Lecture 15 first law of thermodynamics

1. Lecture 15
First law of thermodynamics.

2. Internal energy
State variables: p, T, V
They only depend on the “state” of the system
They do NOT depend on the “history”
Total energy of the molecules is also a state variable:
Internal energy: U = KE all + PE all
KE : kinetic energy of atoms, random motion
PE : interactions, e.g. attraction between molecuses
For ideal gases, no interaction: U = KE all

3. Internal energy for some systems
Nk = nNAk = nR
Monoatomic ideal gas:
3

U = KE all = N  kT ÷
2

3
U = nRT
2
Diatomic ideal gas:
5

U = KE all = N  kT ÷
2

U =
Monoatomic solid crystal: U = KE all + PE all
3

= N  kT ÷× 2
2

5
nRT
2
U = 3nRT
In general, internal energy can also depend on p, V. But it never
depends on the history of the system.

4. Changes in internal energy
Temperature change is associated to heat transfer.
Change in internal energy
transfer of energy
We already know that:
When heat is absorbed/released by a system, its internal
energy increases/decreases.
Other ways of transferring energy?
Work!! (we learned it in 221)

5. Work done by gas volume change
A gas with pressure p expands by
pushing a piston by a distance dx
p
Force by gas on piston F = pA
(A = area of the piston)
dx
Work by gas:
dW = pAdx = pdV
As volume goes from Vi to Vf
Vf
W = ∫ pdV
Vi

6. Work is the area under pV curve
But be careful with the sign!
Expansion
(Work done by gas) > 0
Compression
(Work done by gas) < 0

7. ACT: Three processes - Work
A gas can go from state 1 to state 2 through three different
processes. In which process does gas do the least amount of work ?
A
C
B
D: It’s the same for
all three

8. Even if the initial and final states are the same, work
depends on the path taken by the process
A
B
W is NOT a state variable.
It is only defined when the state changes.
C

9. ACT: Three processes – Internal energy
In which case does the internal energy of the gas change the most ?
A
D: It’s the same for
all three
C
B
U is a state function and
does not depend on the
process.
ΔU = U1 − U2 for all
processes.

10. In-class example: Work in closed cycle
Which of these processes represents the most work done by
the system per cycle?
P
P
P
V
V
A
C
V
E
P
P
Work = area inside cycle.
Note that WE < 0
V
V
B
D
CW ⇔ W > 0
CCW ⇔ W < 0

11. First law of thermodynamics
If heat is absorbed by a system,
Q>0
ΔU > 0
If heat is released by a system,
Q<0
ΔU < 0
W>0
ΔU < 0
If work is done on the gas (compression), W < 0
ΔU > 0
If work is done by the gas (expansion),
∆U = Q −W
Remember the sign convention!
W > 0 work done by system
Q > 0 heat absorbed by system

12. Example: Cyclic process
A gas (not necessarily ideal) goes through the cycle shown in the pV
diagram below.
A
Process 2
3
Process 1
B
Pr
oc
es
s
P
Data:
C
VA = 2.0 m3
VC = 4.0 m3
PA = 1.0 × 105 Pa
PC = 2.0 × 105 Pa
a) Determine the work done by the gas in
each of the three parts of the cycle.
V
1 (A to B): W1 = 0 (constant volume)
5
2 (B to C): W2 = p ∆V = pB (VC −VB ) = 2 × 10 J
5
3 (C to A): W2 = − ( yellow area ) = −3 × 10 J

13. P
A
C
3
Process 2
Pr
oc
es
s
Process 1
B
Data:
V
VA = 2.0 m3
VC = 4.0 m3
PA = 1.0 × 105 Pa
PC = 2.0 × 105 Pa
b) For the entire cycle, what are the work
done by the gas, the change in internal
energy of the gas and the heat exchanged
with the surroundings? Is this heat
absorbed or released by the gas?
Wcycle = W1 +W2 +W3 = −1 × 105 J
∆Ucycle = Uf − Ui = UA − UA = 0
ΔUclosed cycle = 0
Qcycle = ∆Ucycle +Wcycle = −1 × 105 J
Q < 0: heat is released by the gas

14. Overall, in this cycle:
Wcycle < 0
Qcycle < 0
work
Work done on the system.
System releases heat.
system
P
B
C
heat
A
We do work on the system, we obtain heat.
(This could be used to warm up a room…)
Reverse cycle (A→C →B):
Wcycle > 0
Work done by the system.
Qcycle > 0
System absorbs heat.
work
system
heat
System absorbs heat and produces work.
(Like some kind of steam motor…)
V
P
B
C
A
V