Week 3 Monday October 8, 2012 page 1
H=constant in Joule-Thompson experiment
T,V,P,U are all changing in Joule-Thompson experiment
𝜕𝑇
𝜕𝑇
𝜇𝐽𝑇 = ( )
𝜇𝐽 = ( )
𝐻
𝜕𝑃
𝜕𝑉 𝑈
∆𝑇
𝐷𝑜𝑖𝑛𝑔 𝑡ℎ𝑒 𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡 𝑜𝑛𝑐𝑒 𝑔𝑖𝑣𝑒𝑠 ( )
∆𝑃 𝐻
Take P1,T1 →P2,T2
P1>P2
Also, P1,T1 →P3,T3 and P1,T2 →P4,T4
so we’re changing the right side
Connecting these points makes an isenthalp, or isenthalpic curve.
Then repeat the experiment with different left side.
Reversible (equilibrium) but P2 > P1
Test
P1’T1’→P2’T2’ P1’>P2’ and P1’T1’→P3’T3’
Each curve has a maximum.
Connect the maximum of each isenthalp to create a new curve.
When μJT=0, allowing the gas to expand will neither cool it nor warm it up.
Most of the time, an engineer wants to cool the gas.
H=H(T,P)
𝜕𝐻
𝜕𝐻
𝑑𝐻 = ( ) 𝑑𝑇 + ( ) 𝑑𝑃
𝑃
𝜕𝑇
𝜕𝑃 𝑇
𝑠𝑙𝑜𝑝𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎
𝜕𝐻
𝜕𝐻
( ) 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑙𝑜𝑝𝑒 𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑃, 𝜇𝑇 = ( ) 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑙𝑜𝑝𝑒 𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑇
𝜕𝑇 𝑃
𝜕𝑃 𝑇
𝜕𝐻
( ) = 𝐶𝑃𝜇𝐽𝑇 = 𝜇𝑇 (𝑛𝑜𝑡 𝑖𝑛 𝑜𝑢𝑟 𝑡𝑒𝑥𝑡𝑏𝑜𝑜𝑘)
𝜕𝑃 𝑇
Axis: H,P, and T
Perfect gas and the first law:
Idea gas: PV=nRT
1. No intermolecular attraction
2. Molecules do not occupy space
Average translational energy (Et͞ rans α T)
Average vibrational energy (E͞vib α T)
Average rotational energy (E͞rot α T)
𝜕𝑈
𝑈𝑛𝑑𝑒𝑟 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑇, ( ) = 0 (𝑔𝑖𝑣𝑒𝑛, 𝑛𝑜 𝑝𝑟𝑜𝑜𝑓)
𝜕𝑉 𝑇
(
𝜕𝑈
) 𝑖𝑠 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒
𝜕𝑉 𝑇
Define perfect gas as one that obeys:
1.
PV=nRT
𝜕𝑈
2. (𝜕𝑉 )
𝑇
=0
U=U(T,V) for any closed system
For perfect gas: U=U(T)
𝜕𝑈
𝑑𝑈
𝐶𝑉 = ( ) → 𝐶𝑉 =
𝑓𝑜𝑟 𝑎 𝑝𝑒𝑟𝑓𝑒𝑐𝑡 𝑔𝑎𝑠
𝑉
𝜕𝑇
𝑑𝑇
dU=CVdT
for a perfect gas
CV=CV(T) since U=U(T) for perfect gas
H=U+PV=U+nRT
since PV=nRT
U is temperature dependent and nRT is temperature dependent, so H=H(T) for a perfect gas
𝜕𝐻
𝑑𝐻
𝐶𝑃 = ( ) → 𝐶𝑃 =
→ 𝑑𝐻 = 𝐶𝑃𝑑𝑇 𝑓𝑜𝑟 𝑎 𝑝𝑒𝑟𝑓𝑒𝑐𝑡 𝑔𝑎𝑠
𝜕𝑇 𝑃
𝑑𝑇
CP=CP(T) for a perfect gas
𝜕𝑈
𝜕𝑉
𝐶𝑃 − 𝐶𝑉 = (( ) + 𝑃) ( )
𝜕𝑉 𝑇
𝜕𝑇 𝑃
𝜕𝑈
( ) 𝑖𝑠 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒, = 0 𝑓𝑜𝑟 𝑎 𝑝𝑒𝑟𝑓𝑒𝑐𝑡 𝑔𝑎𝑠, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒:
𝜕𝑉 𝑇
𝐶𝑃 − 𝐶𝑉 = 𝑃 (
𝜕𝑉
) 𝑓𝑜𝑟 𝑎 𝑝𝑒𝑟𝑓𝑒𝑐𝑡 𝑔𝑎𝑠
𝜕𝑇 𝑃
Using PV=nRT gives 𝑉 =
𝑛𝑅𝑇
𝑃
(
𝑛𝑅
𝐶𝑃 − 𝐶𝑉 = 𝑃 ( ) = 𝑛𝑅
𝑃
CP,m-CV,m=R
𝜕𝑉
𝑛𝑅
) =
𝜕𝑇 𝑃
𝑃
𝑅 = 8.314
𝐽
𝑚𝑜𝑙 𝐾
𝑛 = # 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠
for a perfect gas
CP,m is molar heat capacity at constant P
CV,m is molar heat capacity at constant V
(
0 = −𝐶𝑉𝜇𝐽 → 𝜇𝐽 = 0
𝜕𝑈
) = −𝐶𝑉𝜇𝐽
𝜕𝑉 𝑇
𝑓𝑜𝑟 𝑎 𝑝𝑒𝑟𝑓𝑒𝑐𝑡 𝑔𝑎𝑠, 𝑛𝑜𝑡 𝑓𝑜𝑟 𝑟𝑒𝑎𝑙 𝑔𝑎𝑠
𝜕𝐻
( ) = −𝐶𝑃𝜇𝐽𝑇
𝜕𝑃 𝑇
0=-CPμJT→μJT=0 since CP is never 0 for a gas
Apply the first law
dwrev=-PdV
dU=CVdT
dU=dq+dw
for reversible volume change
for perfect gas
first law
CVdT=dq+dw=dq-PdV
CVdT=dq-PdV perfect gas, reversible process, PV work only
1. Reversible isothermal process in a perfect gas
∆U=0 since it’s isothermal
so q+w=0 and q=-w
𝑉2
𝑉2
𝑤 = − ∫ 𝑃𝑑𝑉 = ∫
𝑉1
𝑉1
𝑉2
𝑛𝑅𝑇
𝑑𝑉
𝑉
𝑑𝑉 = 𝑛𝑅𝑇 ∫
= −𝑛𝑅𝑇𝑙𝑛 2
𝑉
𝑉
𝑉1
𝑉
1
𝑉
𝑤 = 𝑛𝑅𝑇𝑙𝑛 1 𝑎𝑝𝑝𝑙𝑖𝑒𝑠 𝑡𝑜 𝑒𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛 𝑜𝑟 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛
𝑉2
V2>V1 →w<0
V2<V1→w>0
Boyle’s Law: P1V1=P2V2 at constant T
𝑉1 𝑃2
𝑃
=
𝑠𝑜 𝑤 = 𝑛𝑅𝑇𝑙𝑛 2 𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑇
𝑉2 𝑃1
𝑃1