CHAPTER 13 LEARNING OBJECTIVES - crypt

```13 MATTER: VERY SIMPLE
The gas laws
• Deduce the
relationships between
pressure, volume and
temperature of a gas
from experimental data
• Combine them to give
the Ideal Gas Law
Boyle’s Law
At a constant temperature the
pressure p and volume V of a
gas are inversely proportional
pV=constant
1627-91
What graph could we draw to test this
relationship?!?
Pressure and volume of gases increasing with temperature
Constant volume
Constant pressure
1
2
4
3
pressure p
45.1
45.1
volume V
T/C
T/C
heat gas:
pressure
increases
–273
heat gas:
volume
increases
0
temperature/C
–273
0
temperature/C
Pressure and volume extrapolate to zero at same temperature –273.16 C
So define this temperature as zero of Kelvin scale of temperature, symbol T
0
273
temperature/K
pressure proportional to Kelvin temperature
273
temperature/K
volume proportional to Kelvin temperature
pT
VT
0
Pressure and volume proportional to absolute temperature
Charles’ Law
V/T=constant
At constant pressure, the volume
V of a gas is directly proportional
to its absolute temperature T.
Pressure Law p/T=constant
At constant volume, the pressure
p of a gas is directly proportional
to its absolute temperature T.
1780’s
Absolute Temperature
Absolute temperature measures how hot an
object is on a scale starting at absolute zero
(lowest possible temperature, when all
particles have minimum possible energy).
Measured in: Kelvin, K
Scale: 1K = 1⁰C
Conversion: K = C + 273
E.g.. 0 ⁰C = 273K, 10 ⁰C = 263K, -273K = 0 ⁰C
1800’s
Testing Charles’ Law and the
pressure Law…
1. Test the laws using the experimental setup
2. Take measurements of volume or pressure
at 4/5 different temperatures (doesn’t
matter what temps as long as good range)
3. Plot a rough graph of your results.
Don’t forget to use absolute temperature…
Conversion: K = C + 273
What is a mole?
One mole has 6.023 x 1023 particles in it.
It has a mass equal to the atomic mass
in grams:
1 mole of Carbon = 12g
I mole of Hydrogen = 1g
One law for all gases
Boyle’s law
p1/V
1
2
4
3
3
2
pressure p
Combine the
relationships into one
volume V
compress gas:
pressure p increases
constant temperature T
Amount law
pressure p
2
4
3
Combine unknown N
and k into measurable
quantity R
combine:
pressure p increases
constant temperature T
pV
pT
1
2
4
3
5
0
pN/V
or
pVN
number N
Pressure law
N m–2
combine:
pN
1
1
4
105
introduce
constant k:
pVNkT
pressure p
Number of molecules
N not known
Nk can be measured:
Nk = pV/T
For one mole, define
R = NAk
For n moles:
pV = nRT
45.1
T/K
heat gas:
pressure p increases
constant volume V
combine:
pVT
Charles’ law
VT
k = Boltzmann constant
(number of molecules per mole)
R = molar gas constant
= 8.31 J K –1 mol–1
measured from pV/T for one mole
45.1
volume V
T/K
heat gas:
volume V increases
constant pressure p
When NA could be measured:
Avogadro number NA = 6.02  1023 particles mol–1
R = molar gas constant = NAk = 8.31 J K–1 mol–1
Boltzmann constant k = 1.38  10–23 J K –1 mol–1
One beautiful equation….
pV = nRT
n is the number of moles
R = molar gas constant (JK-1mol-1)
Energy the gas has
per Kelvin per mole
Boltzmann constant is derived thus:
k = R/NA
JK-1per
particle
Energy the gas has per
Kelvin per particle
pV = nNAkT = nRT
Summary….
pV = constant (if T is constant)
V/T = constant (if p is constant)
p/T = constant (if V is constant)
Combine to give the Ideal Gas law…
pV = nRT or pV = NkT
Where…
n = number of moles (amount of gas)
R = Molar gas constant (8.31J mol-1K-1)
k = Boltzmann constant (1.38x10-23JK-1)
N = number of particles
Ideal Gases
Develop problem solving skills
involving the gas laws.
Derive the relationship density
= PM/RT
A sealed aluminium alloy flask contains air at
atmospheric pressure and a temperature of 27
oC. The alloy’s melting point is 620 oC, and the
flask will burst if the pressure exceeds 2.9
atmospheres.
Will the flask melt before it bursts, or burst
before it melts?
Determining the density of air
A plastic vessel (40 cm x 40 cm x 40 cm)
containing air at atmospheric pressure is
weighed. Its mass is recorded as 371.2 g.
The air in the vessel is pumped out and the
vessel re-sealed. Its mass is now 300.3 g.
From these results, work out the density of air
in kg m-3.
ρ = mass/volume kgm-3
ρ = density
Upthrust…
Where does the force
come from that causes a
hot air balloon to rise?
Archimedes:
Any object in a fluid
displaces a volume of the
fluid that weighs more
than the object does, will
float upwards.
Bye Bye Kitty………..
A sadistic physicist decides to find out how many helium party
balloons need to be attached to a kitten to make it take off. He
knows that the upthrust force on a helium balloon is given by
the weight of air displaced by it.
Mass of kitten = 0.25 kg
Volume of a party balloon = 0.027 m3
Mass of balloon (excluding gas in it) = 0.005 kg
Air density = 1.1 kg m-3
Helium density = 0.15 kg m-3
Work out how many balloons are needed to make kitty fly!
ρ = m/V kgm-3 ρ = density
Hint…………………..
For one balloon, work out the net
upward force
by calculating the
Hint: For one balloon, work out the net
force by calculating the up thrust force
upthrustupward
force
and the weight of
and the weight of balloon plus contents. Hint:
For one balloon, work out the net upward
balloon plus
contents.
force by calculating the up thrust force and the
weight of balloon plus contents.
Derive…
ρ = pM/RT
ρ = density
p = pressure
M = molar mass
R = universal gas constant
T = Absolute Temperature
n = number of moles
m = mass
Using the following….
pV=nRT, ρ=m/V, n=m/M
Gas density
Instead, you are going to derive an equation for gas density in terms of
the intensive properties of pressure, molar mass and temperature, all
of which are easy to determine.
1.
Write down the ideal gas equation (molar form)
2.
Write an expression for n, the number of moles in terms of the mass
of gas, m, and the molar mass M.
3.
What is the standard expression for the density of a substance?
4.
Substitute equation 2 into equation 1 and rearrange to give the
correct expression for density on one side.
5.
Use your new equation to calculate the density of air at atmospheric
pressure (105 Pa) and 25 oC.
Take the molar mass of air as 28.6 g mol-1.
Kinetic theory of gases
• Derive PV = nRT from a kinetic
theory standpoint
• Derive expressions for kinetic
energy and molecular speed
from the resulting equations
Kinetic theory of gases
Law
Boyle’s Law
Pressure Law
Charles’ Law
“Amount”
Law
Relationship in Kinetic theory explanation
terms of
variables
P α 1/V
Kinetic Theory
A model that attempts to explain the gas laws.
First some assumptions (of an ideal gas)
• Gas contains large number of particles
• Molecules move randomly at random speeds.
• All collisions between wall and particles are
elastic.
• Low density so space occupied by molecules
is zero
• Energy of motion large enough so that
attractive forces can be ignored.
Momentum Recap…
p = mv
F= Δp/Δt
Impact of a ball on a wall (Qs 50S) (3mins)
A ball of mass 2 kg, moving at 12 ms–1, hits a
1. How much momentum did the ball have
before impact? 24 kgms-1
2. How much momentum does the ball have
when it has stopped, after impact? 0 kgms-1
3. How much momentum did the ball lose
during impact? 24 kgms-1
4. Assume that Newton’s third law is correct
and applies to this case. How much momentum
did the wall gain? 24 kgms-1
Force due to a stream of elastic balls
Suppose the wall is hit by a stream of 2 kg balls. In this
case each ball arrives with a speed of 12 m s–1 and
bounces straight back with an equal speed of 12 m s–1
in the opposite direction. As previously, 1000 balls
arrive at the wall in 10 s.
8. Calculate the change of momentum when
48 kg m s–1
one ball arrives at the wall and bounces away.
9. Calculate the change in momentum for all
the 1000 balls. 48000 kg m s–1
10. Calculate the average force on the wall
during that 10 s period. 4.8 kN
Deriving the ideal gas equation from kinetic theory
Consider a cuboid, dimensions x,y,z, containing a single gas
molecule of mass m travelling at speed v along the X
direction. The molecule collides elastically with wall YZ and
then travels in the opposite direction towards the other YZ
wall.
Q1. Show that the time between collisions of the molecule
with the same wall is t = 2x/v.
Q2. What is the rate at which the molecule collides with the
wall?
Q3. The molecule collides elastically with the wall and
rebounds in the opposite direction. What is its change in
momentum?
Q4. Newton’s second law can be stated as Force = rate of
to the previous questions to show that the average force on
YZ due to the molecule colliding with it is F = mv2/x .
Q5. Use the basic definition of pressure (pressure = force/area)
expression for the pressure P exerted on wall YZ by the
single gas molecule.
Kinetic model of a gas
To start: one molecule in a box
Use change of momentum
end wall of box
momentum +mv before
impulse on wall
v
y
momentum –mv after
z
x
round trip-time between collisions t = 2x /v
collisions per second = v /2x
wall has change in
momentum +2mv
ball has change in
momentum –2mv
momentum 2mv given to wall at each collision
Force = rate of change of momentum
force on wall =
momentum per collision  collisions per second
2mv
v/ 2x
t
force
time
force on wall = mv 2 / x
impulse each time molecule returns
Calculate pressure = force on wall/ area of wall
pressure p
y
force on wall = mv2/x
2
t
force on wall =
momentum per collision  collisions per second
2mv
v/ 2x
force
time
force on wall = mv 2 / x
impulse each time molecule returns
Calculate pressure = force on wall/ area of wall
force on wall = mv2/x
pressure p
y
pressure = mv2/xyz
(area = yz)
xyz = volume V
z
x
pressure p = mv 2 /V
area of wall = yz
add many molecules all doing the same
(V = xyz)
improve model
force
N times as many collisions per second
pressure p = Nmv2/V
N molecules
allow molecules to move in random directions
time
improve model
force
time
1/3 as many collisions per second
Deriving the ideal gas equation from kinetic theory
Now we need to extend the model to a more realistic situation where many
gas molecules are colliding with the walls.
Q1. If there are N molecules in the gas, how many, on average, will be travelling in
the x direction?
to the previous question, show that the pressure due to all molecules is given
by P = Nmv2/3V .
Q3. We normally write the equation from the previous question as
PV = 1/3 Nmv2, where we now allow the molecules to travel at a range of
speeds characterised by a mean square average speed. Can you see any
similarities with the experimental ideal gas law?
Q4. Write down an expression for the kinetic energy of a gas of N molecules of
mass m.
Q5. Combine the expression from the previous 2 questions to show that the
kinetic energy of the gas is given by KE = 3/2 PV.
Q6. The KE of a substance is proportional to its temperature T. Use this to show
that PV α T as in the experimental ideal gas law.
Q7. Show that, for molar quantities, KE = 3/2 RT = 1/2 Mv2, where M is molar
mass.
Q8. Use the equation from the previous question to estimate the speed of nitrogen
molecules in the room around you.
N times as many collisions per second
pressure p = Nmv2/V
N molecules
allow molecules to move in random directions
time
improve model
force
time
1/3 of molecules
in each direction,
on average
allow molecules to move at random speeds
1/3 as many collisions per second
pressure p =
1
3
Nmv2/V
improve model
force
time
average impulse stays the same
take average
over v2
The kinetic theory of gases predicts that pV = 13 Nmv 2
1
pressure p = 3 Nmv2/V
p = 1/3
From the 3
dimensions of
the box
2
Nmv /
V
Average of square
of speeds of
molecules
v1
Mass of one
molecule
v3
v7
v2
v4
v6
v5
Compare….
pV =
2
1/3Nm
&
pV = NkT
Average KE of a molecule = 3/2kT
Total KE of N molecules = 3/2NkT
Kinetic energy of one mole of molecules
U = 3/2RT
r.m.s speed,

(Root Mean Square Speed)
mean speed
most probable
speed
root mean
square speed
0.10
0.08
0.06
0.04
0.02
0.00
speed in metre per second
A way of averaging
speeds.
The distribution of
speeds of
molecules in
nitrogen at 300 K is
shown here.
r.m.s speed,

Remember…
Average KE of a molecule = 3/2kT
So… 1/2m 2 = 3/2kT
Average of square of speeds of molecules
Therefore….
2
=
3

And …. vr.m.s =
3
=

3

Kinetic Theory
(Part 2)
Develop problem solving skills
associated with kinetic theory of
gases, including the calculation
of r.m.s speeds.
Starter:
Estimate the speed of air
molecules in this room
Speed of a nitrogen molecule
Assume warm room temperature T = 300 K
mass of 1 mole of N2 = 28  10–3 kg mol–1
kinetic energy of a molecule
Avogadro constant NA = 6  1023 particles mol–1
from dynamics
Boltzmann constant k = 1.38  10–23 J K–1
mass m of N2 molecule
1
2
from kinetic model
3 kT
v2 = m
mv2
calculate speed
mass of 1 mole of N2
m=
3  1.4  10–23 J K–1  300 K
v =
4.7  10–26 kg
28  10–3 kg mol–1
m=
6  1023 mol–1
v2 = 2.7  105 J kg–1 [same as (m s–1)2]
m = 4.7  10–26 kg
2
v = 500 m s–1 approximately
Air molecules (mostly nitrogen) at room temperature go as fast as bullets
3
2
kT
Diffusion of gases and liquids
• Observe and explain diffusion effects in liquids and
gases
• Derive and apply the Einstein equation for diffusion
Evidence for m oving molecule s
Brom ine expand ing in to a vacuum
remove air from
tube using vacuum
pump
attach bromine
capsule in
sealed tube:
break capsule
open tap:
bromine
instantly fills
whole tube
bromine
to vacuum
pump
tap closed
air in tube
tap open
vacuum
bromine in tube
Brom in e diffu sing in to air
bromine very
diffuses up
the tube
attach bromine
capsule in
sealed tube:
break capsule
bromine
tap closed
air in tube
tap open
bromine diffuses into tube
Diffusion shows that molecules move. Rapid diffusion into a vacuum demonstrates high molecular speeds
Random walk in 1 dimension
Suppose a molecule can either move one step to the right (R) or one step to the left (L) as a result
of collisions with neighbouring molecules. There is a 50 % chance of stepping to the right as a
result of a collision, and a 50 % chance of moving to the left. Assume the distance moved in a
step is always the same. Consider a molecule that starts in a certain place and then experiences
10 collisions.
Q1. How many different outcomes are there for a sequence of 10 steps?
(Hint: one such outcome is LRLLLRRLRL)
Q2. What is the maximum distance from the start, measured in number of steps,
that a molecule could end up after 10 collisions? How likely is this outcome?
Q3. Consider a molecule that ends up at 8R after 10 steps. How many different ways
are there of reaching this position in 10 steps? Enumerate them.
Q4. What about a molecule that ends up at 6R after 10 steps? How many different
ways are there of reaching this position in 10 steps?
Q5. Which is the most likely place for a molecule to end up after 10 steps? Explain your answer.
Q5. Can you sketch a graph to show the distribution of molecules about the starting position after 10
steps? Explain the shape of the graph you have drawn.
Random walk distribution after 10 steps
300
250
200
n
150
100
50
0
10L
8L
6L
4L
2L
0L
2R
4R
6R
8R
10R
Gaussian or Normal distribution with standard deviation = √N = √10 = 3.2
In general, if N steps are taken, estimated distance diffused D = √N x d, where d is
the “step length” or mean free path .
Diffusion of perfume across a room
How far do molecules of perfume diffuse in 2
seconds?
Take the rms speed of the molecules
as 500 ms-1, and the mean free path
(step length or average distance between
collisions) as 10-7 m.
Hint:
Work out the distance gone in 2 seconds, and
then work out how many collisions are
experienced in travelling this distance.
A sixth form student is still recovering from a big night out and
randomly walks around not knowing where he is going.
Sixth form centre
Estimate how many
steps will it take for
a student to
randomly walk from
common room to
this lab
Lab
Diffusion of photons through the radiative zone of the Sun
Photons that are created during
fusion in the core of the Sun can
take a very long time to reach the
surface and be emitted.
This is because they undergo a vast
number of collisions as they are
scattered by protons, electrons
and other particles in the
estimate how long it takes a photon
to traverse the radiative zone as it
undergoes many randomising
collisions.
Diffusion in the Sun
Data:
Depth of radiative zone: 109 m
Mean free path of photon: 0.01 m (distance between successive
collisions)
Speed of photon: 3 x 108 ms-1
1 year = 3 x 107 seconds
Q1. Using the random walk model, how long does it take a photon to
Q2. Will more photons take a greater or less time to transit the radiative
zone than the value calculated in Q1?
Q3. What factors will affect the mean free path of a photon, and how will
they affect it?
Q4. Predict and explain the effect of increasing the mean free path on
the time taken to transit the radiative zone.
Internal energy and specific thermal capacity
• Explain and use the
equation E = mc
• Explain the consequences
of the anomalous STC of
water
• Explain the pattern in molar
STC values for metals
• Apply the First Law to
thermodynamic problems
Starter: What is the kinetic energy of 1 mole of an ideal gas
at 300 K? Use R = 8.3 J K-1 mol-1.
How much energy would you need to supply to raise the
temperature of 1 mole of an ideal gas by 1 degree K?
Compare….
pV =
2
1/3Nm
&
pV = NkT
Average KE of a molecule = 3/2kT
Total KE of N molecules = 3/2NkT
Kinetic energy of one mole of molecules
U = 3/2RT
Molar Specific Thermal Capacity
U = 3/2 RT (internal energy)
Molar STC = dU/dT = 3/2 R
“The molar specific thermal capacity (C) of a
substance is the amount of energy needed to
raise the temperature of 1 mol of substance by
1K (1⁰C)”
Unit: J mol-1 K-1
Specific Thermal Capacity
“The specific thermal capacity (c) of a
substance is the amount of energy needed to
raise the temperature of 1kg of substance by
1K (1⁰C)”
Unit: Jkg-1K-1
How much energy to have a bath?
What do we need to know?
- Mass of water
- STC of water
- Temp change
ΔE=mcΔθ
(at constant volume only)
STC of metals
Metal
Aluminium
Copper
Iron
Mercury
Magnesium
Silver
Platinum
Gold
Molar mass
(kg mol-1)
Specific thermal Specific thermal
capacity
capacity
(J kg-1 K-1)
(J mol-1 K-1)
900
390
130
450
140
1020
230
130
130
Complete the table and comment on the result.
Do you see any connection with the molar STC values for monatomic gases?
E = mc
Q1. The STC of water is 4200 J kg-1 K-1, much higher than the STC values
of most other materials (see p111). Convert this STC into molar units
and compare it with the molar STC of ethanol, 112 J K-1 mol-1.
Q2. The anomalous STC of water can explain many phenomena. Do the
following exercise to predict and explain which way the wind blows
at the sea side on a sunny day.
(a) Solar radiation of intensity 1000 W m-2 is incident on a 1 m3 “block”
of sea water for 1 hour. Calculate the temperature rise in the sea
water, assuming all the energy is absorbed. Density of water = 1000
kg m-3.
(b) Consider a 1 m3 block of land rock. Calculate the corresponding
temperature rise in this block of rock, assuming the same intensity
and heating time. Density of rock = 2700 kg m-3, STC of rock = 880 J
kg-1 K-1.
E = mc
(c) If the air above the land and the sea is heated
efficiently by thermal contact with those bodies, can
you explain the origin of breezes coming off the sea
during the day?
(d) Sail powered fishing boats used to leave harbour at
night. Explain why.
(e) Can you explain why, even in the absence of the Gulf
Stream, the British Isles would have a more temperate
climate than other countries on the same latitude such
STARTER
Q1. Here are the specific thermal capacities for some
gases at 300 K, in units of J g-1 K-1. Use their molar
masses to convert them into units of J mol-1 K-1.
Ar 0.31
02 0.66
I2 0.145
N2 0.74
CO 0.74
Q2. Can you see any patterns in the results?
Q3. Can you explain the trends/patterns you see?
Hint: What is the STC for an ideal gas (with
translational KE only)?
The First Law of Thermodynamics
• State and explain the First Law of Thermodynamics
• Use the First Law to solve problems where gases do work
First law of Thermodynamics
Energy cannot be created destroyed. However
it can be transferred from one place to
another.
ΔU = W + Q
ΔU – Change in internal energy of a material
W – work done on material
Q – Energy transferred thermally
The First Law of Thermodynamics
In what follows, we will consider ideal gases, but the First Law applies to all substances.
Q1. Energy can be supplied to a gas in two ways. Describe them.
Q2. Write down the equation that summarises the First Law, identifying the terms clearly. What property of
the gas gives an indication of its internal energy?
Q3. If a gas expands and does work against its surroundings, and no heat energy is lost or supplied, does the
gas lose or gain internal energy?
Q4. If heat energy is lost from a gas and it does no work, what can you say about its final internal energy?
Q5. For the following changes, predict and explain whether the final temperature of the gas will be higher or
lower than at the start:
(a) 2000 J of work are done on the gas and 3000 J of thermal energy are supplied.
(b) The gas does 2000 J of work and 1500 J of thermal energy are supplied to it.
(c) 4000 J of work are done on the gas and 3000 J of thermal energy are supplied to it.
Q6. (a) What is the internal energy of 100 moles of argon gas at 25 oC? (R=8.3 JK-1mol-1)
(b) What volume would the gas occupy at 100 000 Pa pressure?
(c) The gas is allowed to push back a piston of area 0.25 m2 over a distance of 15 cm. Calculate the work
done by the gas. (Hint: Work = force x distance. Ignore any pressure change.)
(d) If 4000 J of heat energy are lost by the gas during the expansion, calculate its final internal energy
and hence its final temperature.
The first law
An ideal gas at 27 oC is contained in a cubic container of side length 50 cm. The gas
pressure is 1x 105 Pa.
Q1. Calculate the number of molecules of gas in the container. (k=1.4 x 10-23 J K-1)
Q2. Estimate the internal energy of the gas in the container.
Q3. The container is compressed on all sides until each side has shortened in length by
10 %. Calculate the new pressure, assuming that the compression happens isothermally
(no change in temperature).
Q4. Compressing the gas involves doing work. Where does this energy go if the
compression of the gas is being carried out isothermally?
Q5. Estimate the work done on the gas when it is compressed.
Hint: Work = force x distance = pressure x volume change
Q6. If the container is instead thermally isolated when the compression takes place,
estimate the final temperature of the gas.
Otto Engine...
Determining the temperature of a
Bunsen flame
• Devise and carry out an
experiment to determine
the temperature of a
roaring Bunsen flame
• Consider critically random
measurement uncertainties
and systematic errors
Starter: A blacksmith heats a 1.1 kg iron horseshoe in a forge. When the shoe is very
hot, he plunges it into a tub of 15 kg of cold water at 10 oC. The temperature of the
water is observed to rise by 10 oC. To what temperature had the horseshoe been
heated?
STC of water= 4200 J kg-1 K-1
ΔE = m c Δ Θ
STC of iron = 450 J kg-1 K-1
What assumptions do you need to make?
ΔE = m c Δ Θ
A blacksmith heats a 1.1 kg iron horseshoe in
a forge. When the shoe is very hot, he
plunges it into a tub of 15 kg of cold water at
10 oC. The temperature of the water is
observed to rise by 10 oC. To what
heated?
STC of water= 4200 J kg-1 K-1
STC of iron = 450 J kg-1 K-1
Uncertainty and Error
Random uncertainties
• Identify sources of
measurement
uncertainty and
quantify them.
• Which contributes the
most to the
uncertainty in the final