Page 133 - Physical chemistry understanding our chemical world

100 ENERGY AND THE FIRST LAW OF THERMODYNAMICS

large volume ensures a rapid exit of steam, so the kettle produces an intense, shrill
whistle sound.
The steam expelled from the kettle must exert a pressure against the air as it leaves
the kettle, pushing it aside. Unless stated otherwise, the pressure of the air surrounding
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the kettle will be 10 Pa, which we call ‘standard pressure’ p . The value of p is
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10 Pa. The steam must push against this pressure when leaving the kettle. If it
does not do so, then it will not move, and will remain trapped within the kettle. This
pushing against the air represents work. Specifically, we call it pressure–volume work,
because the volume can only increase by exerting work against an external pressure.
The magnitude of this pressure–volume work is w, and is expressed by
w =− (pV ) (3.12)

where (pV ) means a change in the product of p × V . Work is done to the gas
when it is compressed at constant pressure, i.e. the minus sign is needed to make
w positive.
Equation (3.12) could have been written as p × V if both the pressure and the
volume changed at the same time (an example would be the pushing of a piston
in a car engine, to cause the volume to decrease at the same time as the pressure
increases). In most of the physicochemical processes we will consider here, either
p or V will be constant so, in practice, there is only one variable. And with one
variable, Equation (3.12) becomes either w = p × V or w = V × p, depending
on whether we hold p or V constant.
Most chemists perform experiments in which the contents of our
For most purposes, a beaker, flask or apparatus are open to the air – obvious examples
chemist can say w = include titrations and refluxes, as well as the kinetic and electro-
p × V. chemical systems we consider in later chapters. The pressure is
the air pressure (usually p ), which does not change, so any pres-
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sure–volume work is the work necessary to push back the atmosphere. For most
purposes, we can say w = p V .
It should be obvious that the variable held constant – whether p or V – cannot be
negative, so the sign of w depends on which of the variables we change, so the sign
of w in Equation (3.12) depends on the sign(s) of p or V . The sign of w will be
negative if we decrease the volume or pressure while performing work.

Worked Example 3.6 We generate 1 mol of water vapour in a kettle by boiling liquid
water. What is the work w performed by expansion of the resultant steam?

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We have already seen how 1 mol of water vapour occupies a volume of 0.031 m (see
SAQ 3.5). This volume of air must be pushed back if the steam is to leave the kettle. The
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external pressure is p ,i.e. 10 Pa.
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The change in volume
V = V (final) − V (initial)
so
V = V (water, g) − V (water, l)