Chapter 11
Thermodynamics
11-1 Heat and Work and Internal Energy
o Heat = Work and therefore can be converted back and forth
o Work ----- heat
if work is done on something, it can generate heat.
• Increasing the KE of the molecules causes movement that can be
turned into work.
Thermodynamic Systems
 Thermodynamics: Study of the effects of work, heat and energy
on a system
 System – region under study which is separated by a boundary
(real or imagined)
Work, Heat, and Matter can go either way across the boundary.
Example: flask, balloon, water, and steam that was heated.
 Surroundings - environment outside the system.
Example: the classroom
Steam does Work on the
Balloon
The steam’s internal energy
has decreased
Recall from Chapter 5 that
W=F x d
units: (N x m)
Remember that Presure = F/A and
Volume = area x displacement
So, W = P x ∆V
Work = pressure x change in volume
Units:
(N/m2 x m3)
(N x m) or
Joules (J)
Work done on (+) or by (-) a Gas
Question: The cross-sectional area of the
piston is 0.20 m2. A 400.0 N weight pushes
the piston down 0.15 m and compresses
the gas in the cylinder. How much pressure
is exerted on the gas?
Formula: P= F/A
Answer: 2.0 x 103 Pa
Question: How much did the gas volume
decrease?
Formula: W = P ●∆V
Answer: 0.030 m3
3 Types of Systems
“Isolated”
“Closed”
“Open”
Internal Energy (U) =
The Energy contained in the system is
∆U = Q + W
Internal Energy = heat + work)
I am defining work by:
 Work done on the system is positive + (gains energy)
 Work done by the system is negative - (loses energy)
Question: What type of “system” is the
balloon and flask?
Answer: Closed System
1. Heat is added to the system
2. Work is done by the system (-)
W = P X ∆V
 As pressure inside the balloon increases –
volume increases.
 When heat is removed, the SYSTEM returns
to its original state of thermodynamic
equilibrium
Example: A balloon is placed in a vacuum chamber
and the air is removed. What kind of system is this?
Answer: (isolated system)
1. Work is done BY the system (expansion)
2. As pressure on the outside of the the balloon
decreases, it is allowed to expand
(adiabatically). The volume of the balloon
increases.
3. According to the combined gas law:
Then,
T1 > T2
4. When pressure is restored then Work is Done
ON the System (compression) T2 > T1
5. This increases the internal temperature in the
balloon and it returns to its original state of
thermodynamic equilibrium
Example 1: A system undergoes a process in which it
absorbs 800 J of heat while doing 500 J of work.
o the system is doing the work so W = neg
What is the system’s change in internal energy?
∆U = Q + W
Q = 800 J (endothermic +) and W = -500 J
∆U = 300 J
Example 2: A cylinder with a movable piston at one
end is filled with gas. 450 J of heat are removed from
the system, while 175 J of work is done on the system.
What is the change in internal energy of the system?
∆U = Q + W
Q = -450 (exothermic) and W = 175 J
∆U= -275 J
Notes 11.2 Thermodynamic Processes
Internal Energy (U) = Heat (Q) + Work (W)
5 Main Thermodynamic Processes
1. Isobaric process = constant pressure
2. isometric process = constant volume (also called isovolumetric
no work done
3. Isothermal = constant temperature
internal energy constant
4. Isentropic process = constant entropy
5. Adiabatic process = no change in energy (no ∆Q)
energy is not transferred to or from the system as heat
An Isobaric Process – constant pressure
Draw graph
Isometric (Isovolumetric)
Constant volume – no work done.
No work can be done, so ∆U = Q
Example: heart valves, calorimeters, Pressure cooker
Draw Graph
In this case, a system is heated or
cooled and the gas would expand
or contract if it could but the rigid
container keeps the volume
constant.
Isothermal Process – constant temperature
Internal energy remains unchanged
∆U = 0
∆T=0
Isothermal process – constant temperature
Draw Diagram
Adiabatic - work is done on or by the system but no
energy is transferred to or from the system as heat.
Q=0
 No heat gained or lost by the system
11-2 Thermodynamic Processes
1st Law of Thermodynamics – energy conservation
a systems internal energy can be changed by transferring
energy as either work, heat, or a combination of the two.
∆U = ∆Q + ∆W
ENERGY DIAGRAM
Qin
Win
Internal
Energy
Qout
Wout
A Cyclic Process
 A thermodynamic process in which a system returns to the same
conditions under which it started.
Final internal energy = initial internal energy
 The change in internal energy of a system is ZERO in a cyclic process
∆U=0
Machines use heat to do work.
A refrigerator is an example of a cyclic process
 A refrigerator performs work to create a temperature difference between its
closed interior and its environment. Refrigeration systems are heat engines,
except running in reverse.
 The basic principle used for refrigeration is: Expansion of a gas, reduces its
temperature.
 Transfers energy from a body at low temperature to one at a high temperature.
Draw Energy Diagram
Cyclic Process - Refrigerator
Heat Engines use heat to do work
 A heat engine takes heat from a hot body, converts a part of it
into work and rejects the rest to a cold body.
 Uses Heat to do Mechanical work – opposite of a refrigerator.
 There are two types of heat engines: external combustion
(Steam Engine) engines and internal combustion engines (gasoline
engines and diesel engines)
Draw Energy Diagram
Steam Engines – external combustion
Larger the difference in temperature the more work done
 Example of a heat pump
2nd law of Thermodynamics - Heat flows naturally
from a hot object to a cold object; heat will not flow
spontaneously from a cold object to a hot object
11-3 Efficiency of Heat Engines
2nd Law of Thermodynamics
 Greater amount of work is gained by taking in more energy at a higher
temperature and giving up less energy at a lower temperature.
 2nd Law of Thermodynamics – no cyclic process that converts heat
entirely into work is possible.
W ≠ Qh in a cyclic process
 Some energy must always be transferred as heat to the system’s
surroundings.
 A measure of how well an engine operates is given by the engines’
efficiency (eff) - Useful energy
Efficiency = net work done by engine/ energy added to engine as heat
Found by 2 methods
Eff = Wout /Qin
Eff = Qin – Qout
Qin
Can be expressed as a decimal or a %
Qin = hot
Qout = cold
Nuclear power plants use controlled fission to produce energy
(heat) then convert it to electricity.
Generator converts mechanical energy into electrical energy
11-4 Entropy
SECOND LAW ( entropy )
 The total entropy of any isolated thermodynamic system
tends to increase over time.
 Increasing disorder reduces the energy available to do work
∆S=Q/T
Entropy ∆ S = the measure of disorder within a System.
 The Universe will become more disordered no matter what you
do, and ENTROPY will increase.
To Change the Entropy in a System - Add or remove heat
energy (Q).
So if,
∆S = - , then Q is negative – (heat energy is lost)
∆S = +, then Q is positive (heat energy is added)
Phase changes – major changes in entropy – isothermal
Melting = Q +, S = +
Freezing = Q - , S = Evaporating = (adding heat energy to the system) = Q+ , S+
Condensing (removing heat from the system ) = Q - , S -
The system’s entropy increases.
The environment (room, beaker,
test tube etc. has less entropy
The 3rd LAW of THERMODYNAMICS
No Thermodynamic process can ever be 100% efficient.
This would require that ALL the HEAT becomes WORK (no Qout), and
that the exhaust would have to be at ABSOLUTE ZERO – impossible!
The first law of thermodynamics = conservation of energy.
U is the internal energy of the gas in question. Q is the heat added to
the gas, and W is the work done on the gas when the gas is
compressed.
The signs of each term in the first law are extremely important. If
heat is added to the gas, Q is positive, and if heat is taken from the
gas, Q is negative. If the gas's internal energy (and, thus, its
temperature) goes up, then ΔU is positive; if the gas's temperature
goes down, then ΔU is negative.
The sign of W is more difficult to conceptualize. Positive W
represents work done on a gas. However, if a gas's volume expands,
then work was not done on the gas, work was done by the gas. In
the case of work done by a gas, W is given a negative sign.