Example 1.2 Using the perfect gas law
In an industrial process, nitrogen is heated to 500 K in a vessel of constant volume.
If it enters the vessel at 100 atm and 300 K, what pressure would it exert at the
working temperature if it behaved as a perfect gas?
Method We expect the pressure to be greater on account of the increase in temperature.
The perfect gas law in the form pV/nT = R implies that, if the conditions
are changed from one set of values to another, then, because pV/nT is equal to a
constant, the two sets of values are related by the ‘combined gas law’
= (1.9)°
This expression is easily rearranged to give the unknown quantity (in this case p2)
in terms of the known. The known and unknown data are summarized in (2).
Answer Cancellation of the volumes (because V1 = V2) and amounts (because
n1 = n2) on each side of the combined gas law results in
=
which can be rearranged into
p2= ×p1
Substitution of the data then gives
p2= ×(100 atm) = 167 atm
Experiment shows that the pressure is actually 183 atm under these conditions, so
the assumption that the gas is perfect leads to a 10 per cent error.
Self-test 1.3 What temperature would result in the same sample exerting a pressure
of 300 atm? [900 K]
The perfect gas law is of the greatest importance in physical chemistry because it is
used to derive a wide range of relations that are used throughout thermodynamics.
However, it is also of considerable practical utility for calculating the properties of
a gas under a variety of conditions. For instance, the molar volume, Vm =V/n, of a perfect
gas under the conditions called standard ambient temperature and pressure
(SATP), which means 298.15 K and 1 bar (that is, exactly 105 Pa), is easily calculated
from Vm = RT/p to be 24.789 dm3 mol−1. An earlier definition, standard temperature
and pressure (STP), was 0°C and 1 atm; at STP, the molar volume of a perfect gas is
22.414 dm3 mol−1.
(b) The kinetic model of gases
The molecular explanation of Boyle’s law is that, if a sample of gas is compressed to
half its volume, then twice as many molecules strike the walls in a given period of time
than before it was compressed. As a result, the average force exerted on the walls is
500 K
300 K
T2
T1
p2
T2
p1
T1
Combined
gas law
p2V2
n2T2
p1V1
n1T1
2
n p V T
Initial
Final
Same
Same
Same
Same
100
?
300
500