Ideal Gasses (1)

8/11/2019 Ideal Gasses (1)

1/14

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


2/14

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


3/14

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


4/14

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


5/14

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


6/14

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


7/14

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


8/14

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.


9/14

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


10/14

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


11/14

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

=


12/14

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


13/14

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= ?


14/14

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= ?

Ideal Gasses (1)

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