6
PressureVolume Work
Let’s consider a simple example of the work done on the system (considered to be a gas) when a
constant external pressure is applied to it.
Work done on
the system = (external force
×
displacement)
External force = external pressure
×
surface area
W = P
×
A
×
Δ
h
(note the minus
sign)
A
×
Δ
h =
Δ
V = V
final
 V
initial
W = P
Δ
V = P(V
final
 V
initial
)
Let’s do a couple of sample problems: Apply an external pressure of 1.0 atmospheres as a gas
expands
from 0.5 L to 1.0 L.
W = 1.0 atm (1.0 L – 0.5 L) = 0.5 Latm.
Not a particularly useful unit.
Since pressure is a
force per unit area, the appropriate mechanical SI unit for pressure is N/m
2
.
Turns out that
atmospheric pressure is 1.013
×
10
5
N/m
2
.
Remember that 1 liter = 1000 cm
3
= 10
3
m
3
.
(Do
you know why?).
Therefore 1 Latm = 1.013
×
10
5
N/m
2
×
10
3
m
3
= 101.3 Nm = 101.3 J
This is an important unit conversion.
For the expansion W = 0.5 Latm
=50.6 J.
Note that
negative
work is done on
the
system.
Note also for a compression
, the final volume is smaller than the initial volume.
So, the
work of compression
on
the system is
positive
.
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7
P
V
q
V
Δ
T
Absorption of Heat
According to the First Law,
Δ
E = q + w = q  P
Δ
V, assuming PV work is the only kind that can occur.
Therefore,
Δ
E = q
V
.
The subscript means that the process occurs at constant volume.
So, if we can control our system and maintain constant volume, we know that if we inject heat
into a constant volume system then we know the change in internal energy.
For an ideal gas, an increase
in E results in a temperature
increase.
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 Spring '10
 farrar
 Thermodynamics, Energy, Heat, ΔH

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