Transcript Ideal gases

Thermal Physics
IB Physics
Topic 3: Ideal gases
Ideal Gases
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Understand and Apply the
following.
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Pressure.
Gas Laws (by name)
PV = nRT
Kinetic Molecular Theory
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Explain Pressure
WilliamThompson
(Lord Kelvin)
Pressure
Click on Me
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Pressure is defined as force
per unit area
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Newtons per square metre or
N/m2
1 Nm-2 = 1 Pa (pascal)
The weight of the person is
the force applied to the bed
and the small area of each
nail tip combines to make an
overall large area.
The result is a small enough
pressure which does not
puncture the person.
F
P
A
Atmospheric Pressure
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Basically weight of atmosphere!
Air molecules are colliding with you right now!
Pressure = 1.013 x105 N/m2 = 14.7 lbs/in2!
Example: Sphere w/ r = 0.1 m
Demo
A = 4 p r2 = .125 m2
F = 12,000 Newtons (over 2,500 lbs)!
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21
Qualitative Demonstration of
Pressure
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Force due to molecules of fluid colliding with container.
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Force α Impulse = p
Average Pressure = F / A
y
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p y  mv y 
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average vertical force  f y 
 
t
t
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Pressure
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Pressure is defined as force per unit area
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The pressure exerted by a gas results from the atoms/
molecules “bumping” into the container walls
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Newtons per square metre N/m2
More atoms gives more bumps per second and higher pressure
Higher temperature means faster atoms and gives more bumps
per second and higher pressure
At sea level and 20°C, normal atmospheric pressure is
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1atm ≈ 1 x 105 N/m2
Gases
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Gases (as we will see) can behave
near perfectly.
NA= 6.02 x 1023 molecules mol-1
Molecules size ~ 10-8 m to 10-10 m
Example: How molecules are there in 6
grams of hydrogen gas?
 We have 3 moles, H2 has 2 grams mol-1
 3 x 6.02 x 1023 = 1.81 x 1024 molecules.
Example
Make a rough estimate of the number of water
molecules in an ordinary glass of water.
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A glass contains about 0.3 L of water, which has a mass
of about 300 g. Since the molar mass of water (H2O) is
18 g mol-1
Hence, 300g/18g mol-1 ~ 17 mol ~ 1025 molecules
The Boyle-Mariotte Law
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Gases (at constant temperature)
decrease in volume with increasing
pressure.
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P =F/A
V = πr2 h
Figure 17-3
Increasing Pressure by Decreasing Volume
The Boyle-Mariotte Law
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Gases (at constant
temperature) decrease
in volume with
increasing pressure.
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Isothermal Process
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PV = constant
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P1V1 = P2V2
Example
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The pressure of gas is 2 atm and its volume 0.9 L
if the pressure is increased to 6 atm at constant
temperature, what is the new volume?
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Answer:
P1V1 = P2V2
2 x 0.9 = 6 x V; hence V = 0.3 L
The volume-temperature law
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Charles & Gay-Lussac
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Isobaric Process
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V/T = constant
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V1 / T1 = V2 / T2
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At absolute zero a gas would take up
zero volume. In reality they liquefy
when they get really cold!
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Gases (at constant
volume) increase in
temperature with
increasing pressure.
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Isochoric Process
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P/T= constant
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P1/T1 = P2/T2
pressure
The pressure-temperature Law
-200 -100 0 100
temp. °C
200
Example
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A bottle of hair spray is
filled to a pressure of
1 atm at 20°C
What is the canister
pressure if it is placed
into boiling water, and
allowed to reach thermal
equilibrium?
P1/ T1 = P2/ T2
or
p1 T2 = p2 T1
1 / 293 = p2 / 373
p2 = 373/293
p2 = 1.27 atm
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Ideal gas has zero volume
Resistance of metal drops to
zero (actually superconductivity
cuts in above 0K)
Brownian motion ceases!
(kinetic energy = 3/2 kT)
-273 °C
But lowest temperature
attained is ≈ 10-9K
pressure
Absolute zero
0
temp. °C
Equations of state
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State, identifies whether solid liquid or gas
Key parameters or state variables
Volume, V (m3)
 Pressure, p (N/m2)
 Temperature, T (K)
 Mass, M (kg) or number of moles, n
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Equation of state relates V, p , T, m or n
Bulk vs molecules
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Consider force between two
molecules
At absolute zero
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Above absolute zero
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No thermal energy
Molecules sit at r0
Some thermal energy
binding
Molecules are at r> r0 (thermalenergy
expansion)
At high temperature
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Thermal energy > binding
energy
Molecules form a gas
r0
repulsion
attraction
r
thermal energy
Equation of state for a gas
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All gases behave nearly the same
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pV = nRT
R = 8.3 J/(mol K) for all gases (as long as they remain a
gas)
 T is in K!!!!!!
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Example
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What is the mass of a
cubic metre of air?
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Molecular weigh of air ≈
32g
pV = nRT
Atmospheric pressure = 105 N/m2
Atmospheric temp. = 300K
For a volume of 1 m3
n = pV/RT = 105 / (8.3 x 300)
= 40 moles
M = 40 x 0.032 = 1.3kg
Constant mass of gas
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For a fixed amount of gas, its mass or number of
moles remains the same
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pV/T = nR = constant
Comparing the same gas under different conditions
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p1V1/T1 = p2V2/T2
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Hence can use pressure of a constant volume of gas to define
temperature (works even if gas is impure - since all gases the same)
Must use T in K!!!!!!
Example
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A hot air balloon has a
volume of 150m3
If heated from 20°C to
60°C how much lighter
does it get?
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Molecular weight of air
≈32g
pV/T = nR
n = pV/RT
Balloon has constant volume and
constant pressure
ncool =105x150 / (8.3 x293) = 61680
nhot =105x150 / (8.3 x333) = 54271
n = 7409 moles
M = 7409 x 0.032 = 237kg
Molecules have finite size
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Cannot reduce volume of gas to zero!
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When you try, it becomes a liquid
Slightly increases the measured volume
Atoms/ molecules always attract each other
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Slightly reduces the measured pressure
Van de Waals forces
p-V diagrams (for gases)
Useful to consider the
pressure/volume
changes at constant
temperature
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Isotherms are p-V
values for a fixed
amount of gas at
constant volume
p a 1/V
Increasing
temperature
Pressure
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volume
Kinetic theory of gases
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A gas consists of a large number
of molecules.
Molecules move randomly with a range of
speeds. (Maxwell's Distribution)
The Volume of the molecule is negligible
compared with the volume of the gas itself.
Collisions are elastic (KE conserved)
No inter-molecular forces.
Collision time with walls are very smal.
Molecules obey Newton’s Laws of Mechanics
Molecules in a gas
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Gas atoms/molecules move in a
straight line
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velocity due to thermal energy
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KE = 1/2 m vx2 ≈ 3/2 kT
KEavg α absolute temp
RMS – (Root mean squared)
Pressure results from collisions
with the walls of the container.
(NOT collisions between
molecules
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Fimpact = Δp/t = (2m vx)/t
Example
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How fast does a typical gas molecule (travel at
room temperature? Lets take Nitrogen-14!
(k = 1.38x10-23J/K)
KE = 1/2 mv2 = 3/2 kT
v = (3kT/m)1/2
v = [(3)(1.38x10-23 x 293/(4.65x10-26)]1/2
v = 511 m/sec
Example
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If it takes 2 mins for
your kettle to begin
boiling how much
longer does it take to
boil dry?
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Assume kettle is 3kW
Starting temp of water
20°C
Work done by kettle = power x time
= 2 x 60 x 3000 = 360 000J
= Work to boil water of mass M
= T x M x cwater
360 000J = 80 x M x 4190
Mass of water = 1.07kg
Energy to boil water = M x Lv (water)
= 1.07 x 2256 x103 = 2420 000J
Time required = Energy /power
= 2420 000/3000 = 808 s ≈ 13mins