Doc Brown's
Chemistry KS4 science GCSE/IGCSE/AS Revision Notes
Part 2 More
advanced topics on the ideal gas
laws, calculations, kinetic particle model–theory (sections 4a to 4b)
More advanced ideas involving
gases gas law calculations involving Boyle's Law, Charles's Law, Gay–Lussac Law, P1V1/T1 = P2V2/T2.
The gas laws relating pressure, volume and
temperature are very important for both GCSE and A level Chemistry
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 ....
Sub–index for Part 2:
Section 4 Ideal gas behaviour and the gas laws: Introduction–the kinetic
particle theory of an ideal gas
* Kelvin temperature scale * 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 *
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.
The Maxwell
Boltzmann distribution of particle kinetic energies is discussed in the
KINETICS pages.
4. Ideal gas behaviour and
the gas laws
Introduction – the kinetic particle model of an ideal gas

The (advanced) kinetic
theory of gases is founded on the following six fundamental postulates:

Gases are composed of
minute discrete particles (usually molecules).

The particles are in
continuous chaotic motion moving in straight lines between very frequent collisions with
each other and the sides of the container (approximately 10^{9}/s).

The bombardment of the
container walls by the particles causes the phenomenon we call pressure
(i.e. force of impacts/unit area). The greater the force of collision and
the more frequent the collisions the greater the gas pressure exerted on the
container surface.

The collisions are
perfectly elastic i.e. no energy loss on collision due to friction.

At relatively low
pressures the average distance between particles is large compared to the
diameter of the particles and therefore the inter–molecular forces between
the particles is negligible.

The average kinetic
energy of the particles is directly proportional to their absolute temperature
on the Kelvin scale (K) i.e. KE(J)
T(K)

This means if you heat up a gas the
average kinetic energy of the particles increases, therefore the average
speed increases too.

The Kelvin scale of temperature is
explained below.

When a gas behaves
according to this model, the gas laws described in sections 4a to 4e are
obeyed.

However in real gases
things are not so simple and this non–ideal behaviour is
discussed in section 5.
The
Kelvin Scale of Temperature
In the past lots of measurements have been
made to investigate how (i) the pressure and volume of a given mass of vary at
constant temperature and how (ii) the pressure and volume gas of a fixed mass of
gas varies with temperature (see the two graphs left and right). This resulted
in the formulation of the laws of gases described in the next section 4a. along
with how to use them in calculations and problem solving.
However, before this, if you look at these
two graphs of gas behaviour when changing pressure or volume with temperature,
one thing becomes clear, when the graph lines are extrapolated back to the x
axis they give a value of –273^{o}C. This gave rise to the idea that
there was a minimum possible temperature of –273^{ o}C and
further experimentation has confirmed this time and time again. At –273^{o}C
all substances are solid and in terms of the kinetic particle theory of matter,
at –273^{ o}C the particles have virtually no motion i.e. ~no
vibration of the atoms in the solid.
Therefore as well as the established
Celsius scale (centigrade scale), a new temperature scale was proposed in
which the lowest value was 0 (known as absolute zero) rather than –273.
This is called the Kelvin scale of temperature or the absolute
temperature scale, denoted by the unit K.
Incidentally you don't say degrees Kelvin like you say degrees Celsius, you just
say Kelvin. The Kelvin temperature scale was also designed so that a 1 K
temperature change or interval, exactly equals 1^{o}C Celsius change
or interval. Therefore you can easily convert between the two temperature scale
by a simple calculation
K = ^{o}C + 273
and ^{o}C = K – 273 and it is the
temperature in K that you must use in gas law calculations (4b.)
Some examples are worked out below and a
practice in reading a Celsius thermometer, which is what you use in the school
or college
laboratory!
–7^{o}C
–7 + 273 = 267 K 
36^{o}C
36 + 273 = 309 K 
77^{o}C
77 + 273 = 350 K 
101.5^{o}C
101.5 + 273.0 = 374.5 K 
132^{o}C
132 + 273 = 405 K 
206^{o}C
206 + 273 = 479 K 
Some familiar temperatures are quoted below
relating the two temperature scales

absolute
zero 
Freezing
point of water 
Body
temperature 
Boiling
point of water 
Celsius
scale 
–273^{
o}C 
0^{ o}C 
37 ^{o}C 
100 ^{
o}C 
Kelvin
scale 
0 K 
273 K 
310 K 
373 K 
It seems a bit weird to say you body has a
temperature of 310, which is why it is always important to state the units too!
Calculations 4a. Boyle's Law for volume and gas
pressure

The particle theory of gas pressure
was explained in Part 1 so this section concentrates on the gas law
calculations involving pressure and volume.

Boyle's Law states that for given mass of gas
at a constant temperature (^{o}C or K), the product of the pressure multiplied by the
volume is a constant.

p x V = constant

Therefore, for initial
values of p_{1}
and V_{1}, which change to final values of p_{2} and V_{2}, the following
equation applies ...

p_{1} x V_{1}
= p_{2}
x V_{2} (for fixed amount of gas at constant
temperature)

or
p_{2}
= p_{1}
x V_{1}/V_{2}_{ }or V_{2}
= p_{1} x V_{1}/p_{2}

The graph shows how the pressure and volume
vary according to Boyles Law at two different temperatures.

At lower temperatures the volume and pressure
values are lower (see next section).

You can use any volume or
pressure units you like as long as both p's and both V's have the
same units.

Using particle theory and
simple arithmetical values to explain Boyles Law.

If a gas is compressed to half its
original volume the concentration or density of the gas is doubled. Therefore
there will be twice as many collisions with the surface causing twice the
impact effect i.e. double the pressure.

If the volume of a gas is
increased by a factor of three, the concentration is reduced by the same
factor, so the chance of particle collision with the container walls is
similarly reduced, so the pressure decreases by a factor of three.

Gases e.g. oxygen for hospitals, can be
stored under high pressure enabling reasonably efficient storage. Because
the internal pressure in the cylinder is so much greater than the external
pressure, on fitting a valve, a large volume of gas can be released to flow
slowly under controlled conditions for a patients respiration.

Examples of Boyle's Law calculations (constant temperature assumed)

Ex. Q4a 1

240cm^{3} of air
at a pressure of 100kPa in a bicycle pump is compressed to a volume of 150cm^{3}.

What is the pressure of
the compressed air in the pump?

p_{1} x V_{1}
= p_{2}
x V_{2} , rearranging to scale up for the new
higher pressure

p_{2}
= p_{1} x V_{1}/V_{2} = 100 x 240/150 = 160 kPa

Ex. Q4a 2

10 m^{3} of butane
gas at 1.2 atm was required to be stored at 6 atm pressure. To what volume
must the gas be compressed to give the required storage pressure?

p_{1} x V_{1}
= p_{2}
x V_{2} , rearranging to scale down for the new
lower volume

V_{2} = p_{1} x V_{1}/p_{2}
= 1.2 x 10/6 = 2.0 m^{3}

Ex. Q4a 3

A 100 cm^{3} gas syringe
containing 80 cm^{3} of gas that was compressed to 60 cm^{3}.
If atmospheric pressure is 101325 Pa, and the temperature remains constant,
what is the pressure of the gas in the syringe after compression.

p x V = constant

p_{1} x V_{1}
= p_{2}
x V_{2}

p_{2}
= p_{1}
x V_{1}/V_{2 }

p_{2}
= 101325 x 80/60 = 135100 Pa

Ex. Q4a 4

In hospital the gas pressure in a 100 dm^{3}
cylinder of oxygen is 5.52 atm (5 x atmospheric pressure). What volume of
gas can be released slowly to a patient on releasing it to an
atmospheric pressure 1.01?

p x V = constant

V_{2}
= p_{1} x V_{1}/p_{2}

V_{2}
= p_{1} x V_{1}/p_{2} = 5.52 x 100/1.01 = 546.5 dm^{3}
Calculations 4b. Charles's Law/Gay–Lussac's Law for pressure/volume and
temperature
and the combined
gas law equation

The particle theory of gas pressure
was explained in Part 1 so this section concentrates on the gas law
calculations involving pressure and volume and their variation with
temperature.

Charles's/Gay–Lussac's Law states that for a fixed mass of gas
...

(i) the volume of a gas is directly
proportional to the absolute temperature (K) at constant pressure

V = constant x T (left graph), or

V/T = constant,
or

V_{1}/V_{2}
= T_{1}/T_{2} for conditions changing from 1 (initial)
to 2 (final),

or V_{1}/T_{1}
= V_{2}/T_{2} for constant pressure

V_{1} x T_{2} = V_{2}
x T_{1}

V_{2} = V_{1}
x T_{2}/T_{1}

or T_{2} =
T_{1} x V_{2}/V_{1}

Kinetic particle reasoning  increasing
the temperature increases the kinetic energy of the molecules giving
more forceful collisions which push out (expand) the gas at constant
pressure.

OR (ii) the pressure of a gas is directly
proportional to the absolute temperature (K) at constant volume,

p = constant x T (right graph),
or

p/T = constant,
or

p_{1}/p_{2}
= T_{1}/T2 for conditions changing from 1 (initial)
to 2 (final),

or p_{1}/T_{1}
= p_{2}/T_{2} for constant volume

p_{1} x
T_{2} =
T_{1} x
p_{2}

p_{2} = p_{1}
x T_{2}/T_{1}

or T_{2} =
T_{1} x p_{2}/p_{1}

Kinetic particle reasoning  increasing
the temperature increases the kinetic energy of the molecules giving
more forceful collisions that increase the pressure if the volume is
constrained (kept constant).

In all calculations, the
absolute or Kelvin scale of temperature must be used for T (K = ^{
o}C + 273).

If all the laws described
in 4a and 4b are combined, you get the following general expression

p x V/T = a constant (for a
given mass of gas).

This can be expressed in
generalised form for calculations based on an
initial set of conditions_{1} (1) changing to a new and final set of
conditions_{2} (2) for a given mass of gas, giving the
combined pressure–volume–temperature gas calculation equation ...

p_{1} x V_{1} 

p_{2} x V_{2} 
–––––––––––––– 
= 
––––––––––––––– 
T_{1} 

T_{2} 

(p_{1} x V_{1})/T_{1}
= (p_{2} x V_{2})/T_{2}
 therefore the three
permutations for problem solving are ...
 p_{2} = (p_{1} x V_{1}
x T_{2})/(V_{2} x T_{1})
 or V_{2} = (p_{1} x V_{1}
x T_{2})/(p_{2} x T_{1})
 or T_{2} = (p_{2} x V_{2}
x T_{1})/(V_{1} x T_{2})

Note:

If the temperature is constant you get Boyle's
Law.

If p or V is constant you
get Charles's/Gay–Lussac's Law.

You can use any volume or
pressure units you like as long as both p's or both V's have the
same units.

The graphs of p or V
versus temperature become invalid once the gas has condensed into a liquid BUT
when extrapolated back all the lines seem to originate from y = 0 (for p or
V), x = –273^{o}C (for T).

This was part of the scientific evidence
that led to the belief that –273^{o}C was the lowest possible
temperature, though there is no theoretical upper limit at all.

This led to the devising of a new thermodynamic absolute
temperature scale or Kelvin scale which starts at OK.

Examples of P–V–T calculations

Ex. 4b Q1

The pressure exerted by
a gas in sealed container is 100kPa at 17^{o}C. It was found that
the container might leak if the internal pressure exceeds 120kPa. Assuming
constant volume, at what temperature in ^{o}C will the container
start to leak?

17^{o}C + 273 =
290K

p_{1}/T_{1}
= p_{2}/T_{2}

rearranging to scale up to the higher
temperature

T_{2} = T_{1}
x p_{2}/p_{1}

T_{2} = 290 x 120/100 = 348 K or 348
– 273
= 75^{o}C
when the container might leak

Ex. 4b Q2

A cylinder of propane
gas at 20^{o}C exerted a pressure of 8.5 atmospheres. When exposed
to sunlight it warmed up to 28^{o}C. What pressure does the
container side now experience?

20^{o}C = 273 +
20 = 293K, 28^{o}C = 273 + 28 = 301K

p_{2} = p_{1}
x T_{2}/T_{1}

p_{2} =
8.5 x 301/293 = 8.73 atm

Ex. 4b Q3

A student was
investigating the speed of reaction between limestone granules and different
concentrations of hydrochloric acid. However after doing a whole series of
experiments at different acid concentrations, there was no time to do the
last planned experiment. The volume of carbon dioxide collected after 5 minutes in a
100cm^{3} gas syringe was used to determine the rate of reaction.
All the experiments were done in one lesson at a temperature of 22^{o}C
except for the last one. This was done in the next lesson, giving a carbon
dioxide volume of 47.0 cm^{3} after 5 minutes, but at a higher
temperature of 27^{o}C (when in Kelvin call this T_{1},
and the other temperature T_{2}).

To make the data
analysis fair, all the gas volumes should be ideally measured at the same
temperature, but a correction can be made for the last experiment.

(a) Calculate the volume
the of 47.0 cm^{3} of gas at 27^{o}C, would occupy at 22^{o}C.

V_{1}/V_{2}
= T_{1}/T_{2}
so V_{2} = V_{1}
x T_{2}/T_{1}

V_{1}
= 47.0 cm^{3}, T_{1} = 273 + 27 = 300K, T_{2} = 273 +
22 = 295K

V_{2} =
47.0 x 295/300 = 46.2 cm^{3}

(b) If the temperature
was ignored, what is the % error in the rate of reaction measurement?

Volume error = 47.0
– 46.2 = +0.8 cm^{3}, therefore ....

% error = 0.8 x 100/47 = +1.7%
(so you would over calculate the reaction rate without this correction)

The % error in the
volume would be the same as calculated for the rate e.g. in cm^{3}/min.

(c) Should the
calculated value for 22^{o}C be used in the rate calculation
analysis? and are this still other sources of error?

The
theoretical–calculated gas volume for 22^{o}C should be used for
calculating the rate, it will improve the accuracy a little, BUT there is
another problem!

If the reaction was
unfortunately carried out at a higher temperature (i.e. 27^{o}C)
there is a second source of error. At a higher temperature the reaction is
faster, so you are bound to get a higher volume of gas formed in five
minutes. Therefore you will calculate a faster rate of reaction e.g. in cm^{3}
gas/minute at 27^{o}C, that would have occurred/been measured at 22^{o}C
and so an unfair comparison with all the other results from the previous
lesson.

So, although you can
correct reasonably well the volume error due to an 'expanded' gas volume at
the higher temperature, the gas volume will still be too high because of
the faster rate of reaction at 27^{o}C and there isn't much you
can do about that error except repeat the experiment at 22^{o}C,
which is the best thing to do anyway!

Note that if the
temperature of a rates experiment was too low compared to all the other
experiments, the 'double error' would occur again, but this time the
measured gas volume and the calculated speed/rate of reaction would be lower
than expected.

(d) Would you need to do
any correction for the volume of acid added to the limestone? Explain your
decision.

No correction needed
for this at all. Although liquids expand/contract on heating/cooling, the volume
changes are far less compared to gas volume changes for the same temperature
change. This is because of the relatively strong intermolecular forces between liquid
molecules, which are almost absent in gases.

Ex. 4b Q4

25 cm^{3} of a gas
at 1.01 atm. at 25^{o}C was compressed to 15 cm^{3} at 35^{o}C.

Calculate the final
pressure of the gas.

p_{1} = 1.01 atm,
p_{2} = ?, V_{1} = 25 cm^{3}, V_{2} = 15 cm^{3},

T1 = 25 + 273 = 298
K, T2 = 35 + 273 = 308 K

(p_{1} x V_{1})/T_{1}
= (p_{2} x V_{2})/T_{2}

p_{2} = (p_{1} x V_{1}
x T_{2})/(V_{2} x T_{1})

p_{2} =
(1.01 x 25 x 308)/(15 x 298) = 1.74 atm

Ex. 4b Q5

Ex. 4b Q6

The fuel and air gases in the cylinders
of a 1200 cm^{3} car engine go from 25^{o}C before
combustion and rise to a peak temperature of 2100^{o}C after
combustion. If normal atmospheric pressure is 101kPa, calculate the peak
pressure reached after combustion. Although the movement of the piston
changes the volume, (i) for the sake of argument assume the volume is
constant.

T_{1} = 25 + 273 = 298 K, T_{2}
= 2100 + 273 = 2373 K, P1 = 101 KPa

p/T = constant

p_{1}/p_{2}
= T_{1}/T2

p_{2} = p_{1}
x T_{2}/T_{1}

p_{2} = 101 x 2373/298 =
804 kPa

(ii) To be more realistic, assume the
initial volume of fuel vapour plus air was 400 cm3, now recalculate the
final pressure.

You now need to use the full PVT
expression.

(p_{1} x V_{1})/T_{1}
= (p_{2} x V_{2})/T_{2}
 p_{2} = (p_{1} x V_{1}
x T_{2})/(V_{2} x T_{1})
 p_{2} = (101 x 400 x 2373)/(1200 x 298) =
268 kPa

Ex. Q7
OTHER CALCULATION PAGES
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

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
