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Ideal Gasses
The question: U(P,T) = ?, r(P,T)= N/V(P,T) = ?
Here: only for special substances: ideal gasses
Interactions between moleculesneglected
Range of interactions: much shorter than distance between molecules
HW
2.5
2.9
Due 9/11
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PV =NRT
R = 8.314 J mol-1K-1
UIG=NU
IG(T)
Nnumber of molecules
UIG(T)kinetic, rotational and vibrational energy of molecules
No potential energyno interactions
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Heating of ideal gas in a container
with rigid walls: V = const
dUIG
=/dQ=NCV
IGdT /dW= Fdx
0
CV
IG
dUIG
dTHeat capacity at constant volume
CVIG
=3R/ 2 (monoatomic gases)Only translational degrees of freedom: no rotational, no vibrational
Each degree of freedom contributes R/2 to Cv:
CvIG
= 5R/2 (linear molecules)
CV
U
T
V
For interacting molecules:
high density, molecules feel each others presence
Units of Cv: J mol-1
K-1
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CP
IG dH
IG
dT=
d
dTU
IG
+PV
=
d
dTU
IG
+RT
= C
V
IG+R
Heating of ideal gas at constant
pressure: P = const
U=Q P V Q= (U+PV
)=
H
dW = -Fdx = -PAdx = -PdV
For any system: CP
=
1
N
H
T
N ,P
=
H
T
P
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Adiabatic expansion of ideal gasses
dU= PdV NCV
IGdT = NPdV = NPd
RT
P
CV
IGdT= PdV= RdT+
RT
PdP
dQ = 0
Three variables: P, T, and V
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Eliminating P
CV
IGdT= PdV C
V
IGdT= RT
VdV C
V
IG dT
T= RdV
V
T
T0
=
V
V0
R
CV
IG
=
V
V0
1
=
V
V0
1
R
V
V
V
V
C
IGV
T
T
IGV
T
T
IGV
V
V
V
VRVR
V
dVR
T
T
T
TCTC
T
dTC
IG
V
00
00
lnln)ln(
lnln)ln(
0
0
00
g= Cp/CV = (CvIG+ R)/CV
IG= 1 + R/CVIG
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Eliminating V
CV
IGdT= RdT+
RT
PdP
(CV
IG+R
)
dT
T= R
dP
P C
P
IG dT
T= R
dP
P
T
T0
=
P
P0
R
CP
IG
=
P
P0
1
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Eliminating T
PdV= CV
IGdT=
CV
IG
Rd(PV)=
CV
IG
R(PdV+VdP)
CV
IG
RVdP=
CV
IG+R
RPdV C
V
IG dP
P=C
P
IG dV
V
P
P0
=
V
V0
=
V
V0
PV = const.
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Summary: Adiabatic expansion of ideal
gasses
T
T0
=
V
V0
R
CV
IG
=
V
V0
1
=
V
V0
1
T
T0
=
P
P0
R
CP
IG
=
P
P0
1
P
P0
=
V
V0
=
V
V0
PV = const.
U= Q +WW=NC
V
IG T
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Example:Nimoles of ideal gas at temperatureTiare held in volume V.
Same gas at temperature Tinis pumped in until Nf.
No heat transfer.Tf= f(Ti, Tin,Nf/Ni,Cp) = ?
Solution:
From first law:
dU = /dQ+/dW+ HindN
inenteringstreams
HoutdNoutleavingstreams
dU = HindNin
d(U N)= HindNin
U dN+ NdU = HindNin = HindN
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Solution: continued
U(T) = Uo(To) + Cv(T To)
H(T) = Ho(To) + Cp(T
To) Ho(To) = Uo(To) + RTo
U and H known to within a constant:
Assume To= 0 and Uo= 0
U(T) = CvT and H(T) = CpT
We get from
U dN+ NdU = HindN
CvTdN + NCVdT = CpTindN
NCVdT = (CpTinCvT)dN)(
lnln
iinf
iinf
iin
fin
i
f
T
T in
N
N
TTN
NTT
TT
TT
N
N
TT
dT
N
dN f
i
f
i
gg
g
g
g
=
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What if Tin= Ti?
f
i
f
i
if
f
iif
iif
iif
N
N
N
NTT
N
NTT
TTN
NTT
gg
gg
gg
)1(
)(
1
1
,10
f
i
f
i
V
V
V
p
f
i
f
i
N
N
N
N
C
RTC
C
C
N
N
N
N
gg
g
gg
If Nf>>Ni,Tf> Ti
IfNfNi, TfTi
Tf> Ti
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Problem 2.9
A container with N2at Pi= 5 bar and Ti= 300 K is quickly filled
from a much larger cylinder with P = 10 bar and T = 300 K;
Pf= 10 bar. Tf= ?
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Air
water
2.10. Water gun. Air: 750 cm3, P = 2.5 bar, T = 30oC
Water pumped at 8 m/s for 5 s through 2 mm ID tube
Tf= ?, Pf= ?