Week 5 Wednesday October 24, 2012 page 1
Negative pressure can be reported by a closed ended manometer because of “gauge pressure.”
Consider states 1,2,3,4 with the path from step 1 to 2 being irreversible, step 2 to 3 being a reversible
adiabat, step 3 to 4 being a reversible isotherm, and step 4 to 1 being a reversible adiabat.
We found q3→4 > 0 is impossible because that would violate the 2nd law.
We found q3→4 = 0 is impossible because then points 3 and 4 would be the same point.
Therefore q3→4 < so, which makes S2 – S1 > 0 and ΔSsys > 0.
ΔSsys > 0 for any irreversible, adiabatic, closed system
How about reversible adiabatic process?
Then ΔSsys = 0 since dqrev = 0.
Reversible isothermal process:
ΔSsys < 0 if q < 0
ΔSsys > 0 if q > 0
For any isolated system:
ΔSsys > 0 for any irreversible, adiabatic, isolated system
ΔSsys + ΔSsurr = ΔSuniv > 0 irreversible process, adiabatic system
So the entropy of the universe constantly increases.
ΔSuniv = 0
for reversible adiabatic process
ΔSuniv > 0
for irreversible adiabatic process
ΔSuniv ≥ 0
for any process
Entropy can be created but not destroyed.
2nd law:
𝑑𝑞𝑟𝑒𝑣
𝑇
is the differential of a state function S such that ΔSuniv ≥ 0 for any process.
Reversible : idealization
Irreversible: real process
Entropy and equilibrium:
Start with T2 on left side of system and T1 on right side of system, with T2 > T1. Eventually, while S
increases then levels off, the system will be in thermal equilibrium, with the whole system at T.
2H2(g) + O2(g) ⇌ 2H2O(g)
Thermodynamics can’t explain how fast we reach equilibrium.
C(diamond) → C(graphite)
Thermodynamics can’t tell rate of change.
What is entropy?
zeroth law: T T α ave molecular kinetic energy
first law: U U α total molecular energy
second law: S
Sα?
Consider molecules e and d in an isolated container. Start with all e on the left side and all d on the right
side. This is state 1, a transition state, and is less probable. Wait a while until d and e are completely
mixed. This is state 2, the equilibrium state, which is more probable.
Consider a container with an inert molecule in it. The probability P of finding the molecule in the
container is 1. Divide the container in half. The probability P of finding the molecule on the left half of
1
2
the container is . If there are two inert molecules in the container, the probability of finding both in the
1 2
2
left half of the container is ( ) . If there are three inert molecules in the container, the probability of
1 3
finding all three of them in the left half of the container is (2) . If there are Avagadro’s number (NA)
1 𝑁𝐴
molecules in the containers, there is P = (2)
≈ 0 probability of finding all the molecules in the left
half of the container.
Increasing volume increases entropy (always).
So S = f(P) f is an unknown function and P is probability
What is f?
Consider a container with two parts. We’ll call the left part 1 and the left part 2. The container has a
rigid, adiabatic, impermeable wall.
S1+2 = total entropy = S1 + S2 since we know S is extensive
h(P1+2) = f(P1) + g(P2)
h(xy) = f(x) + g(y)
S = klnP + a
just like log(a*b) = log(a) + log(b)
k,a are both constants
a can’t be determined, so thermodynamics can find ΔS but not S.
h(xy) = kln(xy) + c
f(x) = kln(x) + a different a from above
g(y) = kln(y) + b
c=a+b
To find value for k1, first take system in unmixed state: T,P (pressure) ,V,S, P1 (probability)
After mixing we have: T,P,V,S, P2
ΔS = S2 – S1 = (klnP2 + a) – (klnP1 + a)
same a’s, so they cancel
𝑃
∆𝑆 = 𝑘𝑙𝑛 ( 2)
𝑃1
𝑃 𝑖𝑠 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦, 𝑛𝑜𝑡 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒
1 𝑁𝑑 1 𝑁𝑒
1 𝑁𝑑+𝑁𝑒
𝑃1 = ( ) ( ) = ( )
2
2
2
𝑁 𝑖𝑠 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠
To simplify, assume Nd = Ne
P2  1
∆𝑆 = 𝑘𝑙𝑛 (
1
1 2𝑁𝑑
(2)
) = 𝑘𝑙𝑛2𝑁𝑑 = 2𝑁𝑑𝑘𝑙𝑛2
ΔS = 2Ndkln2 call this equation I
ΔS = -Rnalnxa + nblnxb from before
Since we assumed Na = Nb then xa = xb = .5
ΔS = -R(naln.5) = 2Rnaln2
ΔS = 2Rnaln2
call this equation II and set I = II
2Ndkln2 = 2Rnaln2
Ndk = Rnd
𝑛
1
𝑘=𝑅 𝑑 =𝑅
𝑁𝑑
𝑁𝐴
R and NA are fundamental constants, so k is too.
𝐽
𝐽
𝐽
𝑚𝑜𝑙
𝐾
𝑘=
= 1.38 ∗ 10−23
= 1.38 ∗ 10−23
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 𝐾
𝑎𝑡𝑜𝑚 𝐾
6.02 ∗ 1023
𝑚𝑜𝑙
8.314
k is Boltzmann’s constant
S = klnw
w is microstates
w α probability