Doc Brown's
Chemistry KS4 science GCSE/IGCSE/AS Revision Notes
Part 2 More
advanced topics on the ideal gas
laws
Various calculations
including molecular mass determination, partial pressures and Graham's law of
diffusion, kinetic particle model–theory
and ideal and non–ideal real gas behaviour (Sections
4c to 4e)
The ideal gas equation PV = nRT, ideal
gas theory, how to determine the relative molecular mass Mr of a volatile liquid,
Dalton's Law of partial pressures, ideal gas behaviour and non–ideal gas
behaviour, Graham's Law of diffusion, Van der Waals equation of state,
compressibility factors, critical pressure, critical temperature. This
page is well above GCSE level but useful for A Level chemistry or A Level
physics.
The basic particle theory and properties of gases,
liquids and solids, state changes & solutions are described on
GCSE/IGCSE notes on particle models of gases–liquids–solids,
describing and explaining their properties and advanced students should be familiar with ALL its contents
before studying this page ....
AND PVT gas law calculations
are on a separate page
Sub–index for Part 2:
Introduction–the kinetic particle
theory of an ideal gas * Kelvin scale of
temperature * 4a. Boyle's Law * 4b.
Charles's–Gay Lussac's Law and the combined gas law
equation * 4c. The ideal gas equation PV=nRT * 4d. Dalton's Law of partial pressures
* 4e. Graham's Law of diffusion * Section 5. Non–ideal real gas behaviour and Van der Waals Equation:
5a. The deviations of a gases from ideal
behaviour and their causes * 5b. The Van der Waals equation of state
* 5c Compressibility factors * 5d
The
Critical Point – The Critical Temperature and Critical Pressure *
For other calculations see
the Calculations Index page
including
Mole definition and
Avogadro Constant
AND molar gas volume
and reacting gas volume ratios,
studying these helps to follow this page too.
The Maxwell
Boltzmann distribution of particle kinetic energies is discussed in the
KINETICS pages.
Doc Brown's
chemistry revision notes: basic school chemistry science GCSE chemistry, IGCSE chemistry, O level
& ~US grades 8, 9 and 10 school science courses or equivalent for ~1416 year old
science students for national examinations in chemistry
4c.
The
Ideal Gas Equation of State pV = nRT AND determining molecular mass of a
volatile liquid

The most 'compact molar' form of all the P–V–T
equations is known as the ideal gas equation and is the simplest possible
example of an 'equation of state' for gases (see also
Van der Waals equation in
section 5b). The explanation of the use of the word 'ideal'
is explained in the 4. Introduction
(separate page).

The equation combines both Boyle's law
and Charles law plus moles of gas involved.

The equation, known as the
ideal gas equation, is
given as:

p = pressure in
pascals (unit Pa)

V = volume in
cubic metres (m^{3})

n = moles of gas
(mol = mass in g / molecular mass of gas M_{r})

R =
ideal gas constant = 8.314 joules per kelvin per mol (J K^{1}
mol^{1})

T = temperature
in kelvin (K)

You must
convert to
these units for a correct calculation using pV = nRT

Make sure can do all the
rearrangements!

V = nRT /p,
T = pV/nR, p = nRT/V, n =
pV/RT and R = pV/nT

The equation is pV = nRT and requires a consistent
set of units, so see below for a comparison of the two most common examples, and take care!,
and SI units are pretty standard now and
my calculation examples primarily use SI
units (but I have left in a few examples in 'old' nonSI units).

pressure p 
volume V 
n = mass g/M_{r} 
Ideal gas constant R
and its units 
temperature T 
Pa Pascal
760mmHg = 1 atm =
101325 Pa
Pa = kPa x 1000
Pa = MPa x 10^{6} 
m^{3}
1m^{3} = 10^{6} cm^{3}
so m^{3} = cm^{3}/10^{6}
or dm^{3}/1000 = m^{3} 
mol
mol = mass (g)/M_{r} 
8.314
J mol^{–1} K^{–1} 
K
(Kelvin =
^{o}C + 273) 
atm atmospheres
760
mmHg = 1 atm
=
101325 Pa 
litre or dm^{3}
1 litre = 1 dm^{3}
= 10^{3} cm^{3}
dm^{3} = cm^{3}/1000 
mol
mol = mass (g)/M_{r} 
0.08206
atm dm^{3} mol^{–1} K^{–1} 
K
(Kelvin =
^{o}C + 273) 

The first set are becoming the 'norm' since
they are the SI units, but the mass does not have to be in kg and can be in
the more 'practical unit' of g as long as M_{r} is in g mol^{–1}.

Examples
of PV = nRT calculations (all calculations assume ideal gas
behaviour)

Ex.
Q4c.1

(a) Describe with the aid of a diagram a
simple gas syringe method for determining the molecular mass of a volatile
liquid.


An emptied 100 cm^{3} gas syringe is mounted in
an oven (ideally thermostated) but a humble bulb will do and the temperature
is quite stable after an initial warming up period via internal convection.

Some of the liquid (whose M_{r} is toe determined), is sucked into a
fine 'hypodermic' syringe (e.g. 0.2cm^{3}) and the syringe
weighed.

Quickly (to avoid evaporation losses), the liquid is injected into
the gas syringe via a self–sealing rubber septum cap and the syringe
re–weighed immediately.

The difference in weighings gives the mass of liquid
injected.

When the gas volume has settled to its maximum value the volume is
read (to the nearest 0.5cm^{3} if possible),

Then note the oven
temperature and barometric pressure (mercury barometer for best accuracy i.e.
in mmHg).

(b) In an experiment using the above
apparatus the following data were recorded and the molecular mass of the
volatile liquid calculated.

Mass of syringe + liquid = 10.6403 g

Mass of syringe after injection of liquid =
10.4227g

When volatilised the liquid gave 67.3 cm^{3}
of gas.

The temperature of the oven = 81^{o}C,
barometric pressure 752 mmHg.

Using the equation PV = nRT, calculate the
molecular mass of the liquid.
 R = 8.314 J mol^{–1} K^{–1}

Mass of liquid injected = 10.6405 –
10.4227 = 0.2176 g

p = 101325 x 752/760 =
100258
Pa, (converting pressure from mmHg to Pa)

V = 67.3/10^{6} =
6.73 x 10^{–5}
m^{3}, T = 273 + 81 = 354 K

PV = nRT, substituting for moles
n
gives PV = m/M_{r}RT

and then rearranging gives ...

M_{r} = mRT/PV

= (0.2176 x 8.314
x 354)/(100258 x 6.73 x 10^{–5}) =
94.9 (95 to 2sf)

(c) If the compound was formed from the
reaction of bromine and a hydrocarbon, suggest a possible molecular formula
for the compound.

(d) State very briefly, a method of
determining the molecular mass of ANY compound that can be vapourised intact.

Ex.
Q4c.2

Ex.
Q4c.3

(a) A 5 litre container
contained 0.5kg of butane gas (C_{4}H_{10}).

Assuming
ideal gas behaviour calculate the pressure of the gas if the cylinder is
stored at 25^{o}C.

M_{r}(C_{4}H_{10})
= (4 x 12) + 10 = 58, 0.5kg = 500g

moles of gas n = 500/58 =
8.621, T = 273 + 25 = 298K

R = 8.314 J mol^{–1}
K^{–1} or 0.08206 atm litre mol^{–1} K^{–1}

(i) Using SI units to calculate the gas
pressure

PV = nRT, P = nRT/V, 5
litre = 5 dm^{3} = 5 x 10^{–3} m^{3}

P = 8.621 x 8.314 x 298/(5
x 10^{–3})

P = 4271830 Pa = 4272
kPa = 4.27 MPa (3
sf)

Ex. Q4c.4

A 100 dm^{3} (100 litre) cylinder of
oxygen gas exerts a pressure of 900 kPa at 20^{o}C.

Calculate the mass of oxygen in the cylinder in
kg.

R = 8.314 J K^{1} mol^{1},
V = 100/1000 = 0.1 m^{3}, p = 900 000 Pa, T = 20 + 273 =
293 K

pV = nRT, n = pV/RT

n mol = pV/RT = (900 000 x 0.1)/(8.314 x
293) = 36.95 mol

mol = mass/M_{r,} M_{r}(O_{2})
= 2 x 16 = 32

mass = mol x M_{r} = 36.95 x 32 =
1182 g = 1.18 kg
(3sf)

Ex. Q4c.5

A cylinder of ethane gas has a volume of 1600 dm^{3}.

The safe limit of storage pressure for the
cylinder is 1.2 MPa.

If the cylinder contains 20 kg of ethane gas,
what is the highest safest storage temperature in ^{o}C?

pV = nRT, T = pV/nR, M_{r}(C_{2}H_{6})
= (2 x 12) + 6 = 30

pressure p = 1.2 x 10^{6} Pa

volume V = 1600/1000 = 1.6
m^{3}

moles n = 20 000/30 = 666.7, R = 8.314 J K^{1} mol^{1}

temperature T = (1.2 x 10^{6} x 1.6)/(666.7 x
8.314) = 346 K (3 sf)

T = 346  273 =
73^{o}C.
(3 sf)

For other gas
calculations see Mole definition and
Avogadro Constant, molar gas volume
and reacting gas volume ratios.
4d. Dalton's law of partial pressures

Dalton's Law of partial
pressures states that at constant temperature the total pressure exerted by a
mixture of gases in a definite volume is equal to the sum of the individual
pressures which each gas would exert if it alone occupied the same total
volume.

For a mixture of gases
1, 2, 3 ... p_{tot} = p_{1} + p_{2} + p_{3}
... where p_{1}, p_{2} etc. represent the partial pressures.

The partial pressure ratio
is the same as the % by volume ratio and the same as the mole ratio of gases
in the mixture.

This means for a component
gas z:

Examples of partial pressure calculations

Ex. Q4d.1

In the
manufacture of ammonia a mixture of nitrogen : hydrogen in a 1 : 3 ratio is
passed over an iron/iron oxide catalyst at high temperature and high pressure.

What are the partial pressures of nitrogen and hydrogen if the total pressure
of the gases is 200 atm prior to reaction?

The 1 : 3, N_{2}
:H_{2} ratio means that nitrogen forms ^{1}/_{4} of
the mixture, therefore

p_{N2} = ^{1}/_{4} x
200 = 50 atm and

p_{H2} = p_{tot}
– p_{N2} =
150 atm (or from ^{3}/_{4} x 200)

Ex. Q4d.2

Methanol can be
synthesised by combining carbon monoxide and hydrogen in a 1 : 2 ratio.

In an experimental reactor
experiment, 300^{o}C at a total pressure of 400kPa, the final
equilibrium gaseous mixture contained 10% carbon monoxide.

(a) Calculate the % of
hydrogen gas and % methanol vapour in the final mixture.

Whatever hydrogen is left,
its % must be double that of carbon monoxide since they were both mixed and
react in a 1 : 2 ratio, so there will 20% hydrogen in the equilibrium
mixture.

Therefore there will be
100 – 10 – 20 = 70% methanol in the final mixture.

(b) Calculate the partial
pressures of the three gases in the mixture.

(c) Calculate the value of
the equilibrium constant, K_{p}, under these reaction conditions
(use Pa pressure units).

–
TOP OF PAGE
4e. Graham's Law of Diffusion

Diffusion, or the
'self–spreading' of molecules, naturally arises out of their constant chaotic
movement of particles in all directions, though on a time average basis, more molecules
will move in the direction of a region of lower concentration down a diffusion
gradient if such a
situation exists e.g. initially 'pouring bromine vapour into air' in gas jars
(see GCSE notes).

Molecules of differing
molecular mass diffuse at different rates.

The smaller the molecular mass, the
greater the average speed of the molecules at constant temperature.

The greater the average speed of the
particles the greater their rate of diffusion.

See notes on the
Maxwell–Boltzmann distribution of molecular velocities.

This conceptually explains Graham's Law
of diffusion, explained below.

It has been shown that,
assuming ideal gas behaviour and constant temperature, the relative rate of
diffusion of a gas through porous materials or a mixture of gases or a tiny hole
(effusion) is inversely
proportional to the square root of its density.

Since the density of an
ideal gas is proportional to its molecular mass, the relative rate of
diffusion of a gas is also inversely proportional to the square root of its
molecular mass*.

* PV = nRT, PV = m/M_{r}RT,
M_{r} = mRT/PV, since d = m/V, then M_{r} is proportional to
density.

r_{1} 

√d_{2} 

√M_{2} 
––– 
= 
–––– 
= 
–––– 
r_{2} 

√d_{1} 

√M_{1} 
 Which is the mathematical ratio
representation of Graham's law of diffusion for comparing two gases of
different molecular masses.

Graham's Law arises from
the fact that the average kinetic energy** of gas particles is a constant for
all gases at the same temperature.

**The formula for kinetic
energy is KE = ^{1}/_{2}mu^{2}, where m = mass
of particle, u = velocity. This means the average mu^{2}
is a constant for constant kinetic energy, so u is proportional to 1/√m and the m can be shown
via the Avogadro Constant to
be proportional to M_{r}, the molecular mass of the gas.

You have to think of the molecules
'hitting' the space of the pore or tiny hole and passing through the. The
greater the speed the more chance the particle has of passing through this
'porous space'.

Examples of diffusion rate calculations

Ex. Q4e.1


Two cotton wool plugs are
separately soaked in concentrated aqueous ammonia and hydrochloric acid
solutions respectively and sealed in a long tube with rubber bungs.

Using a simple chemical
equation and Graham's Law of diffusion, account for (a) the appearance of a
'white smoke ring' and (b) the fact the smoke ring occurs about ^{2}/_{3}rds
along from the ammonia end of the tube.

(a) The aqueous
ammonia will give off ammonia fumes and the conc. hydrochloric acid gives off
hydrogen chloride fumes which will diffuse down the tube towards each other.
When the meet an acid base reaction gives fine crystals of the salt ammonium
chloride.

(b) M_{r}(NH_{3})
= 17, M_{r}(HCl) = 36.5

If r is the relative rate
of diffusion the following ratio applies,

r_{NH3} 

√M_{r}HCl
= √36.5 

6.04 
–––––––– 
= 
–––––––––––––––––––– 
= 
–––––––– 
r_{HCl} 

√M_{r}NH_{3}
= √17 

4.12 

and this shows that
ammonia will diffuse about 50% faster than hydrogen chloride so the smoke
ring will appear much nearer the HCl end of the tube.

Ex. Q4e.2

Zeolites are silicate
minerals that are porous at the molecular level and they are used as catalysts
and 'molecular sieves' in the petrochemical industry in processes such as
cracking and subsequent molecule separation.

(a) Calculate the relative
rates of diffusion of pentane CH_{3}CH_{2}CH_{2}CH_{2}CH_{3},
hexane CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}CH_{3}
and 2–methylpentane (CH_{3})_{2}CH_{2}CH_{2}CH_{2}CH_{3}
into a zeolite mineral.

Atomic masses: C = 12, H =
1

Hexane and 2–methylpentane
are structural isomers of C_{6}H_{14} with the same molecular
mass.

r_{C5H12} 

√M_{r}C_{6}H_{14}
= √86 

9.27 
––––––––– 
= 
––––––––––––––––––– 
= 
––––
= 1.09 
r_{C6H14} 

√M_{r}C_{5}H_{12}
= √72 

8.49 
 Relative rate of diffusion is 1.09 : 1.00 for
pentane : hexane/2–methylpentane

(b) In practice
2–methylpentane does not diffuse into the zeolite as fast as hexane or maybe
not at all. Suggest a reason for this behaviour.

Ex. Q4e.3

Enriching uranium means to
increase the relative ratio of ^{235}U/^{238}U to produce
uranium metal suitable for use as fuel rods in nuclear reactors. It is the
^{235}U isotope that is very fissile (readily undergoes fission) but
only occurs as a small % in uranium ores in which most uranium is the
non–fissile ^{238}U. To produce 'enriched' uranium metal it is first
extracted by reduction from uranium oxide and then converted into gaseous
uranium(VI) fluoride (uranium hexafluoride). The ^{235}UF_{6}
is concentrated by a diffusion process in huge gas centrifuges before being
converted back to uranium metal. Atomic mass of F = 19

(a) Calculate the relative
rates of diffusion of the hexafluorides of the two uranium isotopes.

r_{235UF6} 

√M_{r}^{238}UF_{6}
= √352 

18.76 
––––––––– 
= 
––––––––––––––––––– 
= 
–––––
= 1.004 
r_{238UF6} 

√M_{r}^{235}UF_{6}
= √349 

18.68 

(b) Suggest why the
process must be repeated many times before enough enrichment has occurred.
5. Non–ideal real gas behaviour and Van der Waals Equation
5a. The deviations of
a gases from ideal
behaviour and their causes

Certain postulates in the
kinetic theory of gases (see
section 4.) are far from true in real gases, particularly at
higher pressures and at lower temperatures.

This can be clearly seen
in the diagram on the right.

If the gases conformed to
the ideal gas law equation PV=nRT, the product PV should be constant with increasing
pressure at constant temperature, clearly this is not the case.

It can also be seen
that the greatest deviation from ideal behaviour always tend to occur at
higher pressures (right diagram) and often at lower temperatures
(see the compressibility factor diagram in 5c) and
both positive and negative deviation occur.

Several points in the
theoretical kinetic particle model cannot be ignored in 'real gases'.

The actual
volume of the
molecules (V_{molecules}) is significant at high pressures i.e. the free space for
random particle movement (V_{ideal}) is less than it appears from volume measurements.

V_{real} = V_{ideal}
+ V_{molecules}

At very high pressures
therefore, the value of PV becomes greater than the ideal value and
presumably outweighs the intermolecular force of attraction factor which would
tend to increase the closer the molecules are and decrease P (see forces
arguments next).

The deviation from ideal
gas behaviour due to the molecular volume factor will generally increase with
(i) the greater the pressure and (ii) the larger the volume of the molecule
(~increasing M_{r}).

Intermolecular
forces
always exist i.e. instantaneous dipole – induced dipole forces (Van der Waals
forces) between ANY molecules and at high pressures the molecules are closer
together and so attraction is more likely to occur. As a particle hits the
container side there is an imbalance of the intermolecular forces which act in
all directions in the bulk of the gas. Just as the particle is about to hit
the surface there will be a net greater attraction towards the bulk of the gas
as the molecule, so reducing its impact force i.e. reduces its 'ideal'
pressure (p_{ideal})
by an amount (p_{reduction}).

p_{real} = p_{ideal}
– p_{reduction}

At lower temperatures
when the KE of the molecules are at their lower values, the intermolecular
forces can have more of an effect in reducing P, so the PV value is less
than the ideal value. The effect becomes less as the temperature increases
(see graph in 5c.) and also as the pressure becomes
much higher when the molecule volume factor outweighs the intermolecular force
factor.

These intermolecular
forces will increase the bigger the molecule (~increasing number of electrons) and the more polar the molecule
where permanent dipole – permanent dipole forces can operate in addition to
the instantaneous dipole – induced dipole forces.

Also, the lower the
temperature, the kinetic energies are lower so its more likely that neighbouring molecules can
affect each other. The reduction in p_{ideal} also increases
with increasing pressure too, since the molecules will be on average closer
together.

There is direct
experimental evidence for the effects of intermolecular forces in gases from
adiabatic expansion or compression situations. Adiabatic means to effect a
change in a system fast enough to avoid heat transfer to or from the
surroundings. e.g.

(i) If a gas at high
pressure is suddenly released through a small nozzle it rapidly cools on
expansion into the lower pressure zone. The reason for the cooling is that
in order to expand the intermolecular forces must be overcome by energy
absorption, an endothermic process. The change is so rapid that the source
of heat energy can only come from the kinetic energy of the gas molecules
themselves, so the gas rapidly cools. This is observed when a carbon dioxide
fire extinguisher is used, just for a second bits of solid CO_{2}
can be seen, which rapidly vaporise. However, it proves that the gas was
rapidly cooled from room temperature to –78^{o}C!

(ii) When you rapidly
pump air into a bicycle tyre the gas warms up because the molecules are
forced closer together so the intermolecular forces can operate more
strongly, this, just like bond formation, is always an exothermic process.

Therefore generally
speaking for any gas the lower its pressure and the higher its temperature,
the more closely it will be 'ideal', i.e. closely obey the ideal gas equation PV=nRT
etc. Also the smaller the molecular mass or the weaker the intermolecular
forces, the gas will be closer to ideal behaviour.

However, for any gas at
a particular P and T, its all a question of what factor outweighs the others.

Note that both positive
and negative deviation from ideal gas behaviour can occur and there will be
situations where the different causes of non–ideal behaviour cancel each other out.

Check out the graphs
at the start of 5a. and
5c.

The measurement and
predictions of gas behaviour is very important in industrial processes and so
many mathematical developments have been devised to accurately describe the
real behaviour of gases. The Van der Waals equation
is one of the earliest and simplest equations to model real gas behaviour.
TOP OF PAGE
5b.
The Van der Waals equation of state

Equations such as the Van
der Waals equation for real non–ideal gases attempt to take into account the volume occupied by the
molecules and the intermolecular forces between them. The idea is to
incorporate 'corrective' terms to reproduce or model real gas P–V–T behaviour
with a modified equation of state.

The Van der Waals equation
for one mole of gas can be most simply stated in (i) as

(i) (p + a') (V – b') = RT

The term a' represents the
extra pressure the gas would exert if it behaved ideally. In real gases the
intermolecular forces are imbalanced at the point of impact on the container
wall, with a net attraction in the direction of the bulk of the gas. In the
bulk of the gas, each molecules is subjected to the same 'time averaged'
attractions in all directions, but heading for the container wall it is
considered to be 'dragged back a bit' by attraction with the bulk of the gas
surrounding it on all sides bar the surface of impact, which is therefore
reduced in force. (see also intermolecular forces
discussion in 5a.)

The term b' represents the
volume that the molecules occupy, so V–b' represents the actual volume of free
space the molecules can move in. (see also
molecule
volume discussion in 5a.)

For n moles of gas the Van
der Waals equation is ...

(ii) [p + (an^{2}/V^{2})] (V
– nb) =
nRT

a and b are the Van der
Waal equation constants.

The factor n^{2}/V^{2} is
related to the gas density, the more dense the gas (i.e. moles/volume), at
higher pressures, the more intense will be
the intermolecular attractive force field effects.

Dividing through by n,
using the (V – nb) term, gives the alternative version ...

(iii) [p + (an^{2}/V^{2})] [(V/n)
– b)]
= RT

(iv) p = [nRT/(V – nb)] – (an^{2}/V^{2})

(v) p = [RT/(V/n – b)] – (an^{2}/V^{2})

For 1 mole of gas the equation
simplifies to

A selection of a and b
Van der Waal's constants are given below.

Data 
Van der Waals constants 
critical values of the gas 
Gas 
a (Pa m^{6} mol^{–2}) 
b (m^{3} mol^{–1}) 
pressure p_{c} (Pa) 
temp.
T_{c} (K) 
air, av M_{r}(mix)
~ 29 
0.1358 
3.64 x 10^{–5} 
3.77 x 10^{6} 
133 K 
ammonia, M_{r}(NH_{3}) = 17 
0.4233 
3.73 x 10^{–5} 
11.3 x 10^{6} 
406 K 
butane, M_{r}(C_{4}H_{10}) =
59 
1.466 
12.2 x 10^{–5} 
3.78 x 10^{6} 
425 K 
carbon dioxide, M_{r}(CO_{2})
= 44 
0.3643 
4.27 x 10^{–5} 
7.39 x 10^{6} 
304 K 
dichlorodifluoromethane, Freon CFC–11, M_{r}(CCl_{2}F_{2})
= 121 
1.078 
9.98 x 10^{–5} 
4.12 x 10^{6} 
385 K 
helium, M_{r}(He) = 4 
0.00341 
2.34 x 10^{–5} 
0.23 x 10^{6} 
5 K 
hydrogen, M_{r}(H_{2}) = 2 
0.0247 
2.65 x 10^{–5} 
1.29 x 10^{6} 
33 K 
nitrogen, M_{r}(N_{2}) = 28 
0.1361 
3.85 x 10^{–5} 
3.39 x 10^{6} 
126 K 
water, M_{r}(H_{2}O) = 18 
0.5507 
3.04 x 10^{–5} 
22.1 x 10^{6} 
647 K 

The constant a
varies considerably from gas to gas because of the wide variety of
intermolecular forces e.g. very low for helium and non–polar hydrogen (2 e's
each, just instantaneous dipole–induced dipole forces), to much higher a
values for larger polar molecules like water or dichlorodifluoromethane (more
electrons and extra permanent dipole–permanent dipole intermolecular forces).

The constant b
varies less, and not unexpectedly, just tends to rise with increase in
molecule size.

Critical values of gas
behaviour.

Critical temperature T_{c}

This is the maximum
temperature at which a substance can exist as a liquid. Above T_{c},
only the gaseous state can exist, however great the density or pressure! It
might be truer to say that above T_{c}, the gaseous and liquid state
become indistinguishable as the meniscus just disappears!

Critical pressure p_{c}

Generally speaking the
critical values for a gas/liquid increase with increase in intermolecular
forces e.g. due to increase in molecular mass or increasing polarity of
molecule.
5c.
Compressibility factors

The compressibility
factor z, is defined as the ratio PV/nRT.

Since PV = nRT for an
ideal gas, then z = 1 for an ideal gas.

z varies with pressure or
temperature for any gas, see the PV versus P graph in
section 5a. which gives an indication of how z might vary with
pressure at a given temperature).
 Clearly from the graph on the right
for methane, z can be at least as high as 2, and, at least as low as 0.6,
showing considerable deviation from ideal gas behaviour, particularly at low
temperatures (influence of intermolecular forces stronger) and high
pressures (where the effect of both actual molecule volume and
intermolecular forces are important). See
more detailed
discussion in 5a.
 As the pressure becomes lower and/or
temperature higher, the gas becomes more ideal in terms of its physical
behaviour and particularly 'ideal' as the pressure tends towards zero.
 Known values of z can be used to
calculate the real P–V values for a non–ideal gas.
 z = pV/nRT, pV = znRT, p = znRT/V
and V = znRT/p
5d The Critical Point –
The Critical Temperature and Critical Pressure
Question! If you increase
the pressure of a gas it can change into a liquid. But, increasing the pressure,
also increases the temperature, so shouldn't the gas remain a gas?

Gases can be converted
to liquids by compressing the gas at a suitable temperature and this is done
commercially at as lower temperature as possible e.g. liquefaction of air to
fractionally distil off nitrogen and oxygen or liquefying petroleum gas.

Gases become more
difficult to liquefy as the temperature increases because the kinetic
energies of the particles that make up the gas also increase and the
intermolecular forces have less influence i.e. more easily overcome.

When you increase the pressure of a gas
you force the molecules closer together and if the extra intermolecular
force is strong enough liquefaction occurs. Remember the force of electrical
attraction is proportional to the numerical +ve charge multiplied by the
–ve charge divided by the distance squared.

However when you
compress a gas it can heat up. This is because heat is generated by the
increased intermolecular interaction (remember bond formation is also
exothermic) but here its just weak molecule association due to the
intermolecular attractive forces.

BUT liquefaction =
condensation and is an exothermic process, so heat must be removed to effect the
state change of gas ==> liquid. If the temperature is low enough and the
heat is dispersed liquefaction can still happen.

If it is too hot it
would stay as a gas. So liquefaction conditions are all about temperature,
pressure and heat transfer i.e. the ambient conditions.

However, above a certain
temperature called the critical temperature (T_{c})
you cannot get a liquid with a 'surface', what you get is an extremely dense
gas that is close to being a liquid but not quite!

The critical temperature
of a substance is the temperature at and above which vapour of the substance
cannot be liquefied, no matter how much pressure is applied.

The critical pressure
(P_{c}) of a substance is the minimum pressure required to
liquefy a gas at its critical temperature i.e. the critical pressure is the
vapour pressure at the critical temperature.

The vapourliquid
critical point denotes the conditions above which distinct liquid and gas
phases do not exist and a meniscus no longer exists!

The point at the critical temperature and critical
pressure is called the critical point of the substance.

–
OTHER USEFUL PAGES
Advanced
notes on gas law calculations, kinetic
model theory of an IDEAL GAS & non–ideal gases
See also for gas calculations
Moles and the molar volume of a gas, Avogadro's Law
Reacting gas volume
ratios, Avogadro's Law
& Gay–Lussac's Law Calculations
All other calculation pages

What is relative atomic mass?,
relative isotopic mass and calculating relative atomic mass

Calculating relative
formula/molecular mass of a compound or element molecule

Law of Conservation of Mass and simple reacting mass calculations

Composition by percentage mass of elements
in a compound

Empirical formula and formula mass of a compound from reacting masses
(easy start, not using moles)

Reacting mass ratio calculations of reactants and products
from equations
(NOT using
moles) and brief mention of actual percent % yield and theoretical yield,
atom economy
and formula mass determination

Introducing moles: The connection between moles, mass and formula mass
– the basis of reacting mole ratio calculations
(relating reacting masses and formula
mass)

Using
moles to calculate empirical formula and deduce molecular formula of a compound/molecule
(starting with reacting masses or % composition)

Moles and the molar volume of a gas, Avogadro's Law

Reacting gas volume
ratios, Avogadro's Law
and Gay–Lussac's Law (ratio of gaseous
reactants–products)

Molarity, volumes and solution
concentrations (and diagrams of apparatus)

How to
do volumetric titration calculations e.g. acid–alkali titrations
(and diagrams of apparatus)

Electrolysis products calculations (negative cathode and positive anode products)

Other calculations
e.g. % purity, % percentage & theoretical yield, volumetric titration
apparatus, dilution of solutions
(and diagrams of apparatus), water of crystallisation, quantity of reactants
required, atom economy

Energy transfers in physical/chemical changes,
exothermic/endothermic reactions

Gas calculations involving PVT relationships,
Boyle's and Charles Laws (this page)

Radioactivity & half–life calculations including
dating materials
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