Problem Set #6, Problem #4
The First law states that dE = † Q − † W , (1) and the trick is to get the sign of the work term correct.
I will regard the tension in the rubber band as positive. Thus, when I stretch the rubber band to increase its length L , I (the surroundings) am doing positive work on the rubber band. If the surroundings do positive work on a system, the system is doing negative work on the surroundings, and the system’s energy must increase.
We must thus have
† W = − F dL , (2) and when we combine this with Eq.(1) we obtain dE = † Q + F dL .
(3)
From Eq.(3) we see that an increase in L results in an increase of E.
Now integrate Eq.(2) from L
1 to L
2 to fi nd
Z
L
2
Z
L
2
W = −
W = − k
2
F dL = − k
L
1 h
( L
2
− L
0
)
2
− (
L
1
L
1
( L
−
−
L
0
)
L
2
0 i
) dL
.
= − k
Z
L
2
− L
0
L
1
− L
0 ydy (4)
(5)
If the work is performed adiabatically, we have † Q = 0 , and it follows from
Eq.(3) that
∆ E = − W = k
2 h
( L
2
− L
0
)
2
− ( L
1
− L
0
)
2 i
.
(6)
From this result, you can easily verify that ∆ E is negative when L
1
> L
2
, i.e., when the rubber band contracts.
This is reasonable.
A rubber band that contracts under tension is doing positive work on the surroundings (e.g., lifting a weight).
Under adiabatic conditions, the rubber band’s energy decreases when it does positive work, W > 0 as seen from Eq.(5).
1