Chem 41 lecture 5 (1)
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Transcript of Chem 41 lecture 5 (1)
8/12/2019 Chem 41 lecture 5 (1)
http://slidepdf.com/reader/full/chem-41-lecture-5-1 1/11
8/12/2019 Chem 41 lecture 5 (1)
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8/12/2019 Chem 41 lecture 5 (1)
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( , )
+ .
+ .
'
T
U U U dV V
∂ = + ∂
' U
U U dT T
∂ = + ∂
8/12/2019 Chem 41 lecture 5 (1)
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&
,
&
T V
U U dU dV dT
V T
∂ ∂ = + ∂ ∂
,
T T
U V
π ∂
= ∂
T V dU dV C dT π = +
V V
U C
T
∂ = ∂
8/12/2019 Chem 41 lecture 5 (1)
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/,
,
, π 0
8/12/2019 Chem 41 lecture 5 (1)
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π
0 0 , Δ 0
, π 0
8/12/2019 Chem 41 lecture 5 (1)
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T V P P
U V
C T T π
∂ ∂ = +
∂ ∂
1
P
V V T
α ∂
= ∂
1T
T
V V p
κ ∂
= − ∂
8/12/2019 Chem 41 lecture 5 (1)
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.
8/12/2019 Chem 41 lecture 5 (1)
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α ,
., π 0
T V P
U V C T
απ ∂ = + ∂
V P
U C T
∂ = ∂
+ +
,
P V P P
C C T T − = − ∂ ∂
P V P P
U U C C nR nR
T T
∂ ∂ − = + − =
∂ ∂ 2
P V T
TV C C
K α − =
8/12/2019 Chem 41 lecture 5 (1)
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,
At const. pressure, ∆H = q P
Let H = H(p,T)For a closed s stem of const. n
Joule-Thomson coefficientP PdH C dp C dT µ
= − +
H
T
p µ
∂=
∂
pT
H H dH dP dT p T
∂ ∂ = + ∂ ∂
8/12/2019 Chem 41 lecture 5 (1)
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Joule-Thomson coefficientT p
µ ∂
= ∂ Joule-Thomson effect
cooling by isenthalpicexpansion