Section III 11 Ideal Gases

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Ideal gases

11. Ideal Gases Content 11.1 Equation of state 11.2 Kinetic theory of gases 11.3 Pressure of a gas 11.4 Kinetic energy of a molecule Learning Outcomes (a) recall and solve problems using the equation of state for an ideal gas expressed as

pV = nRT. (n = number of moles)

(b) infer from a Brownian motion experiment the evidence for the movement ofmolecules.

(c) state the basic assumptions of the kinetic theory of gases. (d) explain how molecular movement causes the pressure exerted by a gas and hence

deduce the relationship, p = Nm/V < c2 > . (N = number of molecules)[A rigorous derivation is not required.]

(e) compare pV = Nm < c2 > with pV = NkT and hence deduce that the averagetranslational kinetic energy of a molecule is proportional to T.


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Microscopic model of a gas

Physics describes and explains the behaviour of varioussystems

In some cases it is impossible to describe what happens to each

component of a system e.g properties of a gas in terms of themotion of each of its molecules

In such cases we describe the average conditions in the gasrather than describing the behaviour of each molecule i.e

macroscopic instead of microscopic

Kinetic theory of an ideal gas is one such example where wemake simple assumptions to derive at a law relating kinetic

energy to temperature


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What is the evidence that gas molecules

are moving around all the time? 1 cm3 of atmospheric air contains approx 3 x 1025 molecules 1827 biologist Robert Brown observed tiny pollen grains

suspended in water with a microscope with a illuminating beam

Although the water was completely still, the grains were alwaysmoving in a jerky, haphazard manner : we call this Brownianmotion

Hence the assumption that the water molecules are in rapid,random motion under the bombardment from all sides of thewater molecules

We can see the same movement of tiny soot particles in smoke

A century later, Mr. Einstein did a theoretical analysis ofBrownian motion and estimated the diameter of a typical atom inthe order of 10-10 m

It has been found that the average spacing of atoms of liquids isup to 2 times that of solids or still in the order of 10-10 m. Forgases it is of the order of 10-9 m


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The gas laws

Experiments in the 17th and 18th centuries showed that thevolume, pressure and temperature of a given sample of gas are

all related


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Boyles Law

Robert Boyle 1627-91

Boyles Law states that thepressureof a given mass of gas atconstant temperatureis inversely proportional to itsvolume.

p 1/V or pV = constant

If p1,V1 are the initial pressure and volume of the gas, and p2,V2are the final values after a change of pressure and volume carried

out at constant temperature, then

p1V1 = p2V2


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Charles Law

Jacques Charles 1746-1823

Charles Law states that the volumeof a given mass of any gas atconstantpressure is directly proportional to its thermodynamictemperatureT

V T or V/T = constant If V1,T1 are the initial volume and temperature of the gas, and

V2,T2 are the final values after a change of volume and

temperature carried out at constant pressure, then

V1/T1 = V2/T2

TK = (273 + ) degrees C


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Pressure Law

Joseph Gay-Lussac 1778-1850

Pressure Law or Gay-Lussacs Law states that thepressureof agiven mass of gas atconstant volumeis directly proportional toits absolute temperature.

p T or p/T = constant

If p1,T1 are the initial pressure and temperature of the gas, and

p2,T2 are the final values after a change of pressure and

temperature carried out at constant volume, then

p1/T1 = p2/T2


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Ideal Gas Equation

In practice, real gases obey the gas laws only at moderate

pressures and at temperatures well above the temperature atwhich the gas would liquefy.

An ideal gas is one which obeys the 3 gas laws and is given by:

pV T or pV/T = constant

i.e. p1

V1

/T1

= p2

V2

/T2

All the above laws relate to a fixed mass of gas

It is also found by experiments that the volume of a gas isproportional to its mass giving a combined equation of

pV mT or pV = AmT

where m is the mass of the gas and A is a constant ofproportionality

As A is different for different gases we express the fixed mass ofgas in terms of the number of moles of gas present


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Avogadros Hypothesis

Avogadros hypothesis states that equal volume of all gases, atthe same temperature and pressure, contain equal number of

molecules.

As such, gases like O2, N2, CH4, CO2, CO, H2 etc all contain the

equal number of molecules at the same temperature and pressure.


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Mole (abbreviated as mol)

A moleis the number of elementary units (atoms ormolecules) of any substance which is equal to the amount

of atoms in 12 g of carbon-12.

The number of molecules per mole for all substances is the

same and is called theAvogadros constantor number

NA

NA is 6.02 x 1023/mol


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Relative atomic mass & atomic mass unit

The relative atomic mass Ar is the ratio of the mass of an atomto one-twelfth of the mass of an atom of carbon-12

The relative atomic mass is numerically equal to the mass ingrams of a mole of atoms

The relative molecular mass Mr is the ratio of the mass of amolecule to one-twelfth of the mass of an atom of carbon-12,and is numerically equal to the mass in grams of a mole ofmolecules

The atomic mass unit u is one-twelfth of the mass of an atomof carbon-12, and has the value 1.66 x 10-27 kg (also called theunified atomic mass constant)

Simple way to find the mass of 1 mole of an element is to takeits nucleon number expressed in grams

e.g nucleon number of argon 40Ar is 40. Therefore, 1 moleof argon then has a mass of 40 g


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Gas laws in terms of moles & molar gas

constant Hence in terms of moles,pV nT

and putting in a new constant of proportionalityR,

pV = nRT

R is called the molar gas constant (or sometimes universal gasconstant as it has the same value for all gases)

R has a value of 8.3 J K-1mol-1

The above equation is expressed in the form pVm = RT whereVm is the volume occupied by 1 mole of the gas

Another version is pV = NkT where N is the number ofmolecules in the gas and k is a constant called the Boltzmannconstant which has a value of 1.38 x 10-23 J K-1

The molar gas constant R and the Boltzmann constant k areconnected through the Avogadro constant NA

R = kNA


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Experiment has confirmed that one mole of all gases ats.t.p. ( 0 degrees C and 760mm Hg or 1.01 x 105 Pa )

occupies 22.4 litres (1 litre is 1000 cm3).

From pV/T = k , substituting the above values, the constant

for one mole R, called the molar gas constant, is therefore

the same for all gases and has a value of 8.31 J/K/mol


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Equation of state

The precise relation between volume, pressure, temperatureand the mass of the given gas in a sample is called the

equation of state of the gas

An ideal gas is one which obeys the equation of state:pV = nRT

For approximate calculations the ideal gas equation can beused with real gases


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Example

Find the volume occupied by 1 mole of air at stp (273K and 1.01x 105 Pa), taking R as 8.3 J K-1 mol-1 for air.

Solution:pV = nRT, so V = nRT/p

= 2.24 x 10-2 m3


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Example

Find the number of molecules per cubic metre of air at stp.

Solution:

from above example, 1 mole of air at stp is 2.24 x 10-2 m3

but 1 mole of air contains NA molecules where NA is the

Avogadro constant.

thus the number of molecules per cubic metre of air is

6.02 x 1023/2.24 x 10-2 = 2.69 x 1025 m-3


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Example

A syringe contains 25 x 10-6 m3 of helium gas at a temperatureof 20 C and a pressure of 5.0 x 104 Pa. The temperature isincreased to 400 C and the pressure on the syringe is

increased to 2.4 x 10

5

Pa.Find the new volume of gas in the syringe.

Solution:

Using p1V1/T1 = p2V2/T2 with T1 = 293 K & T2 = 673 K

V2 = 12 x 10-6 m3


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Example

Oxygen gas contained in a cylinder of volume 1 x 10-2

m3

has atemperature of 300 K and a pressure 2.5 x 105 Pa. Calculate the mass ofthe oxygen used when the oxygen pressure has fallen to 1.3 x 105 Pa.

Solution

Let the initial and final number of moles of oxygen be n1 and n2respectively.

n1 = p1V/(RT), n2 = p2V/(RT)

no. of moles used = n1 n2 = (p1 p2)V/(RT)= [(2.5 1.3) x 105 x 10-2]/(8.31 x 300)= 0.48 moles

= 0.48 x 32 x 10-3 kg = 0.015 kg.


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The kinetic theory

Robert Boyle in 17th century developed an explanation of how a

gas exerts pressure Daniel Bernoulli in 18th century explained it in greater detail

The basic idea was that gases consist of atoms or moleculesmoving about at great speed (later visualised by Robert Brown)

and that the gas exerts a pressure on the walls of its container

because of the continued impacts of the molecules with thewalls

Its value is the total rate of the momentum change of themolecules per unit area of the wall.

The assumptions of the kinetic theory of an ideal gas are: All molecules behave as identical, hard, perfectly elastic spheres

The volume of the molecules is negligible compared with the volume of the

containing vessel

There are no forces of attraction or repulsion between molecules

There are many molecules, all moving randomly so that average behaviour can be

considered


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Derivation of pressure of gas

Assume a gas ofN molecules each of mass m are contained in

a cube of side l.Let the velocity of one molecule be c at anyinstant where its component along the x, y and z axis are u, v

and w respectively.

Therefore resultant c2 = u2 + v2 + w2.

Consider the force F exerted on the face X of the cube due to u

for 1 molecule

momentum change on impact = mu - (- mu) = 2mu

time of travel before next collision for one impact = 2l/unumber of collisions per second = 1/(2l/u) = u/(2l)momentum change per second = [u/(2l)] x 2mu = mu2/lforce on X, F = mu2/lpressure, p on X due to one molecule= F/A = F/l2= mu2/l3


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ForN molecules, mean square speed of all molecules is u2

whereu2 = (u1

2+u22+u3

2+ .+uN2)/N p = (Nmu2)/l3 = (Nmu2)/Valso u2 = v2 = w2 u2 = < c2>

hence p = (N/V)(m) giving

pV = Nm = nRT = NkT

or p = where is density

or = 3p/

c2 = (3p/) where c2 is root mean square(rms) In deriving the above, we have made use of p = F/A, Newtons

2ndlaw F = rate of change of momentum and Newtons third law


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Example

The speed of 7 molecules in a gas are numerically equal to2,4,6,8,10,12 and 14 units.

Find the numerical values of (a) the mean speed (b) the

mean speed squared 2

(c) the mean square speed (d)the rms speed.

a) = (2+4+6+8+10+12+14)/7 = 8 units

b) 2 = 82 = 64 units2

c) = (4+16+36+64+100+144+196)/7 = 80 units2

d) rms speed = 80 = 8.94 units


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Worked Example

Calculate the rms speed of hydrogen gas at s.t.p. where itsdensity is 0.09 kg m-3.

Solution

At s.t.p., p = 1.013 x 105 Pa

Using p = c2c2 = (3 x 1.013 x 105)/0.09 = 3.37 x 106 ms-2

c2 = (3.37 x 106) = 1840 m/s


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Mean Energy of Molecules

The kinetic energy of a molecule moving at an instant witha speed v is mv2

The average kinetic energy of translation of therandom motion of the molecule of a gas is therefore m.

SincepV = Nmsubstituting forwe arrive at a link between our kinetic theory equation,energy and temperature

pV = N( m) = N = NkT

i.e = m =3/2 kT

= (3kT/m)


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Conclusion

The r.m.s. speed or velocity of the molecules of a gas T. The r.m.s. speed of the molecules of different gases at the same

temperature 1/M, so gases of higher molecular mass havesmaller r.m.s. speeds

Hence the rms speed is proportional to the square root of thethermodynamic temperature of the gas, and inverselyproportional to the square root of the mass of the molecule.

Thus at a given temperature, less massive molecules movefaster, on average, than more massive molecules i.e the higherthe temperature, the faster the molecules move.

The molecules of air at normal temperatures and pressureshave an average velocity of the order of 480 m/s


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Example

The r.m.s. speed of hydrogen gas at s.t.p. is 1840 m/s.Calculate its new r.m.s. speed at 1000 C and the same pressure.

What is the r.m.s. speed of oxygen at s.t.p.?

Let cr= r.m.s. speed of hydrogen at 1000

Ccs = r.m.s. speed of oxygen at s.t.p.

cr/1840 = (373/273)

cr= 1840 x (373/273) = 2150 m/s

cs/1840=(2/32) (H2 is 2, O2 is 32) cs=1840 x(2/32) = 460 m/s


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Example

Find the total kinetic energy of the molecules in one mole of anideal gas at standard temperature.

Solution:Average kinetic enrgy of 1 molecule = 3/2 kT

The energy for 1 mole Ek i.e for NA molecules is

Ek = 3/2 NAkT or 3/2 RT = 3400 J mol-1


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Example

Find the rms speed of the molecules in nitrogen gas at 27 C.The mass of a nitrogen molecule is 4.6 x 10-26 kg.

Solution:

crms = (3kT/m) where k is the Boltzmann constant

= 520 ms-1

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Section III 11 Ideal Gases