October 12, 2012

advertisement
Week 3 Friday October 12, 2012 page 1
๐‘…
๐‘‡2
๐‘‰
= ( 1)๐ถ๐‘‰,๐‘š
๐‘‡1
๐‘‰2
๐‘ƒ1๐‘‰1 ๐‘ƒ2๐‘‰2
=
๐‘“๐‘œ๐‘Ÿ ๐‘๐‘’๐‘Ÿ๐‘“๐‘’๐‘๐‘ก ๐‘”๐‘Ž๐‘ 
๐‘‡1
๐‘‡2
๐‘…
๐‘‡2 ๐‘ƒ2๐‘‰2
=
๐‘‡1 ๐‘ƒ1๐‘‰1
๐‘…
๐‘ƒ ๐‘‰
๐‘‰
๐‘‰ ๐ถ๐‘‰ ๐‘š
๐‘ ๐‘œ: 2 2 = ( 1)๐ถ๐‘‰,๐‘š = 1 ๐‘…,
๐‘ƒ1๐‘‰1
๐‘‰2
๐‘‰2๐ถ๐‘‰,๐‘š
Assume CV,m is constant.
๐‘…
๐‘ƒ1๐‘‰1๐ถ๐‘‰,๐‘š
๐‘…
๐ถ๐‘‰ ๐‘š
,
+1
๐‘…
= ๐‘ƒ2๐‘‰2๐ถ๐‘‰,๐‘š
+1
๐‘… + ๐ถ๐‘‰ ๐‘š ๐ถ๐‘ƒ ๐‘š
, =
,
๐‘ ๐‘–๐‘›๐‘๐‘’ ๐ถ๐‘ƒ ๐‘š − ๐ถ๐‘‰ ๐‘š = ๐‘…
,
,
๐ถ๐‘‰ ๐‘š
๐ถ๐‘‰ ๐‘š
,
,
๐ถ
๐›พ = ๐‘ƒ, ๐‘š ๐›พ ๐‘–๐‘  ๐‘”๐‘Ž๐‘š๐‘š๐‘Ž (๐‘›๐‘œ๐‘ก ๐‘Ÿ๐‘Ž๐‘‘๐‘–๐‘Ž๐‘ก๐‘–๐‘œ๐‘›)
๐ถ๐‘‰ ๐‘š
,
+1 =
P1V1แตž=P2V2แตž
Adiabatic process of increasing volume ends with lower pressure since the final temperature is lower.
Lower T → Lower P
q = 0 since adiabatic so โˆ†U=w
โˆ†๐‘ˆ = ๐ถ๐‘‰ ∫ ๐‘‘๐‘‡ = ๐‘ค
๐‘‡2
โˆ†๐ป = ∫ ๐ถ๐‘ƒ๐‘‘๐‘‡ = ๐ถ๐‘ƒ(๐‘‡2 − ๐‘‡1)
๐‘‡1
๐‘Ž๐‘ ๐‘ ๐‘ข๐‘š๐‘–๐‘›๐‘” ๐‘๐‘œ๐‘›๐‘ ๐‘ก๐‘Ž๐‘›๐‘ก ๐ถ๐‘ƒ
โˆ†๐ป = โˆ†๐‘ˆ + โˆ†(๐‘ƒ๐‘‰) = โˆ†๐‘ˆ = ๐‘›๐‘…โˆ†๐‘‡ ๐‘๐‘’๐‘Ÿ๐‘“๐‘’๐‘๐‘ก ๐‘”๐‘Ž๐‘ 
1.
Cyclic process
a. โˆ†H=0 โˆ†T=0 โˆ†U=0 โˆ†P=0 โˆ†V=0
b. โˆฎUdU means cyclic integral
c. โˆฎUdU=0 since U is a state function
d. โˆฎdq 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=-Pโˆ†V
f.
โˆ†๐‘‰ ∝
๐‘š
๐œŒ
g. โˆ†H=qP
h. โˆ†U=qP+w
3. Constant pressure heating with no phase change
2
a. ๐‘ค = ๐‘ค๐‘Ÿ๐‘’๐‘ฃ = − ∫1 ๐‘ƒ๐‘‘๐‘‰ = −๐‘ƒโˆ†๐‘‰
๐‘š
๐œŒ
๐‘›๐‘…๐‘‡
๐‘ƒ
b. โˆ†๐‘‰ ∝
c. ๐‘‰ =
๐‘“๐‘œ๐‘Ÿ ๐‘ ๐‘œ๐‘™๐‘–๐‘‘ ๐‘œ๐‘Ÿ ๐‘™๐‘–๐‘ž๐‘ข๐‘–๐‘‘
๐‘“๐‘œ๐‘Ÿ ๐‘๐‘’๐‘Ÿ๐‘“๐‘’๐‘๐‘ก ๐‘”๐‘Ž๐‘  ๐‘ ๐‘œ โˆ†๐‘‰ = ๐‘‰2 − ๐‘‰1
๐‘‡
d. โˆ†๐ป = ๐‘ž๐‘ƒ = ∫๐‘‡ 2 ๐ถ๐‘ƒ(๐‘‡)๐‘‘๐‘‡
๐ถ๐‘ƒ(๐‘‡)๐‘–๐‘  ๐‘ ๐‘๐‘’๐‘๐‘–๐‘“๐‘–๐‘ ๐‘ก๐‘œ ๐‘ ๐‘ข๐‘๐‘ ๐‘ก๐‘Ž๐‘›๐‘๐‘’ ๐‘Ž๐‘›๐‘‘ ๐‘โ„Ž๐‘Ž๐‘ ๐‘’
1
e. โˆ†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 โˆ†V=0 w=0 โˆ†U=qV
๐‘‡
b. โˆ†๐‘ˆ = ∫๐‘‡ 2 ๐ถ๐‘‰๐‘‘๐‘‡ = ๐‘ž๐‘‰
1
c. โˆ†H=โˆ†U+โˆ†(PV)=โˆ†U+Vโˆ†P=โˆ†U+nRโˆ†T only for perfect gas
5. Perfect gas change of state (T1V1 → T2V2)
a. dU=CVdT perfect gas
๐‘‡
b. โˆ†๐‘ˆ = ∫๐‘‡ 2 ๐ถ๐‘‰๐‘‘๐‘‡
1
c. ๐‘‘๐ป = ๐ถ๐‘ƒ๐‘‘๐‘‡
๐‘‡
d. โˆ†๐ป = ∫๐‘‡ 2 ๐ถ๐‘ƒ๐‘‘๐‘‡
1
๐‘‡
e. ๐‘ค = −๐‘ƒ๐‘‘๐‘‰ = −๐‘›๐‘… ∫ ๐‘‰ ๐‘‘๐‘‰
๐‘™๐‘–๐‘›๐‘’ ๐‘–๐‘›๐‘ก๐‘’๐‘”๐‘Ÿ๐‘Ž๐‘™, ๐‘›๐‘’๐‘’๐‘‘ ๐‘‡ = ๐‘‡(๐‘‰), ๐‘‡&๐‘‰ ๐‘›๐‘œ๐‘ก ๐‘๐‘œ๐‘›๐‘ ๐‘ก๐‘Ž๐‘›๐‘ก
f. q=โˆ†U-w
6. Reversible isothermal process of a perfect gas
a. T=constant
b. โˆ†U=0 since U=U(T) for a perfect gas
c. โˆ†H=0
2
๐‘‰
๐‘‰
d. ๐‘ค = − ∫1 ๐‘ƒ๐‘‘๐‘‰ = −๐‘›๐‘…๐‘‡๐‘™๐‘› ๐‘‰2 = ๐‘›๐‘…๐‘‡๐‘™๐‘› ๐‘‰1
1
e. q=-w
7. Reversible adiabatic process of a perfect gas
a. q=0 โˆ†U=w
๐‘‡
b. โˆ†๐‘ˆ = ∫๐‘‡ 2 ๐ถ๐‘‰๐‘‘๐‘‡
๐‘‡
๐‘‡1
1
๐‘‰ ๐ถ๐‘‰,๐‘š
๐‘…
๐‘‰2
c. ( 2) = ( 1)
2
๐‘‡
โˆ†๐ป = ∫๐‘‡ 2 ๐ถ๐‘ƒ๐‘‘๐‘‡
1
๐‘Ž๐‘ ๐‘ ๐‘ข๐‘š๐‘–๐‘›๐‘” ๐ถ๐‘‰ ๐‘š ๐‘–๐‘  ๐‘๐‘œ๐‘›๐‘ ๐‘ก๐‘Ž๐‘›๐‘ก
,
8. Adiabatic expansion of a perfect gas into a vacuum (free expansion)
a. w=0 q=0 โˆ†U=0 first law
b. โˆ†H=โˆ†U+โˆ†(PV)=โˆ†U+nRโˆ†T=0
c. also U=U(T) and โˆ†U=0 so โˆ†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
๐‘ฅ:
1
๐‘…๐‘‡
2
๐‘ฆ:
1
๐‘…๐‘‡
2
๐‘ง:
1
๐‘…๐‘‡
2
๐‘… = 8.314
๐ฝ
๐‘š๐‘œ๐‘™ ๐พ
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)
๐œ•๐‘ˆ
๐œ•
(3.5๐‘…๐‘‡) = 3.5๐‘…
๐ถ๐‘‰ ๐‘š = ( ๐‘š ) =
,
๐œ•๐‘‡ ๐‘‰ ๐œ•๐‘‡
CO2 linear molecule but not diatomic
Um=1.5RT+RT+4RT=6.5RT (translation + rotation + vibration)
CV,m=6.5R
๐ถ๐‘‰ ๐‘š
, = 6.5
๐‘…
๐‘๐‘œ๐‘›๐‘ ๐‘ก๐‘Ž๐‘›๐‘ก ๐‘๐‘ฆ ๐‘๐‘™๐‘Ž๐‘ ๐‘ ๐‘–๐‘๐‘Ž๐‘™ ๐‘โ„Ž๐‘ฆ๐‘ ๐‘–๐‘๐‘ 
Download