Constant Volume Heat Capacity
179
Let’s get crazy..
How crazy?
Let’s look at a protein.
Lysozyme, for
example. It has 129
amino acids.
You are out of
control girl…
,
,
∆
Constant Volume Heat Capacity
At pH 4.5, the heat capacity is
about 4 kcal/mol/K
This is 16 kJ/mol/K, or about
2,000 times R
180
Constant Volume Heat Capacity
Let’s get crazy..
Let’s look at a protein.
Lysozyme, for
example. It has 129
amino acids.
181
How crazy?
You are out of
control girl…
The more ways that a molecule has of taking up energy, the higher the heat capacity.
Measuring changes in the Cp of proteins is a useful way of following denaturation and
other conformational changes.
Constant Volume Heat Capacity
182
,
From “The Molecules of Life” by J. Kuriyan et al.
Constant Volume Heat Capacity
183
This area is the amount of heat
you need to unfold one mole of
protein at constant pressure
,
,
.
This is the enthalpy of unfolding.
From “The Molecules of Life” by J. Kuriyan et al.
Describing the state of a system: m, V, n, T, P… Units!
Equations of state: The ideal gas, the van der Waals gas.
Types of Energy: kinetic, potential. Units.
Work: mechanical work against a constant or variable force.
Thermodynamics : definitions
Energy exchanges: work and heat.
pV work
Heat: Heat capacity.
Internal energy: molecular interpretation
Internal energy: First principle
Heat capacity
Enthalpy: Definition, Reversible processes.
184
A new state function: Enthalpy
185
By definition:
H = U + pV
It can be shown that the change in enthalpy of a system can be identified with the
heat transferred to it at constant pressure, and in the absence of work other than
pV*:
H = qp
This is interesting… q is a
path variable, but qp = H and
qV = U are sate functions
* This means that the system performs only pV work, and
does not do other types of work, such as electrical work.
Ok…
A new state function: Enthalpy
the change in enthalpy of a system can be
identified with the heat transferred to it at
constant pressure
186
Remember the “selfish” sign
convention:
•
negative heat means the system
releases heat to the surroundings.
H = qp
•
positive heat: the system absorbs
heat from the surroundings.
More definitions
• An exothermic process releases energy as heat to the surroundings.
• An endothermic process absorbs energy as heat from the surroundings.
A new state function: Enthalpy
the change in enthalpy of a system can be
identified with the heat transferred to it at
constant pressure
H = qp
187
Remember the “selfish” sign
convention:
• negative heat means the system
releases heat to the surroundings.
• positive heat: the system absorbs
heat from the surroundings.
H > 0
Endothermic process
H < 0
Exothermic process
•
Cooking an egg
•
mixing water and strong acids
•
Melting ice cubes
•
Combustion (burning something
•
Producing sugar by photosynthesis
in the presence of oxygen)
A new state function: Enthalpy
H > 0
Endothermic process
188
H < 0
Exothermic process
Will the final temperature be
greater or smaller than 25°C?
25°
NH4NO3 (s) + H2O(l) → NH4+(aq) + NO3- (aq)
water 25°
A new state function: Enthalpy
willincrease
willnotchange
189
willdecrease
Thereisnowaytotell
Will the final temperature be
greater or smaller than 25°C?
25°
NH4NO3 (s) + H2O(l) → NH4+(aq) + NO3- (aq)
water 25°
A new state function: Enthalpy
H > 0
Endothermic process
190
H < 0
Exothermic process
A new state function: Enthalpy
193
Assume you have 100 mL of H2O at 23°C. You now add 5 g of NH4NO3 and stir. What will be the
final temperature of the resulting solution?
Coffee cup calorimetry
Insulated coffee cup
Stirring rod
thermometer
This is a constant pressure calorimeter because
the cork lid will not allow the pressure to build up
inside the cup (that is, the pressure inside will be
the same as outside, which is the atmospheric
pressure)
•
This is a process at constant
pressure, so q = ΔH
•
The cup is insulated, so there’s no
heat being transferred from the
inside of the cup to the outside
•
The reaction is endothermic, so it
will absorb heat, which will result in
a decrease in temperature (because
the cup is insulated)
A new state function: Enthalpy
194
Assume you have 100 mL of H2O at 23°C. You now add 5 g of NH4NO3 and stir. What will be the
final temperature of the resulting solution?
Define the system clearly:
Everything inside the coffee cup.
The heat absorbed by the reaction (q1 > 0) is
taken from the solution (q2 < 0).
The cup is insulated, so there’s
no heat being transferred from
the inside of the cup to the
outside:
q =0
q 1 + q2 = 0
A new state function: Enthalpy
195
Assume you have 100 mL of H2O at 23°C. You now add 5 g of NH4NO3 and stir. What will be the
final temperature of the resulting solution?
80g/mol
The heat absorbed by the reaction (q1 > 0)
is taken from the solution (q2 < 0).
q1 + q2 = 0
How much heat does the reaction absorb?
.∆
5 80
Because the process is
at constant pressure
25.69
1.61
We have the molar value (per mol),
so we need to multiply by the
number of moles.
A new state function: Enthalpy
196
Assume you have 100 mL of H2O at 23°C. You now add 5 g of NH4NO3 and stir. What will be the
final temperature of the resulting solution?
80g/mol
The heat absorbed by the reaction (q1 > 0)
is taken from the solution (q2 < 0).
q1 + q2 = 0
How much heat does the reaction absorb?
.∆
5 80
25.69
1.61
This amount of heat comes from cooling down the solution
We already saw that:
.
∆
where m = 100 g (density of water is 1 g/mL), and c is 4.184 J.g-1.°C-1
1.61
.
∆ =0
A new state function: Enthalpy
197
Assume you have 100 mL of H2O at 23°C. You now add 5 g of NH4NO3 and stir. What will be the
final temperature of the resulting solution?
80g/mol
The heat absorbed by the reaction (q1 > 0)
is taken from the solution (q2 < 0).
q1 + q2 = 0
How much heat does the reaction absorb?
.∆
5 80
25.69
1.61
-1,610
This amount of heat comes from cooling down the solution
100 . 4.184
.
.
23
= 19.2°C
We already saw that:
.
∆
where m = 100 g (density of water is 1 g/mL), and c is 4.184 J.g-1.°C-1
1.61
.
∆ =0
Energy and Enthalpy
198
OK, that is a lot of stuff.
Let’s try a problem. We did
most of the hard work
when calculating U some
slides ago
Totally. We just need to
incorporate two new
equations:
H = U + pV
H = qp
Internal Energy
199
Example: Calculate U, q and w and H for the process depicted in the figure.
Assume one mole of a gas that behaves ideally. CV = 12.5 J/(K.mol)
1
3
1
2
3p0
3p0 w
9
2
V0, 3p0
q
3V0, 3p0
U = q + w

Easy: -6p0V0
U - w = q

,
.
,
.∆
Internal Energy
200
Example: Calculate U, q and w and H for the process depicted in the figure.
Assume one mole of a gas that behaves ideally. CV = 12.5 J/(K.mol)
1
3
1
2
3p0
3p0 w
9
2
V0, 3p0
3V0, 3p0
q
U = q + w

Easy: -6p0V0
U - w = q
H = qp 

,
.
,
.∆
Energy and Enthalpy
201
Maybe something more
challenging?
Let’s see….
H = U + pV
H = qp
Internal Energy
202
Example: Calculate U, q and w and H for the process depicted in the figure.
Assume one mole of a gas that behaves ideally. CV = 12.5 J/(K.mol)
∆
3p0
3p0
3p0
V0, 3p0
V0, 3p0
V0, p0
0 if volume does not change
3
1

2
,
.
,
.∆
How can I reduce the pressure
while keeping the volume
constant?
We need to cool down the gas.
We need to cool down the gas. T2 < T1  we are lowering the internal energy of the
gas, 
< 0, w = 0
Internal Energy
203
Example: Calculate U, q and w and H for the process depicted in the figure.
Assume one mole of a gas that behaves ideally. CV = 12.5 J/(K.mol)
∆
3p0
3p0
3p0
V0, 3p0
V0, 3p0
V0, p0
0 if volume does not change
3
1

,
.
,
.∆
H = U + pV
2
H = U + (pV) =
U + (pfVf- piVi) =
U + (p0V0- 3p0V0) =
U - 2p0V0
Energy and Enthalpy
204
Still pretty easy…
Maybe something
with liquids?
We did ALL the hard work
when calculating U some
slides ago
H = U + pV
H = qp
Energy and Enthalpy
205
Energy and Enthalpy
Initial state:
1 mol of H2O(l) at 0°C and 1 bar
Final state:
1 mol of H2O(l) at 100°C and 10 bar
Goal: calculate U and H
206
Energy and Enthalpy
207
Initial state:
1 mol of H2O(l) at 0°C and 1 bar
Final state:
1 mol of H2O(l) at 100°C and 10 bar
Goal: calculate U and H
What do you mean “choose
a path”? Can I choose
anything?
That is the beauty of
state functions like U and
H. Their changes do not
depend on the path, just
on the initial and final
states
Energy and Enthalpy
208
Initial state:
1 mol of H2O(l) at 0°C and 1 bar
Final state:
1 mol of H2O(l) at 100°C and 10 bar
Goal: calculate U and H
We know how to deal with constant pressure, constant temperature and constant
volume, so let’s imagine a path that combines reversible processes that keep one
variable constant.
Initial state:
final state:
T = 273K
T = 373K
P= 1 bar
P= 10 bar
HTotal
UTotal
Energy and Enthalpy: Dependence on p, V, and T
209
Initial state:
1 mol of H2O(l) at 0°C and 1 bar
Final state:
1 mol of H2O(l) at 100°C and 10 bar
Goal: calculate U and H
We know how to deal with constant pressure, constant temperature and constant
volume, so let’s imagine a path that combines reversible processes that keep one
variable constant.
Constant T
Constant p
Initial state:
T = 273K
P= 1 bar
H1
U1
Intermediate state:
T = 273K
P= 10 bar
H2
U2
HT = H1 +H2
UT = U1 + U2
final state:
T = 373K
P= 10 bar
Energy and Enthalpy: Dependence on p, V, and T
210
Initial state:
1 mol of H2O(l) at 0°C and 1 bar
Final state:
1 mol of H2O(l) at 100°C and 10 bar
Goal: calculate U and H
We should get the same result if hold p constant first and T constant second
Constant p
Initial state:
T = 273K
P= 1 bar
H3
U3
Constant T
Intermediate state:
T = 373K
P= 1 bar
H4
U4
HT = H3 +H4
UT = U3 + U4
final state:
T = 373K
P= 10 bar
Energy and Enthalpy: Dependence on p, V, and T
Initial state:
1 mol of H2O(l) at 0°C and 1 bar
Final state:
1 mol of H2O(l) at 100°C and 10 bar
Goal: calculate U and H
Watch the whole solution in the screencast
211