Page 134 - Physical chemistry understanding our chemical world

ENTHALPY 101

Inserting numbers yields 3
Each aliquot of 1 cm
represents a volume of
3
−6
( 0.031 18 × 10 −6 ) 1 × 10 m .
∆V = − m ≈ 0.031 m 3
3
volume per mole volume per mole
of steam of liquid water

Inserting values into Equation (3.12), w = p V :

5
w = 10 Pa × 0.031 m 3
3
w = 3.1 × 10 J

so the pressure–volume work is 3100 J.

We often encounter energies of the order of thousands of joules.
An energy expressed
As a shorthand, we often want to abbreviate, so we rewrite the with a letter ‘k’ in the
answer to Worked Example 3.6, and say w = 3.1 × 1000 J. Next, answer means ‘thou-
we substitute for the factor of 1000 with an abbreviation, generally sands of joules’. We
choosing a small letter ‘k’. We rewrite, saying w = 3.1kJ. say kilojoules.The
choice of ‘k’ comes
SAQ 3.6 What is the work done when the gas from a from the Greek word
party balloon is released? Assume the inflated balloon for thousand, which is
3
has a volume of 2 dm , and a volume of 10 cm 3 when kilo.
O
deflated. Assume there is no pressure change, so p = p .
3
[Hint: 1 l = 1 × 10 −3 m ].
Worked Example 3.7 What is the work performed when inflating a car tyre from p O
3
to 6 × p . Assume the volume inside the tyre stays constant at 0.3m .
O
Firstly, we calculate the change in pressure, from an equation like Equation (3.1), p =
O
p (final) − p (initial) ,so p = (6 − 1) × p ,i.e.
p = 5 × p O
5
p = 5 × 10 Pa

Then, inserting values into Equation (3.5), w = p × V :

5
w = 5 × 10 Pa × 0.3m 3

yielding
w = 15 000 J = 15 kJ