first law of thermodynamics

1. First Law of
Thermodynamics

2. Work Done by a Gas
Suppose you had a piston filled with a specific amount of gas. As you add heat, the
temperature rises and thus the volume of the gas expands. The gas then applies a
force on the piston wall pushing it a specific displacement x. Thus it can be said that a
gas can do WORK.
PA
F 

The negative sign is explained on the next page.
F
x
Remember Area x Depth = Volume

3. P V Graphs
)
(
)
(
)
(
)
(















positive
W
negative
W
gas
the
ON
done
is
Work
decrease
volume
V
gas
the
BY
done
is
Work
increase
volume
V
V
P
W
on
by
The ‘negative’ sign in the
equation for WORK is often
misunderstood. Since work
done BY a gas has a positive
volume change we must
understand that the gas itself
is USING UP ENERGY or in
other words, it is losing
energy, thus the negative
sign shows a loss of energy.
When work is done ON a gas the change in volume is negative. This cancels out the negative
sign in the equation. This makes sense as some EXTERNAL agent is ADDING energy to the gas.
i = initial,
f = final



 V
P
W graph
pv
under
area
P
Pi = Pf
Vi Vf V
V
P
W 

area
W 
ΔV
Pressure against Volume Graphs
W = Work done

4. Internal Energy (ΔU) and Heat Energy (Q)
All of the energy inside a system is
called INTERNAL ENERGY, ΔU. It
usually consists of the sum of the
potential and kinetic energies of the
working gas molecules.
When you add HEAT(Q), you are
adding energy and the internal
energy INCREASES.
Both are measured in joules. But
when you add heat, there is usually
an increase in temperature
associated with the change.
0
,
0
,










U
T
if
U
T
if
T
and
U




released
absorbed
Q
Q

5. First Law of Thermodynamics
“The internal energy of a system tends to increase
when HEAT is added and work is done ON the
system.”
by
Add
on
Add
W
Q
U
or
W
Q
U
W
Q
U










Suggests a CHANGE could be + or -
You are really adding a negative here!
The bottom line is that if you ADD heat then transfer work TO the
gas, the internal energy must obviously go up as you have MORE
than what you started with.

6. Example 1
Sketch a PV diagram and find the work done
by the gas during the following stages.
a) A gas is expanded from a volume of 1.0 L to
3.0 L at a constant pressure of 3.0 atm.
b) The gas is then cooled at a constant volume
until the pressure falls to 2.0 atm



 V
P
WBY
- 600 J
0

V
since
P
V
5
10
3
3 

atm N/m2 =
3
10
300 N/m2
3
3
3
10
1
001
.
0
1 m
m
L 



3
3
3
10
3
003
.
0 m
m
L
3 



1 2
0 4
3
100
200
300
400
x 10-3
x
10
3



 )
001
.
0
003
.
0
(
10
300 3
Work Out (Area under Graph)
5
10
2
2 

atm N/m2 =
3
10
200 N/m2



 V
P
WBY
0 (No change in the Work Done)
(b)
(a)

7. Example 1 (continued)
c) The gas is then compressed at a
constant pressure of 2.0 atm from
a volume of 3.0 L to 1.0 L.
d) The gas is then heated until its
pressure increases from 2.0 atm
to 3.0 atm at a constant volume.
P
V
1 2
0 4
3
100
200
300
400
x 10-3
x
10
3
5
10
2
2 

atm N/m2 =
3
10
200 N/m2



 V
P
WON + 400 J
3
3
3
10
1
001
.
0
1 m
m
L 



3
3
3
10
3
003
.
0 m
m
L
3 






 )
003
.
0
001
.
0
(
10
200 3
Work In (Area under Graph)
(d)
(c)
(b)
(a)
0

V
since



 V
P
WBY
0 (No change in the Work Done)

8. What is the NET WORK?
The area inside the shape
General Rule : If the system rotates Clockwise, the NET work is negative.
If the system rotates Anticlockwise, the NET work is postitive.
1 2 4
3
100
200
300
400
x 10-3
x
10
3
(d)
(c)
(b)
(a)
Example 1 (continued)
- 600 J + 400 J = - 200 J
NET work is the
area inside the
shape

9. Example 2
A series of thermodynamic processes is shown in the PV - diagram.
In process a) 150 J of heat is added to the system, in process b)
600J of heat is added and in process c) 300J of heat is lost. No heat
is lost or gained in process d). Fill in the chart.
P
V
1 2
0 4
3
20
40
60
80
x 10-3
x
10
3
(b)
(a)
5
30
(c)
(d)
Conclusion – Heat energy has increased by 450J
Work has decreased by 150J
Internal Energy has increased by 300J
Step Q W ΔU = Q +
W
a
b
c
d
Total
Step Q W ΔU = Q +
W
a +150 0
b +600
c -300
d 0
Total
Step Q W ΔU = Q +
W
a +150 0 150 J
b +600 -240
c -300 0
d 0
Total
Step Q W ΔU = Q +
W
a +150 0
b +600 -240
c -300 0
d 0 +90
Total
Step Q W ΔU = Q +
W
a +150 0
b +600 -240
c -300 0
d 0 +90
Total +450
Step Q W ΔU = Q +
W
a +150 0
b +600 -240
c -300 0
d 0 +90
Total +450 -150
Step Q W ΔU = Q +
W
a +150 0 150 J
b +600 -240
c -300 0
d 0 +90
Total +450 -150
Step Q W ΔU = Q +
W
a +150 0 150 J
b +600 -240 360 J
c -300 0
d 0 +90
Total +450 -150
Step Q W ΔU = Q +
W
a +150 0 150 J
b +600 -240 360 J
c -300 0 -300
d 0 +90
Total +450 -150
Step Q W ΔU = Q +
W
a +150 0 150 J
b +600 -240 360 J
c -300 0 -300
d 0 +90 +90
Total +450 -150
Step Q W ΔU = Q +
W
a +150 0 150 J
b +600 -240 360 J
c -300 0 -300
d 0 +90 +90
Total +450 -150 +300 J
Net Work
Done
-150J

10. Example 3
A series of thermodynamic processes is shown in the PV - diagram.
In process q) 100 J of heat is lost by the system, in process r) 400J
of heat is lost and in process s) 300J of heat is added to the system.
No heat is lost or gained in process t). Fill in the chart.
P
V
1 2
0 4
3
20
40
60
80
x 10-3
x
10
3
(s)
(t)
5
(r)
(q)
Conclusion – Heat energy has decreased by 200J
Work energy has increased by 160J
Internal Energy has decreased by 40J
Step Q W ΔU = Q +
W
q
r
s
t
Total
Step Q W ΔU = Q +
W
q -100
r -400 0
s +300
t 0 0 0
Total
Step Q W ΔU = Q +
W
q -100 -80
r -400 0
s +300
t 0 0 0
Total
Step Q W ΔU = Q +
W
q -100 -80
r -400 0
s +300 +240
t 0 0
Total
Step Q W ΔU = Q +
W
q -100 -80
r -400 0
s +300 +240
t 0 0 0
Total -200
Step Q W ΔU = Q +
W
q -100 -80
r -400 0
s +300 +240
t 0 0 0
Total -200 +160
Step Q W ΔU = Q +
W
q -100 -80 -180 J
r -400 0
s +300 +240
t 0 0 0
Total -200 +160
Step Q W ΔU = Q +
W
q -100 -80 -180 J
r -400 0 -400 J
s +300 +240
t 0 0 0
Total -200 +160
Step Q W ΔU = Q +
W
q -100 -80 -180 J
r -400 0 -400 J
s +300 +240 +540
t 0 0 0
Total -200 +160
Step Q W ΔU = Q +
W
q -100 -80 -180 J
r -400 0 -400 J
s +300 +240 +540
t 0 0 0
Total -200 +160 -40 J
Net Work
Done
160J

11. First Law of Thermodynamics
Question
1) A 2L sample of oxygen gas has an initial temperature and pressure of 200K and
1 atm. The gas undergoes four processes. Find the net work completed.
ab - heated at constant volume to 400K - ISOCHORIC
bc - heated at constant pressure to 800K - ISOBARIC
cd - cooled at constant volume back to 1atm - ISOCHORIC
da -cooled at constant pressure back to 200K - ISOBARIC
P
V
2
0
1
x
10
5
a (da) d
c
b
(cd)
(bc)
(ab)
x 10-3
PV Diagram

12. Using Gas Laws
Process ab – Constant Volume (ISOCHORIC) – Pressure Law
Add to graph
Process cab – Constant Pressure (ISOBARIC) – Charles Law
Add to graph

13. Using Characteristic Gas Law
Process cd – Constant Volume (ISOCHORIC) – Pressure Law
Using Gas Laws

14. P
2
0
1 a (da) d
c
b
(cd)
(bc)
(ab)
PV Diagram
2
4 V
x
10
5
x 10-3
Process ab – Constant Volume (ISOCHORIC)
Add to Table

15. Process bc – Constant Pressure (ISOBARIC)
Add to Table

16. Thermodynamic Processes - Isothermal
To keep the temperature constant both
the pressure and volume change to
compensate.
(Volume goes up, pressure goes down)
i.e. BOYLES’ LAW
P
V
0
2
1
Isotherm
Isothermal T1 = T2 (Constant Temperature)
V1 V2
T1 = T2
Q
-W
0

T 0


 U


V
Since
By
W
Q
U 


By
W
Q 

0
By
W
Q 
Net
Work
Done

17. Thermodynamic Processes - Isobaric
Heat is added to the gas which
increases the Internal Energy (U).
Work is done by the gas as it changes
in volume.
The path of an isobaric process is a
horizontal line called an isobar.
0
2
1
V1 V2
Isobar
T1
T2
T2>T1
Q
-W
Isobaric P1 = P2 (Constant Pressure)
By
W
Q
U 


P
V
Net Work
Done

18. Thermodynamic Processes – Isovolumetric
(Isochoric)
0
2
1
V1= V2
Isomet
T1
T2
T2>T1
P
V
Isovolumetric V1 = V2 (Constant Volume)
Q W = 0
0

V
Since
By
W
Q
U 


0


 Q
U
Q
U 


19. Thermodynamic Processes - Adiabatic
ADIABATIC- GREEK
(adiabatos- "impassable")
In other words, NO HEAT can leave
or enter the system, it is fully
insulated.
0 V1 V2
P
V
Adiabatic = Nothing is Constant
T1
T2
T1>T2
2
1 Adiabat
Q = 0
-W
0

Q
Since
By
W
Q
U 


By
W
U 

 0
By
W
U 



20. In Summary
0
P
V
(b)
(a)
(c)
(d)
Label Process Important Points Gas Law
a Isothermal Constant T, ΔU = Q , Q = WBy Boyles Law
b Isovolumetric (isochoric) Constant V, W = 0, ΔU = Q Charles Law
c Isobaric Constant P, ΔU = Q - WBy Pressure Law
d Adiabatic Nothing is Constant, Q = 0, ΔU = -WBy Combined Gas Law
Isobaric
Isothermal
Isochoric
Adiabatic