PES 2130 Fall 2014, Spendier
Lecture 3/Page 1
Lecture today: Chapter 18 and 19: Thermodynamics and gases
1) Heat Flow processes
2) Ideal gas law
3) Work done by gases
Announcements:
- HW 1 due next lecture (Monday)
Last lecture:
Heat can have three different effects on objects:
1) Thermal Expansion.
2) Change in temperature.
3) Change in phase.
(Latent heat of sublimation Ls = Lf + Lv, sublimation is the transition of a substance
directly from solid to the gas phase.)
Heat Flow Processes
When we talked about the Zeroth law of thermodynamics, we talked about temperature
being transferred from a hotter to a colder object. This method of heat transfer is called
conduction (similar to electric conduction)
In general, heat can move from one object to another in one of three main processes:
1) Conduction, 2) Convection, and 3) Radiation
1) Conduction:
Conduction occurs between bodies in contact. In this case heat is transferred by
transfer of kinetic energy via collisions.
Some materials are better "thermal conductors" as others. Good thermal conductors have
a high value of thermal conductivity.
Don’t need to know stuff in red box! Did not have time to cover in class.
Rate of heat transfer, conduction rate Pcond, for a slab of face area A and thickness L,
whose faces are maintained at temperatures TH and TC by a hot reservoir and a cold
reservoir is:
Pcond 
T  TC
Q
 kA H
t
L
Thermal conductivity constant: k = [W/(Km)]
Some examples for k:
steel:
14 W K-1 m-1
copper:
401 W K-1 m-1
glass:
1.0 W K-1 m-1
wood:
0.12-0.04 W K-1 m-1
Just as metals are the best electrical conductors, they are also the best thermal conductors.
PES 2130 Fall 2014, Spendier
Lecture 3/Page 2
DEMO 1 (did not have time): Time it takes to melt ice on block of wood and metal at
room temperature.
2) Convection
Convection (convective heat transfer) is the transfer of heat from one place to
another by the movement of fluids.
When you look at the flame of a candle or a match, you are watching thermal energy
being transported upward by convection. Such energy transfer occurs when a fluid, such
as air or water, comes in contact with an object whose temperature is higher than that of
the fluid. The temperature of the part of the fluid that is in contact with the hot object
increases, and (in most cases) that fluid expands and thus becomes less dense. Because
this expanded fluid is now lighter than the surrounding cooler fluid, buoyant forces (we
will talk bout this force at the end of the semester) cause it to rise. Convection is also in
action when you boil water (Create so called convection currents.)
3) Radiation
Radiation is the transfer of heat by electromagnetic waves, such as visible light or
infrared. (Transfer of heat through vacuum.)
Don’t need to know stuff in red box! Did not have time to cover in class.
The rate Prad at which an object emits energy via electromagnetic radiation:
Prad   AT 4
depends on the object’s surface area A and the temperature T of that area in kelvins. Here
σ = 5.6704 x 10-8 W/m2 K4 is called the Stefan–Boltzmann constant
The symbol ε represents the emissivity of the object’s surface, which has a value
between 0 and 1, depending on the composition of the surface. The emissivity of the
surface of a material is its effectiveness in emitting energy as thermal radiation. A
surface with the maximum emissivity of 1.0 is said to be a blackbody radiator, but such a
surface is an ideal limit and does not occur in nature. For example the value of ε is larger
for a black shirt than a white shirt.
The rate Pabs at which an object absorbs energy via thermal radiation:
Pabs   ATenv 4
from its environment, which we take to be at uniform temperature Tenv (in kelvins).
Note: A black shirt will absorb more sun heat than a white shirt
Because an object will radiate energy to the environment while it absorbs energy from the
environment, the object’s net rate Pnet of energy exchange due to thermal radiation is
Pnet  Pabs  Prad   A Tenv 4  T 4 
Example using all three heat transfer mechanism: Thermos Flask
Vacuum between metal layers stops conduction and convection, and silver metal surface
inside reflects radiative heat back inside the flask.
PES 2130 Fall 2014, Spendier
Lecture 3/Page 3
(Chapter 19 - note we will come back and cover the rest of chapter 18 later)
1) How can we keep track of energy as it is transferred from one system to another?
2) How can we calculate the amount of internal energy - a quantity that seems to be
hidden within the very "guts" of matter?
3) What is the difference between temperature and heat, and between heat and work?
We will now focus on the first of two central thermodynamic principles: the
conservation of energy, or, as it is sometimes called, the first law of thermodynamics.
The second basic principle, which deals with the inevitable increase of a quantity called
entropy, is the subject of chapter 20, Second Law and Entropy. These two abstract
principles, plus a few other concepts, some of which we talked about in lecture 2, are the
entire content of thermodynamics.
In today's lecture, our approach is macroscopic - that is, we shall deal with systems that
are approximately of human scale in size and mass (thermometers, blocks of ice, heat
engines).
observable quantities: pressure, volume, and temperature to describe the behavior of
these systems.
Next lecture, we will talk about the microscopic approach, which regards the behavior
of the atoms and molecules as fundamental.
observable quantities: molecular velocities, energies, and momenta
The macroscopic approach should be seen as supplementary to the microscopic approach.
Values for macroscopic observables are derived from the microscopic picture.
Thermodynamics process:
Is a process where a system changes thermodynamic state (pressure, volume,
temperature) slowly so that at all times it is in a state of thermodynamic equilibrium.
Changes in the thermodynamic state can occur through expansion or contraction, or by
adding or taking away heat (by contact with another substance at a different temperature).
Work Done During Volume Changes
The work done by a system can be understood in one of two ways:
- On a microscopic level the atoms bounce off of the walls of the container (or the piston
as in the case of the figure), and either
- Lose kinetic energy by doing positive work on the piston (volume increase) or
- Gain kinetic energy by doing negative work on the piston (volume decrease).
PES 2130 Fall 2014, Spendier
Lecture 3/Page 4
Pressure:
When the molecules in a gas or liquid collide with its container, they exert a force on the
container. Collisions can cause an outward force over the entirety of the container’s
surface area.
Pressure (p)= “average” force (F) per area (A)
p=F/A unit: N/m2 = Pa (Pascal)
Other pressure units: atmospheres (atm), millimeters of mercury (mmHg = torr), and
pounds per square inch (psi)
1 atm = 101300 Pa = 101.3 kPa = 760 mm Hg = 14.7 psi
THE GAS LAWS
1) Avogadro's number: How many atoms or molecules are in a sample?
A mole is a number like a dozen, just a lot bigger. A dozen means 12;
1 mole = 6.022 x 1023.
That number, NA = 6.022 x 1023, is known as Avogadro's number after Italian scientist
Amedeo Avogadro (1776–1856), who suggested that all gases occupying
the same volume under the same conditions of temperature and pressure contain
the same number of atoms or molecules.
1 mole = number of atoms in a 12 g sample of carbon-12 (6 protons and 6 neutrons)
We use Avogadro's number to determine how many atoms or molecules there are in a gas
sample.
The number of moles n contained in a sample of any substance is equal to the ratio of the
number of molecules N in the sample to the number of molecules NA in 1 mol:
n = N/NA
We will see that, the mass of a molecule is not important in determining its
macroscopic behavior! (Ideal gas Law).
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Lecture 3/Page 5
2) Ideal Gases:
–Non-interacting molecules (interactions between molecules can be ignored and all
collisions are elastic)
–Point particles (volume of molecules is small compared to total volume)
3) The Ideal Gas Law:
Our goal is to explain the macroscopic properties of a gas—such as its pressure and its
temperature—in terms of the behavior of the molecules that make it up. However, there is
an immediate problem: which gas? Should it be hydrogen, oxygen, or methane, or
perhaps uranium hexafluoride? They are all different. Experimenters have found, though,
that if we confine 1 mol samples of various gases in boxes of identical volume and hold
the gases at the same temperature, then their measured pressures are almost the same, and
at lower densities the differences tend to disappear. Further experiments show that, at low
enough densities, all real gases tend to obey the relation
pV = NkBT
N = number of molecules
kB = 1.38 × 10−23 J/K = Boltzmann’s constant
If forced to use moles: n = number of moles: N = n × NA
pV = nRT
R= kB × NA = 8.31 J/mole · K = gas constant
Use whichever version is convenient!
Also note that:
The molar mass M of 1 mol is the product of the mass m of one molecule and the number
of molecules NA in 1 mol.
M = mNA
PES 2130 Fall 2014, Spendier
Lecture 3/Page 6
Example 1:
The air pressure in Manitou Springs is 815 mb (millibar) and the air pressure on the top
of Pikes Peak is 612 mb. 1 mb of pressure is equal to exactly 100 Pa.
Assuming that each breath of air you take fills your lungs with the same volume of gas,
and air has the same composition everywhere, what is the reduction in oxygen intake
from Manitou to Pikes Peak? Assume temperature is the same at both places.
Notice:
1) If we know 3 thermodynamic quantities (p,V, N, or T) then we automatically know the
fourth.
2) Both the left and right hand sides of the ideal gas law have units of energy!
- kBT is a thermal energy and appears in many equations (microscopic behavior)
- By changing the volume of a gas by ∆V, work is done by/on the gas
W = p∆V
DEMO (Ideal gas apparatus):
The Ideal Gas Law and describes a relationship between the pressure, temperature and
volume of a gas that is constant. One can verify this relationship using air and this unique
apparatus. Twist the handle to compress the air, and simply read the volume, temperature
and pressure from the scales. Repeat and show that the relationship stays the same
PES 2130 Fall 2014, Spendier
Lecture 3/Page 7
WORK DONE BY AN IDEAL GAS AT CONSTANT TEMPERATURE
As we mentioned when gases expand, they do work. When they contract they have work
done to them.
Volume changes occur when the gas exerts a large enough force on its container to move
it by some distance.
Here we have two forces:
Fgas : force exerted by the gas due to its collisions with the piston.
and Fext : external force due to air if piston is not connected to anything
When the gas expands it does work. Here d is the distance
traveled by the piston. A is cross-sectional area of piston.
For a constant force: W = Fd
 Fgas 
Wgas  
dA is an incremental change in volume
  dA
 A 
Wgas   p  V  Work done by gas if pressure is constant
So what happens if the pressure is not constant? Then we need to integrate and the work
done by the gas, Wgas is the area under the p-V diagram.
Here the pressure and volume are both
changing in the transformation from
thermodynamic state 1 (initial state) to
thermodynamic state 2 (final state).
Therefore the total work done by the gas is
an integral that is not always easy to solve.
Vf
Wgas 
 pdV > 0 (expansion, increasing volume)
Vi
The amount of work done does depend on the path taken from the initial thermodynamic
state to the final thermodynamic state. The work done by the expanding gas depends on
the path of the expansion.
If the piston compresses the gas back to state 1, then the work is done on the gas:
Vf
Wgas 
 pdV
Vi
< 0 (compression, decreasing volume)
PES 2130 Fall 2014, Spendier
Lecture 3/Page 8
Some examples for constant temperature ("isothermal expansion"):
a) Constant pressure volume expansion ("isobaric process")
Wa 
Vf
Vf
Vi
Vi
 pdV  p  dV  p(V
f
 Vi )
b) Constant volume change in pressure ("isochoric or isovolumetric process")
Wb = 0
c) Both p & V change along the whole path: first constant pressure then constant volume
Wc  Wa  Wb  p(V f  Vi )  0  p(V f  Vi )
d) Both p & V change along the whole path (curved path, constant T)
Wd 
Vf
Vf
Vi
Vi
 pdV  
Vf
nRT
dV
V
dV  nRT 
 nRT ln V V f  nRT ln V f  ln Vi 
i
V
V
Vi
V f 
Wd  nRT ln  
 Vi 
Both paths for processes described in c) and d) may end start and stop at the same
thermodynamic states. However, the area underneath the p-V diagram in c) is larger than
in d). Hence the gas in process c) does more work than the gas in process d).