Kinetic Theory of Gases I

Kinetic Theory of Gases I

Ideal GasThe number of molecules is large

The average separation between molecules is large

Molecules moves randomly

Molecules obeys Newton’s Law

Molecules collide elastically with each other and with the wall

Consists of identical molecules

The Ideal Gas Law

PV nRTn: the number of moles in the ideal gas

n N

NA

total number of molecules

Avogadro’s number: the number ofatoms, molecules, etc, in a mole ofa substance: NA=6.02 x 1023/mol.

R: the Gas Constant: R = 8.31 J/mol · K

in K

Pressure and TemperaturePressure: Results from collisions of molecules on the surface

P F

APressure:

Force

Area

Force: F dp

dtRate of momentumgiven to the surface

Momentum: momentum given by each collisiontimes the number of collisions in time dt

Only molecules moving toward the surface hitthe surface. Assuming the surface is normal to the x axis, half the molecules of speed vx movetoward the surface.

Only those close enough to the surface hit it in time dt, those within the distance vxdt

The number of collisions hitting an area A in time dt is 1

2

N

V

Avx dt

The momentum given by each collision to the surface 2mvx

Average density

dp 2mvx 1

2

N

V

Avxdt

Momentum in time dt:

Force: F dp

dt 2mvx 1

2

N

V

Avx

Pressure: P F

A

N

Vmvx

2

Not all molecules have the same average vx2

P N

Vmvx

2

vx

vx2 =

1

3v2

1

3vx

2 vy2 vz

2

Pressure: P 1

3

N

Vmv2

2

3

N

V

1

2mv2

Average Translational Kinetic Energy:

K 1

2mv2

1

2mvrms

2

vx2 =

1

3v2

1

3 vrms

2

vrms v2 vx

2 + vy2 + vz

2

3

is the root-mean-square speed vrms

Pressure: P 2

3N

VK

PV 2

3N KFrom and PV nRT

Temperature: K 3

2nRT

N

3

2kBT

Boltzmann constant: kB R

NA1.3810 23 J/K

PV 1

3N mvrms

2From

and PV nRT N

NART

vrms 3RT

M

Avogadro’s number

N nNA

M mN A

Molar mass

Pressure Density x Kinetic Energy

Temperature Kinetic Energy

Internal Energy

For monatomic gas: the internal energy = sumof the kinetic energy of all molecules:

Eint N K nNA 3

2kBT

3

2nRT

Eint 3

2nRT T

(d) mass of gas n

pV nRT nConstant temperaturep

T

nR

V nConstant volume

V

T

nR

p nConstant pressure

HRW 16P (5th ed.). Consider a given mass of an ideal gas. Compare curves representing constant-pressure, constant volume, and isothermal processes on (a) a p-V diagram, (b) a p-T diagram, and (c) a V-T diagram. (d) How do these curves depend on the mass of gas?

p p

V

constant pressure

isothermal

constant volume

T

constant pressure

constant volume

isothermal V

T

constant volume

constant pressure

isothermal

PV nRT

(b) TB pBVB

nR1.8 103 K

(c) TC pCVC

nR6.0 102 K

(d) Cyclic process ∆Eint = 0

Q = W = Enclosed Area= 0.5 x 2m2 x 5x103Pa = 5.0 x 103 J

(a) n pAVA

RTA1.5 mol.

HRW 18P (5th ed.). A sample of an ideal gas is taken through the cyclic process abca shown in the figure; at point a, T = 200 K. (a) How many moles of gas are in the sample? What are (b) the temperature of the gas at point b, (c) the temperature of the gas at point c, and (d) the net heat added to the gas during the cycle?

Volume (m3)

a

b

c

1.0 3.0Pr

essu

re (

kN/m

2 )

2.5

7.5

PV nRTEint Q W

(a)

vrms 3RT

M

3 8.31 J/molK 293 K 28.0 10-3 kg/mol

511 m/s

for 0.5 vrms T 0.52T 73.3K = -200C

(b) Since vrms Tv rms2

vrms2

T

T

for 2 vrms T 22T 1.17 103K = 899C

HRW 30E (5th ed.).(a) Compute the root-mean-square speed of a nitrogen molecule at 20.0 ˚C. At what temperatures will the root-mean-square speed be (b) half that value and (c) twice that value?

vrms 3RT

M

(a)K 3

2kBT 3

21.3810 23 J/K 1600K

3.3110 20 J

(b) 1 eV = 1.60 x 10-19 J

K 3.3110 20 J

1.60 10 19 J/eV0.21 eV

HRW 34E (5th ed.). What is the average translational kinetic energy of nitrogen molecules at 1600K, (a) in joules and (b) in electron-volts?

K 3

2kBT

Kinetic Theory of Gases II

Mean Free Path

Molecules collide elastically with other molecules

Mean Free Path : average distance between

two consecutive collisions

1

2d2N / V

the more moleculesthe more collisions

the bigger the moleculesthe more collisions

Molar Specific Heat

Definition:

For constant volume: Q nCVT

For constant pressure: Q nCpT

The 1st Law of Thermodynamics:

Eint 3

2nRT Q W (Monatomic)

Q cmT

Eint Q W

Eint 3

2nRT

Constant Volume

W PdV 0

3

2nRT Q W

(Monatomic)

Q nCVT

3

2nRT nCVT

CV 3

2R

Eint nCVT

Eint 3

2nRT

Constant Pressure

W PV nRT

3

2nRT Q W

(Monatomic)

Q nCpT

3

2nRT nCpT nRT

CV Cp R

Cp

CV

Cp 5

2R

5

3

dEint dW pdV

Adiabatic Process

(Q=0)

dEint dQ dW

pV nRT

1st Law

Ideal Gas Law

Eint nCVT

nCVdT

pdV Vdp nRdT nR pdV

nCV

Divide by pV:

dV

V

dp

p

Cp CV

CV

dV

V(1 )

dV

V

Cp CV R

Cp

CV

dV

V

dp

p(1 )

dV

V

dp

p

dV

V0

ln p lnV ln(pV ) = const.

pV = const.

pV nRTIdeal Gas Law

(nRT

V)V = const.

TV 1 = const.

Equipartition of Energy

The internal energy of non-monatomic molecules includes also vibrational androtational energies besides the translational energy.

Each degree of freedom has associated with

it an energy of per molecules.1

2kBT

Monatomic Gases

3 translational degrees of freedom:

Eint 3

2kBT nNA

3

2nRT

CV 1

ndEint

dT

3

2R

Eint nCVT

Diatomic Gases

3 translational degrees of freedom

Eint 5

2nRT CV

5

2R

Eint nCVT

2 rotational degrees of freedom

2 vibrational degrees of freedom

HOWEVER, different DOFs require differenttemperatures to excite. At room temperature,only the first two kinds are excited:

(a) Constant pressure: W = p∆V

Eint Q W Q pV

20.9 1.0 105 Pa 100 cm3 50cm3 10 6 m3/cm3 15.9 J

HRW 63P (5th ed.). Let 20.9 J of heat be added to a particular ideal gas. As a result, its volume changes from 50.0 cm3 to 100 cm3 while the pressure remains constant at 1.00 atm. (a) By how much did the internal energy of the gas change? If the quantity of gas present is 2.00x10-3 mol, find the molar specific heat at (b) constant pressure and (c) constant volume.

Eint Q W

pV nRT(b) Cp Q

nT Q

n pV / nR R

p

Q

V

8.31 J/molK 20.9J 1.0 105 Pa 50 10 6 cm3 34.4 J/molK

(c) CV Cp R

34.4 J/molK 8.31 J/molK = 26.1 J/molK

HRW 63P (5th ed.). Let 20.9 J of heat be added to a particular ideal gas. As a result, its volume changes from 50.0 cm3 to 100 cm3 while the pressure remains constant at 1.00 atm. (a) By how much did the internal energy of the gas change? If the quantity of gas present is 2.00x10-3 mol, find the molar specific heat at (b) constant pressure and (c) constant volume.

Q nCpT

ln / 5

3ln /

i f

f i

p p

V V

(a) Adiabatic piVi p f Vf

pi / p f Vf / Vi

Monatomic

HRW 81P (5th ed.). An ideal gas experiences an adiabatic compression from p =1.0 atm, V =1.0x106 L, T = 0.0 ˚C to p =1.0 x 105 atm, V =1.0x103 L. (a) Is the gas monatomic, diatomic, or polyatomic? (b) What is its final temperature? (c) How many moles of gas are present? (d) What is the total translational kinetic energy per mole before and after the compression? (e) What is the ratio of the squares of the rms speeds before and after the compression?

pV = const.

Cp

CV

(b)piVi

Ti

p f Vf

Tf

pV nRT

Tf pf Vf

piViTi

1.0 105 atm 1.0 103 L 273 K

1.0 atm 1.0 106 L 2.7 104 K


n piVi

RTi4.5104 mol.(c) (Pay attention to the units)

K 3

2nRT

N

Ki 3

2RTi 3.4 103 J

(d) For N/n = 1

K f 3

2RTf 3.4 105 J


pV nRT

K 1

2mvrms

2 3

2kBT

(e)vrms,i

2

vrms, f2

Ti

Tf0.01


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Kinetic Theory of Gases I