Chapter 6: Thermodynamics the First Law
• Everything in the universe happens because of
energy
• In this chapter, we’re going to look at the
transformation of energy from one form to
another. This is called Thermodynamics
• First, we’ll need to define some terms and
concepts…
Systems
• To study energy, we need to make a distinction
between the region/volume we want to study and the
world outside it.
• The System is the volume/region we are interested in
– The flask, beaker, cylinder, etc.
• The Surroundings is the volume/region immediately
around the system where heat can be observed
coming into or out of the system
Systems
•
Systems can be open, sealed or isolated
•
An Open System can take both matter and energy from the
surroundings
– Engines (fuel & electricity)
– Your body (food & heat)
•
A Closed System can only exchange energy with the surroundings
– Glow sticks
– Thermal Packs
•
An Isolated System exchanges nothing with the surroundings
– Thermos bottles (insulated walls and sample cylinder in the middle)
Work and Energy
• Work is defined as motion against an opposing force
– When we push a large block, we are doing work
against gravity
– Batteries move current through a circuit
• We can use a simple test to determine if something does
work:
– Can the process be harnessed to raise a weight?
• Electric circuit --> Motor --> raise weight
• Gas expansion --> Piston --> raise weight
• Food digestion --> Muscle --> raise weight
Work
Remember from earlier discussions:
Work = Force x Distance
1J = 1 kgm2s-2
Energy
The capacity of a system to do work is called its internal
energy, U
We cannot readily compute U for a system, but we can
measure the changes in U
If a system does 15J of work, then we know that the internal
energy of the system has decreased by 15J or:
U = - 15 J
• Always use the appropriate sign, either ‘+’ or ‘-’ when describing
energy
Provided no other changes are occurring, we can set the
change in internal energy equal to the work done by the
system as:
U = w
Types of Work
• Expansion Work: Work that changes the volume of
a system
– Car engines are an example of work done by expansion
• Non-expansion Work: Work that doesn’t involve a
change in volume
– The electron transport chain in cells is a perfect example
of nonexpansion work
Expansion Work
Lets Look at a Cylinder…
We can relate the work done by
expanding the volume of a
cylinder to the pressure by:
Work = F x d
But in expanding a cylinder against
atmospheric pressure, F=PA
w = PA(d)
But Area x distance moved is equal to V
w = PV
Now, since the system is doing work, it
is losing energy, so we must use the
correct sign convention.
w = -PV
Let’s Look at the Equation for
Expansion Work
w = -PV
What does this tell us?
1) If there is no pressure to push against, there’s no work
•
2)
This is called free expansion
Let’s check the units. Pressure is measured in Pa or 1kgm1s-2 and volume is in m3
1 kgms-2 (m3) = 1kgm2s-2 = 1 J
ALWAYS REMEMBER YOUR UNITS!!!!!!!

Special Type of Expansion: Reversible
Isothermal Expansion
If we have a piston in a water bath, the piston slowly
expands against atmospheric pressure due to the
constant input of thermal energy
We can reverse this process by removing heat from
the cylinder
Work
P, T
Heat
In this type of
setup, we have
a reversible
isothermal
expansion
Reversible Isothermal Expansion
• The work done in this type of expansion
of an ideal gas is given by :
Vfinal 
w  nRT ln 

Vinitial 


Heat
•
Heat has a special definition in Thermodynamics
Heat is the energy transferred as a result of a
temperature difference
•
Energy flows as heat from regions of high
temperature to regions of low temperature
–
•
For hundreds of years, human thought that heat was a
tangible thing like a liquid, however, despite the terms we
use it isn’t!
Let’s think about what happens when we bring a
population of high temperature gas molecules into
contact with a population of low temperature gas
molecules…
Heat
•
•
We’ll abbreviate the heat transferred to a system
as ‘q’
When the only change to the internal energy of a
system is in the form of heat:
U = q
•
Conventions:
– When heat enters the system from the surroundings,
we us a ‘+’ sign
•
This is an endothermic reaction
– When heat leaves the system and goes to the
surroundings, we use a ‘-’ sign.
•
This is an exothermic reaction
Measuring Heat
• We can use a thermometer to measure the
change in temperature caused by the transfer
of heat
– However, to convert this measurement to heat, we
need to know the relationship between change in
temperature and heat supplied
• The Heat Capacity, C, of a system is the ratio
of heat supplied and change in temperature
q
C
T
A large heat capacity means that
for a given amount of heat, you’ll
only see a small change in
temperature
Heat Capacity
• Once we know the heat capacity, C, we
can calculate the heat necessary to
raise the temperature of a substance
q = CT
• Heat capacity is an extensive property:
The more sample you have, the more
heat would be required to raise the
temperature of the sample one degree
Heat Capacity
•
Because Heat Capacity is an
extensive property, we can use 2
terms to describe it in a standardized
manner.
1. Specific Heat Capacity:
Cs = C/m = JK-1g-1
2. Molar Heat Capacity:
Cm = C/n = JK-1mol-1
Heat Capacity
q = CT = mCsT
nCmT
Calorimetry
•
Practical measurements of heat
transfers are done with calorimeters
Two types:
1) Coffee cup
calorimeter:
Constant
Pressure
2) Bomb
calorimeter:
Constant
Volume
CALORIMETRY
Measures the energy needed or
produced in a chemical reaction
A calorimeter allows the measurement
of this energy
A constant pressure calorimeter allows
a direct measurement of the
enthalpy change during a reaction
Coffee Cup Calorimeter
The reaction under study
is carried out in
solution
An exothermic reaction
causes heat to be
released into the
solution and the
solution temperature
increases
We’ll measure ΔT, so if
we know mass and Cp
we can calculate ΔH
Coffee Cup Calorimeter
System: Solution and chemicals that react
Surroundings: Cup and the world around it!
Assumptions: We use 2 cups to prevent energy
transfer to the surroundings (we assume that
it works as designed)
Expected Changes:
i) As the chemical reaction occurs, the
potential energy in the reactants will be
released as heat or the solution can supply
heat to allow formation of a product with a
higher potential energy
ii) The solution will abosorb or release energy
during the reaction. We will see this as a
temperature change
qr + qsolution = 0
Constant Pressure Calorimetry: An
Example
We place 0.05g of Mg chips in a coffee cup
calorimeter and add 100 mL of 1.0M HCl,
and observe the temperature increase
from 22.21°C to 24.46°C. What is the
ΔH for the reaction?
Mg(s) + 2HCl (aq) --> H2(g) + MgCl2(aq)
Assume: Cp of the solution = 4.20 J/gK
Density of HCl is 1.00 g/mL
Constant Pressure Calorimetry: An
Example
To solve this:
ΔT = (24.46°C – 22.21°C) = (297.61K – 295.36K)=2.25K
Mass of solution =
1g 

100ml x
  0.05g  100.05 g
ml 

Now, let’s calculate qsolution:
qsolution = mCmΔT = (100.05g)(4.20 J/gK)(2.25K)
= 945.5 J
Now, let’s calculate qr:
qr = -qsolution = -945.5 J
Constant Pressure Calorimetry:
Another Example
A piece of chromium metal weighing
24.26 g is heated in boiling water to
a temperature of 98.3°C and then
dropped into a coffee cup
calorimeter containing 82.3g of
water at 23.3°C. When thermal
equilibrium is reached, the final
temperature is 25.6°C. Calculate
the Cm of chromium.
Constant Volume Calorimetry: Using
Bombs in the Lab
• Technique can be used to obtain the
heat content of combustion of
compounds
• Used in the food, fuel and
pharmaceutical industries to know
how much energy would be released
by completely consuming the
compound
• Uses a BOMB Calorimeter
CALORIMETRY
•Place sample of known
mass inside the bomb
•Place oxygen in the
sample chamber and
immerse bomb into water
•Ignite the bomb and
measure temperature of
water
•Since the volume
doesn’t change, no P-V
work is done, so the qr is
a measurement of the ΔU
Calorimetry
Some heat from reaction warms
water
qwater = CmH2O(water mass)(∆T)
Some heat from reaction warms
“bomb”
qbomb = (heat capacity, J/K)(∆T)
Total heat evolved = qtotal = qwater + qbomb
Measuring Heats of Reaction
CALORIMETRY
Calculate energy of combustion (∆U)
of octane.
2C8H18 + 25O2 --> 16CO2 + 18 H2O
• Burn 1.00 g of octane
• Temp rises from 25.00 to 33.20 oC
• Calorimeter contains 1200. g water
• Heat capacity of bomb = 837 J/K
Measuring Heats of Reaction
CALORIMETRY
Step 1 Calc. energy transferred from reaction to
water.
q = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 J
Step 2 Calc. energy transferred from reaction to
bomb.
q = (bomb heat capacity)(∆T)
= (837 J/K)(8.20 K) = 6860 J
Step 3 Total energy evolved
41,200 J + 6860 J = 48,060 J
Energy of combustion (∆U) of 1.00 g of octane
= - 48.1 kJ
The First Law of Thermodynamics
• Up until now, we have only considered the
changes in the internal energy of a system as
functions of a single change: either work or
heat
• However, these changes rarely occur singly,
so we can describe the change in internal
energy as:
U = q + w
(The 1st Law)
• The change in internal energy is dependent
upon the work done by the system and the
heat gained or lost by the system
The First Law: Put Another Way
1st Law of Thermodynamics
A system can store energy. A change in the
energy of a system means that there must be
a change in the heat or the work done BY or
TO the system.
OR
The Total Energy of the Universe is Constant
The First Law
• Internal energy is an
example of a state
function
• A state function is a
property that only
depends on the current
state of the system and is
independent of how that
state was reached
• Pressure, Volume,
Temperature and density
are all examples of state
functions
State Functions and Things that Aren’t
• Work and heat ARE
NOT state functions
• The amount of work
done depends on how
the change was
brought about
Change in Internal Energy
(Implications of a State Function)
• It doesn’t matter what path we take to get to the
final point, the change in internal energy is only
dependent on where we started and where we
finished
• Let’s think about this on a molecular level…
– If we expand an ideal gas isothermally, the molecules will
have the same kinetic energy and will move at the same
speed
– Despite the fact that the volume has increased, the
potential energy of the system remains the same because
there are no forces between molecules (KMT)
– Since neither the kinetic nor potential energy has
changed, the change in internal energy is…
Zero!
U = 0 for the isothermal expansion of an ideal gas