Pressure on the wall with surface area
A
:
P =
F
A
=
N m v
V
2 x
=
1
3
N m v 2
V
L
PV =
2
3
N (
1
2 m v 2 ) =
2
3
N K
The average kinetic energy of a molecule
The ideal gas law (experimental fact!)
K =
1
2 m v 2
PV = N kT
Therefore: NkT =
2
3
N K K =
3
2 kT

We assumed that the molecules did not change their motion between their collisions with the right and left walls.
L
In reality, the molecules collide with each other and may also collide with upper/lower and front/back walls.
What do we do about it?

Collisions do not change the situation as long they are elastic…
Colliding particles exchange momenta and velocities.
Collision with a wall – the velocity components along the wall stay the same.
The situation after a collision is not different from if molecules would just go through each other.

Kinetic energy is the only form of molecular energy that is important and it is preserved in the collision events.

Physical meaning of the absolute temperature – it is a measure of the average kinetic energy of gas molecules.
K =
1
2 m v 2
=
3
2 kT
The average speed of a molecule – thermal speed: v th
=
3 kT m

Probability to find molecular speed between 400 and 401 m/s vs.
between 1000 and 1001 m/s.

PV = nRT
1. The molecules occupy a significant fraction of the volume.
Collisions are more frequent.
There is less volume available for molecular motion.
2. There are long range attractive – Van der Waals forces between the molecules, which become more important as the density grows.
Introduces corrections to the ideal gas law to take into account some of these effects.
n
V
2
2 a
( P + n
V
2
2 a
)( V !
nb ) =
Attractive force between couples of molecules.
Goes as
( n / V ) 2 square of the concentration.
V !
nb Less volume available for motion, because of n moles of the gas.
nRT
T = 0 !
V = nb P = anything

We start from a piece of ice at -30 °C, supply heat at a constant rate and monitor the temperature…
Straight segments with constant slopes – warming up of ice, water and steam.
The slopes are inversely proportional to their respective specific heats.
Horizontal lines – no temperature change over long time intervals.
Phase transitions – melting and vaporization.

It takes some amount of energy to melt and vaporize substances.
Those energies per unit mass of the materials are called heat of fusion
L f and heat of vaporization,
L v
.
Q = Lm L =
Q m
Since no temperature change occurs during the phase transitions, those energies are sometimes called latent heats of fusion/vaporization.

Q = Lm L =
Q m
Positive energy is required to melt/vaporize. The same amount of energy, however, is released when the substance solidifies/condenses.

Paraffin compress – I hope you never had one applied to you!
Typical melting temperature is from 49 to 71 °C (120 to 160 °F).
So, if you melt it, wrap it in a cloth and apply to yourself (or your kid) it is going to stay at its melting temperature, around 140 °F until its heat of fusion is transferred to your body…
Q = L f m L f
!

So, if you melt it, wrap it in a cloth and apply to yourself (or your kid) it is going to stay at its melting temperature, around 60 ºC (140 ºF) till its heat of fusion is transferred to your body…
Q = L f m L f
!
For water cooling down:
Q = cm !
T c = 4 .
2 kJ/kg !
° C
To give away (reject) the same amount of heat, 1 kg of water must cool down by ~35 ºC, from 60 ºC to 25 ºC (140 to 80 ºF).

Heat of vaporization of water is huge, L v
= compared with its specific heat of c =
,
It takes a kettle 5 minutes to heat 1 kg of water from the room temperature of 20 °C to the boiling temperature of 100 °C.
How much time it would take the same kettle to evaporate the same amount (1 kg) of water at the same rate of energy supply
(neglecting all energy losses)?
E h
Energy to heat to 100 °C is
= mc
T = 1 kg
4 .
2 kJ kg ° C
( 100 ° C
20 ° C ) = 336 kJ
Energy to vaporize the water is
E v
= mL v
= 1 kg !
2257 kJ kg
= 2257 kJ
So the total evaporation will take t v
= t h
!
E v
E h
= 5 min !
2257 kJ
336 kJ
= 33 min