Week 4 Friday October 19, 2012 page 1
∮
𝑑𝑞𝑟𝑒𝑣
=0
𝑇
𝑓𝑜𝑟 𝐶𝑎𝑟𝑛𝑜𝑡 𝑐𝑦𝑐𝑙𝑒 (𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑏𝑙𝑒)
mnsrm is a cycle (just like Carnot cycle)

2 reversible adiabats and 2 reversible isotherms
𝑞𝑎𝑏 𝑞𝑑𝑐
+
=0
𝑇𝑚𝑛 𝑇𝑠𝑟
divide cycle into infinite number of adiabats
Ta ≈ Tb for infinite number of adiabats
likewise for Tmn ≈ Tab and Tsr ≈ Tdc
𝑑𝑞𝑎𝑏 𝑑𝑞𝑑𝑐
+
=0
𝑇𝑎𝑏
𝑇𝑑𝑐
∮
𝑑𝑞𝑟𝑒𝑣
=0
𝑇
𝑇 𝑖𝑠 𝑡𝑒𝑚𝑝 𝑎𝑡 𝑤ℎ𝑖𝑐ℎ ℎ𝑒𝑎𝑡 𝑖𝑠 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟𝑟𝑒𝑑
being = 0 means it’s a state function
𝑑𝑆 =
𝑑𝑞𝑟𝑒𝑣
𝑇
𝑆 𝑖𝑠 𝑒𝑛𝑡𝑟𝑜𝑝𝑦
𝑇 𝑖𝑠 𝑖𝑛 𝐾𝑒𝑙𝑣𝑖𝑛𝑠
𝑞
∆𝑆 = 𝑟𝑒𝑣
𝑇
Why q instead of w?
w: organized motion in surroundings, orderly motion of atoms or molecules in the surroundings
q: disorganized motion, disorderly motion of atoms or molecules in the surroundings
T takes into account entropy that’s already in the system.
Adding heat at low temperature is analogous to a sneeze in a quiet library:
100𝑘𝐽 ∗ 103
∆𝑆 =
𝐽
𝑘𝐽
273𝐾
= +366
𝐽
𝐾
Adding heat at high temperature is analogous to a sneeze in a noisy city:
100𝑘𝐽 ∗ 103
∆𝑆 =
373𝐾
𝐽
𝑘𝑗
= +268
𝐽
𝐾
2
∆𝑆 = 𝑆2 − 𝑆1 = ∫
1
𝑑𝑞𝑟𝑒𝑣
𝑇
𝑐𝑙𝑜𝑠𝑒𝑑 𝑠𝑦𝑠𝑡𝑒𝑚, 𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑏𝑙𝑒 𝑝𝑟𝑜𝑐𝑒𝑠𝑠
Any isolated system is closed so it applies to an isolated system too.
∆Sirrev = ∆Srev
note: S is an extensive state function
prove: partitioned container
𝑙𝑒𝑓𝑡 𝑝𝑎𝑟𝑡𝑖𝑡𝑖𝑜𝑛: 𝑑𝑆1 =
𝑑𝑞1
𝑇
𝑟𝑖𝑔ℎ𝑡 𝑝𝑎𝑟𝑡𝑖𝑡𝑖𝑜𝑛: 𝑑𝑆2 =
𝑑𝑞2
𝑇
dS = dS1 + dS2
∆S = ∆S1 + ∆S2
S = S1 + S2
so S is extensive
𝑆
𝐽
𝑐𝑎𝑙
𝑢𝑛𝑖𝑡𝑠: 𝑆𝑚 =
𝑖𝑛
𝑜𝑟
𝑛
𝑚𝑜𝑙 𝐾
𝑚𝑜𝑙 𝐾
𝑆 𝑖𝑛
𝐽
𝑐𝑎𝑙
𝑜𝑟
𝐾
𝐾
Calculation of entropy changes:
𝑑𝑆 =
𝑑𝑞𝑟𝑒𝑣
𝑇
2
∆𝑆 = ∫ 𝑑𝑞𝑟𝑒𝑣
1
𝑇
∆𝑆𝑟𝑒𝑣 = ∆𝑆𝑖𝑟𝑟𝑒𝑣
1. Cyclic process ∆S=0 System (not surroundings) is always implied
2. rev adiabatic process dqrev=0 so ∆S=0
3. rev phase change at constant T,P
2 𝑑𝑞𝑟𝑒𝑣
a. ∆𝑆 = ∫1
𝑇
1
= ∫ 𝑑𝑞𝑟𝑒𝑣 =
𝑇
𝑞𝑟𝑒𝑣
𝑇
=
𝑞𝑃
𝑇
=
∆𝐻
𝑇
b. ∆H is latent heat of transition
c. If ∆H>0 then ∆S>0. If ∆H<0 then ∆S<0.
4. rev isothermal process
a. T is constant
1
b. ∆𝑆 = 𝑇 ∫ 𝑑𝑞𝑟𝑒𝑣 =
c. ∆𝑆 =
𝑞𝑟𝑒𝑣
𝑇
𝑞𝑟𝑒𝑣
𝑇
𝑆 𝑖𝑠 𝑛𝑜𝑡 𝑣𝑒𝑟𝑦 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑠𝑒𝑛𝑠𝑖𝑡𝑖𝑣𝑒
5. rev change of state of a perfect gas (V1T1 → V2T2)
a. dqrev=dU-dwrev
1st law
b. 𝑑𝑞𝑟𝑒𝑣 = 𝐶𝑉𝑑𝑇 + 𝑃𝑑𝑉 = 𝐶𝑉𝑑𝑇 +
c.
𝑑𝑞𝑟𝑒𝑣
𝑇
=
𝐶𝑉𝑑𝑇
𝑇
+
𝑛𝑅
𝑑𝑉
𝑉
𝑛𝑅𝑇
𝑑𝑉
𝑉
𝑡ℎ𝑒𝑛 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑒
𝑡ℎ𝑒𝑛 𝑑𝑖𝑣𝑖𝑑𝑒 𝑏𝑦 𝑇
d. ∫
𝑑𝑞𝑟𝑒𝑣
𝑇
e. ∆𝑆 =
f.
𝐶𝑉(𝑇)𝑑𝑇
𝑛𝑅
+ ∫ 𝑉 𝑑𝑉
𝑇
2 𝐶 (𝑇)
𝑉
∫1 𝑉𝑇 𝑑𝑇 + 𝑛𝑅𝑙𝑛 𝑉2
1
=∫
Now assume CV is constant over temperature range:
𝑇
𝑉
g. ∆𝑆 ≈ 𝐶𝑉𝑙𝑛 𝑇2 + 𝑛𝑅𝑙𝑛 𝑉2
1
1
h. So expansion increases entropy and heating increases entropy.
i. ∆S↑ when T↑ or when V↑
6. irrev change of state of a perfect gas
a. ∆Sirrev = ∆Srev
but qrev ≠ qirrev since T plays role
7. constant pressure heating (no phase change)
a. dqrev=dqP=CPdT=TdS
b. 𝑑𝑆 =
c. ∆𝑆 =
𝑑𝑞𝑟𝑒𝑣
𝐶 𝑑𝑇
= 𝑃𝑇
𝑇
𝑇 𝐶 (𝑇)
∫𝑇 2 𝑃𝑇 𝑑𝑇
1
d. For constant CP:
𝑇 𝑑𝑇
𝑇
= 𝐶𝑃𝑙𝑛 𝑇2
𝑇
1
1
𝑇
𝐶𝑃𝑙𝑛 𝑇2 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑃 𝑎𝑛𝑑
1
e. ∆𝑆 = 𝐶𝑃 ∫𝑇 2
f.
∆𝑆 =
𝐶𝑃, 𝑛𝑜 𝑝ℎ𝑎𝑠𝑒 𝑐ℎ𝑎𝑛𝑔𝑒, 𝑟𝑒𝑣 𝑜𝑟 𝑖𝑟𝑟𝑒𝑣
8. general change of state (P1T1 → P2T2)
a. ∆S=? wait until chapter 4
b. For problem 31, use PV=nRT
9. irreversible phase change
a. We want to go from liquid H2O at -10˚C and 1atm straight to ice at -10˚C and 1 atm,
which is irreversible.
b. Instead, we’ll add two intermediate steps:
i. liquid H2O at 0˚C and 1 atm
ii. solid H2O at 0˚C and 1 atm
c. ∆Srev is the reversible step from liquid H2O at -10˚C and 1atm to liquid H2O at 0˚C and 1
atm.
d. ∆S’rev is the reversible step from liquid H2O at 0˚C and 1 atm to solid H2O at 0˚C and 1
atm.
e. ∆S”rev is the reversible step from solid H2O at 0˚C and 1 atm to ice at -10˚C and 1 atm.
f. Since S is extensive: ∆S=∆Srev + ∆S’rev + ∆S”rev
𝑇
g. ∆𝑆𝑟𝑒𝑣 = 𝐶𝑃𝑙𝑛 𝑇2
1
𝐶𝑃 𝑓𝑜𝑟 𝑙𝑖𝑞𝑢𝑖𝑑 𝑤𝑎𝑡𝑒𝑟
h. ∆𝑆′𝑟𝑒𝑣 =
=
i.
∆𝑆"𝑟𝑒𝑣
∆𝐻𝑓𝑟𝑒𝑒𝑧𝑖𝑛𝑔
𝑇
𝑇
= 𝐶𝑃𝑙𝑛 2
𝑇1
−∆𝐻𝑚𝑒𝑙𝑡𝑖𝑛𝑔
𝑇
𝐶𝑃 𝑓𝑜𝑟 𝑠𝑜𝑙𝑖𝑑 𝑤𝑎𝑡𝑒𝑟