Week 3 Friday October 12, 2012 page 1
𝑅
𝑇2
𝑉
= ( 1)𝐶𝑉,𝑚
𝑇1
𝑉2
𝑃1𝑉1 𝑃2𝑉2
=
𝑓𝑜𝑟 𝑝𝑒𝑟𝑓𝑒𝑐𝑡 𝑔𝑎𝑠
𝑇1
𝑇2
𝑅
𝑇2 𝑃2𝑉2
=
𝑇1 𝑃1𝑉1
𝑅
𝑃 𝑉
𝑉
𝑉 𝐶𝑉 𝑚
𝑠𝑜: 2 2 = ( 1)𝐶𝑉,𝑚 = 1 𝑅,
𝑃1𝑉1
𝑉2
𝑉2𝐶𝑉,𝑚
Assume CV,m is constant.
𝑅
𝑃1𝑉1𝐶𝑉,𝑚
𝑅
𝐶𝑉 𝑚
,
+1
𝑅
= 𝑃2𝑉2𝐶𝑉,𝑚
+1
𝑅 + 𝐶𝑉 𝑚 𝐶𝑃 𝑚
, =
,
𝑠𝑖𝑛𝑐𝑒 𝐶𝑃 𝑚 − 𝐶𝑉 𝑚 = 𝑅
,
,
𝐶𝑉 𝑚
𝐶𝑉 𝑚
,
,
𝐶
𝛾 = 𝑃, 𝑚 𝛾 𝑖𝑠 𝑔𝑎𝑚𝑚𝑎 (𝑛𝑜𝑡 𝑟𝑎𝑑𝑖𝑎𝑡𝑖𝑜𝑛)
𝐶𝑉 𝑚
,
+1 =
P1V1ᵞ=P2V2ᵞ
Adiabatic process of increasing volume ends with lower pressure since the final temperature is lower.
Lower T → Lower P
q = 0 since adiabatic so ∆U=w
∆𝑈 = 𝐶𝑉 ∫ 𝑑𝑇 = 𝑤
𝑇2
∆𝐻 = ∫ 𝐶𝑃𝑑𝑇 = 𝐶𝑃(𝑇2 − 𝑇1)
𝑇1
𝑎𝑠𝑠𝑢𝑚𝑖𝑛𝑔 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝐶𝑃
∆𝐻 = ∆𝑈 + ∆(𝑃𝑉) = ∆𝑈 = 𝑛𝑅∆𝑇 𝑝𝑒𝑟𝑓𝑒𝑐𝑡 𝑔𝑎𝑠
1.
Cyclic process
a. ∆H=0 ∆T=0 ∆U=0 ∆P=0 ∆V=0
b. ∮UdU means cyclic integral
c. ∮UdU=0 since U is a state function
d. ∮dq isn’t necessarily 0 since q isn’t a state function
2. Reversible phase change at constant T and P
a. Latent heat (of evaporation or melting…)
b. q>0 if it’s melting
c. q is latent heat (experiment, not from thermodynamics)
d. q<0 if it’s freezing
e. w=-P∆V
f.
∆𝑉 ∝
𝑚
𝜌
g. ∆H=qP
h. ∆U=qP+w
3. Constant pressure heating with no phase change
2
a. 𝑤 = 𝑤𝑟𝑒𝑣 = − ∫1 𝑃𝑑𝑉 = −𝑃∆𝑉
𝑚
𝜌
𝑛𝑅𝑇
𝑃
b. ∆𝑉 ∝
c. 𝑉 =
𝑓𝑜𝑟 𝑠𝑜𝑙𝑖𝑑 𝑜𝑟 𝑙𝑖𝑞𝑢𝑖𝑑
𝑓𝑜𝑟 𝑝𝑒𝑟𝑓𝑒𝑐𝑡 𝑔𝑎𝑠 𝑠𝑜 ∆𝑉 = 𝑉2 − 𝑉1
𝑇
d. ∆𝐻 = 𝑞𝑃 = ∫𝑇 2 𝐶𝑃(𝑇)𝑑𝑇
𝐶𝑃(𝑇)𝑖𝑠 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑡𝑜 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒 𝑎𝑛𝑑 𝑝ℎ𝑎𝑠𝑒
1
e. ∆U=qP+w
f. These formulas work as long as P1=P2 , not just constant P.
4. Constant volume heating with no phase change
a. V=constant ∆V=0 w=0 ∆U=qV
𝑇
b. ∆𝑈 = ∫𝑇 2 𝐶𝑉𝑑𝑇 = 𝑞𝑉
1
c. ∆H=∆U+∆(PV)=∆U+V∆P=∆U+nR∆T only for perfect gas
5. Perfect gas change of state (T1V1 → T2V2)
a. dU=CVdT perfect gas
𝑇
b. ∆𝑈 = ∫𝑇 2 𝐶𝑉𝑑𝑇
1
c. 𝑑𝐻 = 𝐶𝑃𝑑𝑇
𝑇
d. ∆𝐻 = ∫𝑇 2 𝐶𝑃𝑑𝑇
1
𝑇
e. 𝑤 = −𝑃𝑑𝑉 = −𝑛𝑅 ∫ 𝑉 𝑑𝑉
𝑙𝑖𝑛𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙, 𝑛𝑒𝑒𝑑 𝑇 = 𝑇(𝑉), 𝑇&𝑉 𝑛𝑜𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
f. q=∆U-w
6. Reversible isothermal process of a perfect gas
a. T=constant
b. ∆U=0 since U=U(T) for a perfect gas
c. ∆H=0
2
𝑉
𝑉
d. 𝑤 = − ∫1 𝑃𝑑𝑉 = −𝑛𝑅𝑇𝑙𝑛 𝑉2 = 𝑛𝑅𝑇𝑙𝑛 𝑉1
1
e. q=-w
7. Reversible adiabatic process of a perfect gas
a. q=0 ∆U=w
𝑇
b. ∆𝑈 = ∫𝑇 2 𝐶𝑉𝑑𝑇
𝑇
𝑇1
1
𝑉 𝐶𝑉,𝑚
𝑅
𝑉2
c. ( 2) = ( 1)
2
𝑇
∆𝐻 = ∫𝑇 2 𝐶𝑃𝑑𝑇
1
𝑎𝑠𝑠𝑢𝑚𝑖𝑛𝑔 𝐶𝑉 𝑚 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
,
8. Adiabatic expansion of a perfect gas into a vacuum (free expansion)
a. w=0 q=0 ∆U=0 first law
b. ∆H=∆U+∆(PV)=∆U+nR∆T=0
c. also U=U(T) and ∆U=0 so ∆T=0
The molecular nature of internal energy U
U=Utran+Urot+Uvib+Uelec+Uintermolecular+Urest mass
Utran+Urot+Uvib are functions of T
Uelec is constant at moderate T (<5000K)
Urest mass is constant
Motion in 3 dimensions: each axis is a degree of freedom
thermodynamics is based on classical mechanics
𝑥:
1
𝑅𝑇
2
𝑦:
1
𝑅𝑇
2
𝑧:
1
𝑅𝑇
2
𝑅 = 8.314
𝐽
𝑚𝑜𝑙 𝐾
N is the # of atoms per molecule
species
Translation
Rotation
Vibration(1500K or
higher)
Atom
1.5 RT
0
0
Linear molecule
1.5 RT
RT
(3N-5)RT
Nonlinear molecule
1.5 RT
1.5 RT
(3N-6)RT
For linear molecule, the rotation only has 2 degrees of freedom because rotation about the nuclear axis
doesn’t contribute.
N2,O2 and other diatomic linear molecules
Um=1.5RT + RT + RT = 3.5RT (translation + rotation + vibration)
𝜕𝑈
𝜕
(3.5𝑅𝑇) = 3.5𝑅
𝐶𝑉 𝑚 = ( 𝑚 ) =
,
𝜕𝑇 𝑉 𝜕𝑇
CO2 linear molecule but not diatomic
Um=1.5RT+RT+4RT=6.5RT (translation + rotation + vibration)
CV,m=6.5R
𝐶𝑉 𝑚
, = 6.5
𝑅
𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑏𝑦 𝑐𝑙𝑎𝑠𝑠𝑖𝑐𝑎𝑙 𝑝ℎ𝑦𝑠𝑖𝑐𝑠