October 15, 2012

Week 4 Monday October 15, 2012 page 1
Chapter 3 (second law of thermodynamics)
First law: ΔU=q+w
ΔUsys + ΔUsurr = 0
closed system
conservation of energy
Places now restriction on transfer of energy
N2 + 3H2 ⥬ 2NH3
equilibrium? Second law answers question
Second law: Entropy (S)
state function
Clausius statement: (a cofounder of thermodynamics):
It is impossible to devise an engine which
working in a cycle shall produce no effect other than the transfer f heat from a colder to a hotter body.
So it takes energy to operate a refrigerator
Kelvin-Plank: It is impossible in a cyclic process to convert heat into work with 100% efficiency.
That would be perpetual motion of the second kind. The first kind would be violating the first law.
Heat engines: tao is any generic temperature scale.
Hot reservoir is τH. Cold reservoir is τC. +qH is heat going into system. –qC is heat leaving system.
Engine has a working substance (any substance)
Takes heat, performs work
Closed system
Could be molten Na in a nuclear reactor
Could be ammonia in refrigerator
For heat engine, qH > 0, w<0, qC<0.
e = efficiency of heat engine
𝑤𝑜𝑟𝑘 𝑜𝑢𝑡𝑝𝑢𝑡 −𝑤 |𝑤|
𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑛𝑝𝑢𝑡 𝑞𝐻 𝑞𝐻
For cyclic process: ΔU = 0 (first law)
q + w = qH + qC + w = 0
qH = -w + (-qC)
𝑠𝑜 𝑒 𝑖𝑠 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒
So the energy from qH is split between –w and –qC.
Heat res →cyclic machine (system) → work output = q
This machine violates the second law.
Experimental and theoretical evidence:
sea water → machine → work (useful)
𝑉 𝑖𝑠 𝑣𝑜𝑙𝑢𝑚𝑒
not possible
𝑙𝑖𝑞𝑢𝑖𝑑 ⥬ 𝑣𝑎𝑝𝑜𝑟
a. 𝑐𝑎𝑛 𝑏𝑒 𝑑𝑒𝑟𝑖𝑣𝑒𝑑 𝑓𝑟𝑜𝑚 𝑠𝑒𝑐𝑜𝑛𝑑 𝑙𝑎𝑤
b. ΔV = Vgas – Vliquid
In short:
1st law says you are not going to get something for nothing (ie you cannot win)
2nd law says you cannot break even
Efficiency of a heat engine:
−𝑤 |𝑤| 𝑞𝐻 + 𝑞𝐶
=1+ 𝐶
𝑞𝑤 𝑞𝐻
< 0 𝑞𝐶 < 0 𝑞𝐻 > 0
so e < 1 for a heat engine
That is base on this assumption: τH and τC will not change during process, τH > τC
Carnot’s principle (1825)
e = efficiency of any engine working in between τH and τC
e ≤ erev
Carnot’s principle
cannot have esuper ≥ erev