# Ideal gas behaviour and the gas laws ```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:
1. Gases are composed of minute discrete particles (usually molecules).
2. 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
109/s).
3. The bombardment of the container walls by the particles causes the phenomenon
we call pressure (i.e. force of impacts/unit area).
4. The collisions are perfectly elastic i.e. no energy loss on collision due to friction.
5. 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.
6. The average kinetic energy of the particles is directly proportional to the absolute
temperature on the Kelvin scale (K).

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
4a. Boyle's Law for volume and gas pressure

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, the product of the pressure
multiplied by the volume is a constant.

p x V = constant

Therefore, for initial values of p1 and V1, which change to final values of p2 and V2, the
following equation applies ...

p1 x V1 = p2 x V2 (for fixed amount of gas at constant temperature)

or p2 = p1 x V1/V2 or V2 = p1 x V1/p2

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.
o
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.
o
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.

Examples of Boyle's Law calculations (constant temperature assumed)

Ex. Q4a.1


o
240cm3 of air at a pressure of 100kPa in a bicycle pump is compressed to a volume
of 150cm3.
o
What is the pressure of the compressed air in the pump?
o
p1 x V1 = p2 x V2 , rearranging to scale up for the new higher pressure
o
p2 = p1 x V1/V2 = 100 x 240/150 = 160 kPa
Ex. Q4a.2
o
10 m3 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?
o
p1 x V1 = p2 x V2 , rearranging to scale down for the new lower volume
o
V2 = p1 x V1/p2 = 1.2 x 10/6 = 2.0 m3
-
4b. Charles's Law/Gay-Lussac's Law for pressure/volume and temperature
and the combined gas law equation

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 ...
o
o
(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

V1/V2 = T1/T2 for conditions changing from 1 (initial) to 2 (final),

or V1/T1 = V2/T2 for constant pressure

V2 = V1 x T2/T1

or T2 = T1 x V2/V1
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

p1/p2 = T1/T2 for conditions changing from 1 (initial) to 2 (final),

or p1/T1 = p2/T2 for constant volume

p2 = p1 x T2/T1

or T2 = T1 x p2/p1

In all calculations, the absolute or Kelvin scale of temperature must be used for T (K = oC +
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 conditions1, changing to a new and final set of conditions2 for a given mass of gas, giving
the combined pressure-volume-temperature gas calculation equation ...

p1 x V1
p2 x V2
--------------
=
T1
--------------T2

(p1 x V1)/T1 = (p2 x V2)/T2

therefore the three permutations for problem solving are ...

p2 = (p1 x V1 x T2)/(V2 x T1)

or V2 = (p1 x V1 x T2)/(p2 x T1)

or T2 = (p2 x V2 x T1)/(V1 x T2)

Note:
o
If the temperature is constant you get Boyle's Law.
o
If p or V is constant you get Charles's/Gay-Lussac's Law.
o
You can use any volume or pressure units you like as long as both p's or both
V's have the same units.
o
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 = -273oC (for T).
o
This was part of the scientific evidence that led to the belief that -273oC was the
lowest possible temperature, though there is no theoretical upper limit at all.
o
This led to the devising of a new thermodynamic absolute temperature scale or
Kelvin scale which starts at OK.

e.g. ice melts at 0oC or 273K and water boils at 100oC or 373K.

Examples of P-V-T calculations

Ex. Q4b.1
o
The pressure exerted by a gas in sealed container is 100kPa at 17oC. It was found
that the container might leak if the internal pressure exceeds 120kPa. Assuming
constant volume, at what temperature in oC will the container start to leak?


o
17oC + 273 = 290K
o
p1/T1 = p2/T2
o
rearranging to scale up to the higher temperature
o
T2 = T1 x p2/p1
o
T2 = 290 x 120/100 = 348 K or 348 - 273 = 75oC when the container might leak
Ex. Q4b.2
o
A cylinder of propane gas at 20oC exerted a pressure of 8.5 atmospheres. When
exposed to sunlight it warmed up to 28oC. What pressure does the container side
now experience?
o
20oC = 273 + 20 = 293K, 28oC = 273 + 28 = 301K
o
p2 = p1 x T2/T1
o
p2 = 8.5 x 301/293 = 8.73 atm
Ex. Q4b.3
o
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
100cm3 gas syringe was used to determine the rate of reaction. All the experiments
were done in one lesson at a temperature of 22oC except for the last one. This was
done in the next lesson, giving a carbon dioxide volume of 47.0 cm3 after 5 minutes,
but at a higher temperature of 27oC(when in Kelvin call this T1, and the other
temperature T2).
o
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.
o
(a) Calculate the volume the of 47.0 cm3 of gas at 27oC, would occupy at 22oC.
o

V1/V2 = T1/T2 so V2 = V1 x T2/T1

V1 = 47.0 cm3, T1 = 273 + 27 = 300K, T2 = 273 + 22 = 295K

V2 = 47.0 x 295/300 = 46.2 cm3
(b) If the temperature was ignored, what is the % error in the rate of reaction
measurement?

Volume error = 47.0 - 46.2 = +0.8 cm3, therefore ....

% error = 0.8 x 100/47 = +1.7% (so you would over calculate the reaction
rate without this correction)

o
The % error in the volume would be the same as calculated for the rate
e.g. in cm3/min.
(c) Should the calculated value for 22oC be used in the rate calculation analysis? and
are this still other sources of error?

The theoretical-calculated gas volume for 22oC 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.
27oC) 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
cm3 gas/minute at 27oC, that would have occurred/been measured at 22oC
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 27oC and there isn't
much you can do about that error except repeat the experiment at 22oC,
which is the best thing to do anyway!

o
(d) Would you need to do any correction for the volume of acid added to the


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.
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. Q4b.4
o
25 cm3 of a gas at 1.01 atm. at 25oC was compressed to 15 cm3 at 35oC.
o
Calculate the final pressure of the gas.
o
p1 = 1.01 atm, p2 = ?, V1 = 25 cm3, V2 = 15 cm3,
o
T1 = 25 + 273 = 298 K, T2 = 35 + 273 = 308 K
o
(p1 x V1)/T1 = (p2 x V2)/T2
o
p2 = (p1 x V1 x T2)/(V2 x T1)
o
p2 = (1.01 x 25 x 308)/(15 x 298) = 1.74 atm
4c. The Ideal Gas Equation of State pV = nRT

A 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 . The
equation is pV = nRT and requires a consistent set of units, so see below for the two most
common examples, and take care!

pressure p
volume V
n = mass g/Mr
Ideal gas
constant Rand
its units
temperature T
K
Pa (Pascal,
m3 (1m3 =
760mmHg = 1 atm 106 cm3 or m3 = mol
= 101325 Pa)
cm3/106)
8.314
J mol-1 K-1
(Kelvin = oC +
273)
litre or dm3 (litre
atm (atmospheres,
= dm3 = 1000
760 mmHg = 1 atm
mol
cm3 and dm3 =
= 101325 Pa)
cm3/1000)
0.08206
K
atm dm3 mol-1 K- (Kelvin = oC +
1
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 Mr is in g mol-1.

Examples of PV = nRT calculations

Ex. Q4c.1
o
(a) Describe with the aid of a diagram a simple gas syringe method for determining
the molecular mass of a volatile liquid.


o
A 100 cm3 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 Mr 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 0.5cm3 if possible), together with 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 cm3 of gas.

The temperature of the oven = 81oC, barometric pressure 752 mmHg.

Using the equation PV = nRT, calculate the molecular mass of the liquid.

o

Mass of liquid injected = 10.6405 - 10.4227 = 0.2176 g

p = 101325 x 752/760 = 100258 Pa,

V = 67.3/106 = 6.73 x 10-5 m3, T = 273 + 81 = 354 K

PV = nRT, substituting for moles n gives PV = m/MrRT

and then rearranging gives ...

Mr = 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 formulafor the compound.

o
bromomethane, CH3Br, Mr = 12 + 3 + 80 = 95
(d) State very briefly, a method of determining the molecular mass of ANY
compound that can be vapourised intact.


R = 8.314 J mol-1 K-1
Ex. Q4c.2
Mass spectrometer, from the molecular ion peak.
o
o

(a) What is the volume of 6g of chlorine at 27oC and 101kPa (approx. 1 atm)?

pV = nRT, V = nRT/p

T = 273 + 27 = 300K,

n = 6/71 = 0.08451 mol chlorine, Mr(Cl2) = 2 x 35.5 = 71

and p = 101 x 1000 = 101000 Pa.

V = 0.08451 x 8.314 x 300/101000 = 0.002087 m3
(b) What is the volume of the chlorine in dm3 and cm3?

1 m3 = 1000 dm3 = 106 cm3

V = 0.002087 x 1000 = 2.087 dm3

V = 0.002087 x 106 = 2087 cm3
Ex. Q4c.3
o
o
o
(a) A 5 litre container contained 0.5kg of butane gas (C4H10). Assuming ideal gas
behaviour calculate the pressure of the gas if the cylinder is stored at 25oC.

Mr(C4H10) = (4 x 12) + 10 = 58, 0.5kg = 550g

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) SI units

PV = nRT, P = nRT/V, 5 litre = 5 dm3 = 5 x 10-3 m3

P = 8.621 x 8.314 x 298/(5 x 10-3)

P = 4271830 Pa = 4272 kPa = 4.272 MPa
(ii) 'old units'

P = 8.621 x 0.08206 x 298/5 = 42.16 atm
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