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Types of Processes

A thermodynamic process is one that takes a system from one state to
another.

There are several types of processes dependent on the conditions for the
change. (We’ve already seen most with the simple gas laws.)

Isothermal process – Temperature is constant
e.g. Boyle’s law

Isobaric process – Pressure is constant
e.g. Charles’ law

Isovolumetric process – Volume is constant
e.g. Gay-Lussac’s Law

Adiabatic process – No heat transfer
PV Diagrams

Pressure vs Volume graphs are the most common
way to describe the state of an ideal gas and the
processes it undergoes.

Isobaric and isovolumetric processes are horizontal
and vertical lines respectively.

Isothermal processes are described by a strict
inverse relationship.

Recall Boyles law PV = constant when T is constant.

The closer a line is to the origin, the lower the
temperature.
Adiabatic Processes

In an adiabatic process, Q = 0. From this, the
relationship between P and V can be determined.

The resulting equation means


Its graph on a PV diagram will be similar to an
isothermal line, but steeper.

Problems similar to Bolye’s law problems can be
solved using this equation if an adiabatic process is
described.
Ex: An ideal gas initially at 105 kPa expands
adiabatically from 3.5 x 10-4 m3 to 8.9 x 10-4 m3. What
is the new pressure?
PV
5
3
 constant
Adiabatic Processes

Notice that for an adiabatic process, P, V and T change.

PV
5
3
The initial and final states of an adiabatic process are located on
different isothermal lines.

Using the ideal gas law to find the VT relationship, solve PV = nRT for
P.

Substituting this into the adiabatic equation and collecting all
constant terms. (n, R and constant)

If a gas expands adiabatically, the temperature drops.

If a gas compresses adiabatically, the temperature rises.
TV
2
3
 constant
nRT
P
V
 constant
PV Work Derived Part 1 – Piston h signs
W  F s

Work is defined as a force applied through a distance.

Recall that work done on a system is positive and work done by a
system is negative.

Therefore, the work done by an external force for a compression will be
positive and the work done by an external force for an expansion will
h=h
h=0
be negative.

Consider a piston on a canister holding a gas. The piston creates an
eternal force. Pick a height = 0 point as the initial piston location.

For an expansion, the h of the gas sample is positive (h – 0)

For a compression, the h of the gas sample is negative (-h – 0)
h=-h
PV Work Derived – Part 2 Work external
W  F s cos 

Workext for expansion is negative

Expansion, h positive Compression, h negative

Recall that Work requires the displacement to be in the same direction as
the force.

Here, our external force is 180 from the direction of h.

Cos 180 = – 1 . Therefore Wext = – F h

Wext, exp = – F h would be negative, Wext,comp = – F h would be positive.

Results as expected above. Verify this.
Workext for compression is positive
Wext   F h
PV Work Derived – Part 3 Gas Pressure

To transform our Wext expression into PV work, we first recognize
that the force exerted by the gas pressure is an equal and
opposite reaction force to the external applied force.

W done by the gas is the opposite of the Wext

To convert force into P and h into V, multiply by A/A as a
form of 1.

Sort the factors and replace by the definitions of P and V.

Ta Da. That’s why the work done by a gas is called PV work.

W for an expansion is positive and W for a compression is
negative
Wext   F h
W  F h
A
W  F h 
A
F
W  h  A
A
W  PV
PV work on a PV diagram

P is the vertical axis and V is the horizontal, so PV
corresponds to the area under the curve for a process.

If the process is not isobaric, the PV work is still the area
under the curve.

Sometimes this is easy to calculate, but for isothermals and
adiabatic processes, it requires calculus.

Area sweeps to the right are positive work (V is positive)

Area sweeps to the left are negative work (V is negative)
Work done for a PV diagram cycle

Many processes that are shown on a PV
diagram are cycles that return the state of
the system to the initial conditions.

The energy change for a cycle is zero
because energy is a state function.

Work is not a state function.

Work is the area within the loop created by
the processes that make up the cycle.

CW loop is pos. W.
CCW loop is neg. W.
First Law of Thermodynamics







When heat is added to a gas, that heat may increase the temperature of the
sample of gas or it may cause the gas to expand doing some PV work, or some
combination of the two options.
Q = U + W
This is the First Law of Thermodynamics.
Or: The change in internal energy of a closed system is equal to the amount of
heat added to the system less the work done by the system.
U = Q – W
Notice in this form, a state function is dependent on two non-state functions.
Also called the Law of Conservation of Energy.
Sign conventions for First Law

W > 0 for expansion

W< 0 for compression

Q > 0 for heat added to system (endothermic)

Q < 0 for heat removed (exothermic)

U > 0 if temperature increases.

U < 0 if temperature decreases.
W = PV
U = 3/2 nRT
U = Q – W
First Law Problems

Remember to think about the sign conventions.

Ex: 5000 J of heat are added to two moles of an ideal monatomic gas,
initially at a temperature of 500 K, while the gas performs 7500 J of work. a)
What is the change in internal energy of the gas? b)What is the final
temperature of the gas?
First Law Consequences of Processes

Isothermal process

Isobaric process W = PV nothing special

Isovolumetric process W = 0
U = Q

Adiabatic process Q = 0
U = – W

T = 0
So U=0
0 = Q – W and Q = W
Second Law of Thermodynamics

The total entropy of an isolated system can only increase over time
or remain constant.

In ideal cases where the system is in a steady state (equilibrium) or
undergoing a reversible process, there is no change in entropy.

The increase in entropy accounts for the irreversibility of natural
processes, and the asymmetry between future and past. The arrow
of time only goes one way.

Entropy is a measure of the amount of disorder in a system or a
counting of the number of possible arrangements of items.
Factors affecting Entropy

State of sample: Gases have more entropy by far than liquids,
which have more entropy than solids.

A sample with a larger number of particles will have a larger
entropy

A sample at a higher temperature will have more entropy
Calculating Entropy Change

Significant entropy change happens during phase changes.

When heat is added or removed from a system at a constant
temperature (a phase change) the change in entropy is given by:

∆𝑺 =

Symbol for entropy: S

Unit for S is J/K

Entropy is a state function – it has a value that is path independent
𝑸
𝑻
Change in entropy signs

If heat is added, Q > 0 and S is positive.

If heat is removed, Q < 0 and S is negative.

For the special cases where a process is
reversible, S = 0

It is possible to calculate a value for entropy
using statistical mechanics with a zero
entropy being a single particle at absolute
zero as a reference. S= kB log W
Boltzmann’s Grave in Vienna
Heat Engines

The purpose of a heat engine is to capture thermal energy
and transform it into mechanical energy.

Converting mechanical energy into thermal energy is easy
and is usually accomplished with friction.

From the laws of thermodynamics, we know that thermal
energy flows from hot objects to cold objects. It is this
thermal energy that a heat engine can capture.

Heat engines usually take the form of a gas within a cylinder
with a piston. The piston moves in response to state changes
of the gas. The piston can do mechanical work, thus
completing the transformation to mechanical energy.
A Heat Engine Cycle Visualized

Heat is added from a hot heat source (first image)
QH

Gas expands doing mechanical work to lift M (second image) W

The gas is cooled isovolumetrically (third image)

Heat is expelled at a lower temperature (fourth image) QC
Refrigerators

The second law states heat will not naturally flow
from a cold object to a hot object

But this transformation can be accomplished if a
refrigerator does work on the system.
Versions of the Second Law

Clausius version:

It is impossible for thermal energy to flow from a cold to a hot object
without performing work.

Kelvin version:

It is impossible, in a cyclic process, to completely convert heat into
mechanical work.

Mechanical work can be completely converted into heat.
Engine Efficiency

Engine Efficiency, , is defined as W/QH x100. That is, what
percentage of the thermal energy extracted from the heat sink can
be transformed into mechanical energy.

Any unconverted energy is vented to a cold sink as QC.

W = QH - QC

The second law requires that there is no such thing a 100% efficient
engine.

The most efficient engine possible is called the Carnot engine and
is a limit on the efficiency of any engine operating between any
two given temperatures.
The Carnot Engine

Four stages of the cycle:

12 Isothermal compression, Qc at Tc leaves

Heat expelled, Qc< 0

23 adiabatic compression, increase T

34 isothermal expansion QH at TH added

Heat added, QH > 0

41 adiabatic expansion, decrease T

W = QH – QC
Carnot Efficiency

S =Q/T for both isothermal steps. S=0 for the adiabatic steps.

Total S is zero because entropy is a state function

0 = QH/TH – QC/TC

QH/TH = QC/TC

QH/QC = TH/TC

 = W/QH = (QH – QC)/QH = 1 – QC/QH = 1 – TC/TH

 = 1 – TC/TH
QC/QH = TC/TH
Efficiency = 100% only if TC = 0 K or TH = 
Other Engines