Kinetic Theory (Essay) All matter is made up of very tiny particles, which are constantly in...

Kinetic Theory (Essay)

All matter is made up of very tiny particles, which are constantly in motion.

The molecules repel other strongly when they are forced too close together.

When they are slightly apart, they attract each other. When they are far apart, they hardly attract or repel each

other at all.

close particles

strong attraction

distant particles

weak attraction

Interpreting temperature by kinetic theory

The motion of the particles in a body depends on its temperature.

When the temperature rises, particles vibrate more rapidly or move faster. Hence, the temperature of the body is a measure of the average kinetic energy of the particles.

Two bodies have the same temperature if particles in each body have the same average kinetic energy.

A gas model (Essay)

Gases have no fixed volume and shape. They can quickly fill any space available. In a gas, particles are very far apart. Therefore, the

attraction among them is weak.

Particles move around freely at high speed

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A gas model

The particles move at random at very high speeds (500 ms-1 at room temperature!).

They collide with each other and bombard the walls of the container.

When gas molecules hit the walls of container, they produce forces on the surface, which give rise to gas pressure.

Particles move around freely at high speed

container

video

Gas pressure

surfaceofArea

surfacethetolarperpendicuForcePressure

A

FP

Pressure is defined as the perpendicular force exerted per unit area.

Pressure is measured in N m-2 or pascal (Pa). 1 Pa = 1 N m-2.Normal atmospheric pressure or 1 standard pressure = 1.01 x 105 Pa

10 cm5 cm

Example 1

Find the pressure exerted on the 5 kg block in each of the following cases.

(a)

Solution:

P = F / A

= 50 / (0.05 x 0.1)

= 10000 Pa

Example 1

Find the pressure exerted on the 5 kg block in each of the following cases.

(b)

Solution:

P = F / A (F = normal reaction)

= 50 cos30o / (0.05 x 0.1)

= 8660 Pa

30o

10 cm

5 cm

Example 2 The figure shows a cylinder with a

piston. If the cross section area of the cylinder is 20 cm2, and find the pressure of the gas. Given that the atmospheric pressure is 1.01 x 105 Pa.

Answer: 2.05 x 105 Pa

0.2 kg

Measuring gas pressure A Bourdon gauge is used to measure

gas pressures.

Measuring gas pressure A bourdon gauge works in the same way as a paper tube whistle.

When a paper tube whistle is blown, it uncoils. In a Bourdon gauge, the gas pressure makes the curved metal

tube stretch out slightly. This causes the pointer to move round the scale. The higher the gas pressure, the more the pointer moves.

Pressure, volume and temperature

When a gas is heated, its pressure P, volume V and temperature T will change. We are going to investigate these relations.

Boyle’s law (P-V relationship)

In 1660, Robert Boyle investigated the relationship between the volume of a fixed amount of gas and its pressure, keeping the temperature of the gas constant.

Video

Experiment Press the air pump gently to c

hange the pressure and the volume of the trapped gas in the glass tube without changing its temperature.

Measure the pressure and the volume of the gas.

Repeat the above step to get more sets of pressure and volume.

Results

P ∝ 1/V or PV = constant or P1V1 = P2V2

Boyle’s law: For a fixed mass of gas at a constant temperature, the product of its pressure and volume is constant.

Example 2 The air pressure inside an inflated ball is 2 x 105 Pa.

The ball is now squeezed to one-third of its original volume. Find the new pressure inside the ball.

Solution: Let V be the original volume of the ball. By P1V1 = P2V2

2 x 105 V = P2(V/3)

P2 = 6 x 105 PaThus, the new pressure is 6 x 105 Pa

Charles’ law (V – T relationship)

In this experiment, we keep the pressure of the gas constant and try to investigate the relation between its volume and temperature.

Heat the water until its temperature is increased by 10oC.

Record the volume and temperature of the trapped gas.

Repeat the above step to get more sets of volume and temperature.

The mercury thread should be able to move up and down the capillary tube freely.

The friction between the mercury thread and the capillary tube should be small so that the pressure of the trapped gas is equal the atmospheric pressure.

Gas column

Experiment

Charles’ law (V – T relationship) Heat the water until its temperature is

increased by 10oC. Record the volume and temperature of

the trapped gas. Repeat the above step to get more sets

of volume and temperature. The mercury thread should be able to

move up and down the capillary tube freely.

The friction between the mercury thread and the capillary tube should be small so that the pressure of the trapped gas is equal the atmospheric pressure.

Gas column

simulation

Absolute zero The straight line cuts the axis at -

273oC which is known as the absolute zero of temperature.

Absolute zero is the theoretical lowest attainable temperature.

Theoretically, the gas at absolute zero occupies zero volume and exerts no pressure. However, in practice, it will liquefy before it can reach this temperature.

Kelvin scale When the temperature is measured in Kelvin (K), the volume of the gas

is directly proportional to its temperature shown in the figure below.

Note: Absolute temperature (K) = Celsius temperature + 273i.e. melting point of ice: 0oC = 273 K

room temperature: 27oC = (27 + 273) K = 300 Kboiling point of water: 100oC = (100 + 273) K = 373 K

volume V

Temperature T (K)

0

volume V

Temperature T (K)

0

or or TV constant

T

V

2

2

1

1

T

V

T

V

Charles’s law: For a fixed mass of gas at a constant pressure, its volume is directly proportional to its absolute temperature.

Example 2An inflated balloon contains 4 x 10-3 m3 of air at 27oC. It is put into a large tank of liquid nitrogen at -173 oC. What is the new volume of the balloon?

Solution: By Charles’ law,

V1/T1 = V2/T2

4 x 10-3 / (27 + 273273) = V2 / (-173 + 273273)

V2 = 1.33 x 10-3 m3

Thus, the new volume of the balloon is 1.33 x 10-3 m3

Pressure law (P – T relationship) Experiment Heat the water unit its

temperature increases by about 10oC. Stir the water thoroughly and wait for several minutes to allow the air in the flask to reach the temperature of the water.

Measure the temperature of the air by the thermometer and its pressure by the Bourdon gauge.

Repeat the above steps to get more sets of pressure and temperature.

Results The pressure of the gas increases linearly with its temperature.

When the temperature is measured in Kelvin (K), the pressure is directly proportional to its temperature shown in the figure below.

pressure P

Temperature T (K)0

TP constantT

P

2

2

1

1

T

P

T

P or or

Pressure law: For a fixed mass of gas at a constant volume, its pressure is directly proportional to its absolute temperature.

Example 3When the flask used in the experiment is put into a bath of melting ice, the pressure is 8.7 x 104 Pa. What is the pressure when the flask is put into a bath of boiling water?

Solution: Temperature of melting ice = 0 oC = 273 K

Temperature of boiling water = 100 oC = 373 K By pressure law, P1/ T1 = P2 / T2

8.7 x 104 / 273 = P2 / 373

P2 = 1.19 x 105 Pa

Thus, the new pressure is 1.19 x 105 Pa

General gas law Boyle’s law: PV = constant if T is constant Charles’ law: V/T = constant if P is constant Pressure law: P/T = constant if V is constant

All the gas laws can be summarized by a general equation:

constantT

PV

2

22

1

11

T

VP

T

VPor

Example 4A weather balloon contains 5 m3 of helium at the normal atmospheric pressure of 100 k Pa and at temperature of 27 C. What will be its volume when it rises to an altitude where the pressure is 80 k Pa and the temperature is 7 C?

Solution:

2

22

1

11

T

VP

T

VPBy

(100 k)(5)/(27 + 273273) = (80 k)(V2) / (7 + 273273)

V2 = 5.83 m3

Thus, its new volume is 5.83 m3

Equation of state and molar gas constantEquation of state and molar gas constant

RT

PV

nRTPVnRT

PV

For 1 mole of gas (i.e. 6.02 x 1023 gas particles),

(same for all gases) where R is called the universal molar universal molar gas constantgas constant. R = 8.31 J mole-1 K-1

For nn moles moles of gas,

Equation of state and molar gas constantEquation of state and molar gas constant

M

mn

If m is the mass of a gas and M is the mass of 1 mole of gas, the number of mole

nRTPV RTM

mPV From , we have

AN

Nn

If N is the number of molecules in a gas and NA = Avogadro

number = 6.02 x 1023, the number of mole

nRTPV RTN

NPV

A

From , we have

.

R = 8.31 J K-1 mol-1

Example 5Show that the volume of 1 mole of gas occupied at s.t.ps.t.p. is 22.4 litres.Note: s.t.p. stands for standard temperature and pressure.

Standard temperature = 0oC or 273 KStandard pressure = 1 atmospheric pressure = 1.01 x 105 Pa

Solution By PV = nRT (1.01 x 105)V = (1)(8.31)(273) V = 0.0224 m-3 or 22.4 litres

Ideal gases An ideal gas is a gas that obeys Boyle’s lawBoyle’s law (PV is consta

nt) for all pressures and temperatures. However, real gases such as hydrogen, oxygen, nitrogen o

r carbon dioxide are not ideal gases but they behave like an ideal gas at high temperaturehigh temperature and low pressurelow pressure.

Example 6 Two insulated gas cylinders A and B are c

onnected by a tube of negligible volume as shown.

Each cylinder has an internal volume of 2 x 10-2 m3. Initially, the tap is closed and cylinder A contains 1.2 mol of an ideal gas at a temperature of 37oC. Cylinder B contains the same ideal gas at pressure 1.2 x 105 Pa and temperature 37 oC.

Solution:Consider the gas in cylinder BBy PV = nRT(1.2 x 105 )(2 x 10-2) = n(8.31)(37 + 273)n = 0.932 The amount of gas in cylinder B is 0.932 mol

(a) Calculate the amount, in mol, of the gas in cylinder B.

Example 6 Two insulated gas cylinders A and B are c

onnected by a tube of negligible volume as shown.

Each cylinder has an internal volume of 2 x 10-2 m3. Initially, the tap is closed and cylinder A contains 1.2 mol of an ideal gas at a temperature of 37oC. Cylinder B contains the same ideal gas at pressure 1.2 x 105 Pa and temperature 37 oC.

Solution:Consider the gas in cylinder ABy PV = nRTP(2 x 10-2) = (1.2)(8.31)(37 + 273)P = 1.55 x 105 Pa The pressure of gas in cylinder A is 1.55 x 105 Pa

(a) Calculate the pressure of the gas in cylinder A.

Example 6 Two insulated gas cylinders A and B are c

onnected by a tube of negligible volume as shown.

Each cylinder has an internal volume of 2 x 10-2 m3. Initially, the tap is closed and cylinder A contains 1.2 mol of an ideal gas at a temperature of 37oC. Cylinder B contains the same ideal gas at pressure 1.2 x 105 Pa and temperature 37 oC.

(b) determine the final pressure of the gas in the cylinder.

1.2 mol 0.962 mol

Solution

By PV = nRT

P(4 x 10-2) = (1.2 + 0.962)(8.31)(37 + 273273)

P = 1.37 x 105 Pa

Example 7There are two containers X and Y filled with the same type of ideal gas as shown. They are connected by a tube. A steady state is obtained with X held at 100 K and Y at 400 K. If the volume of X is half that of Y and the mass of gas in X is m, what is the mass of gas in Y, in terms of m? Solution:

Let P be the pressure of the gas and V be the volume of container XFor the gas in container Y, P(2V) = nYR(400) --- (1)For the gas in container X, P(V) = nXR(100) --- (2)

(1)/(2):2 = 4nY/nX

nY = ½ nX

Mass of gas in Y = ½ mass of gas in X = ½ m

Example 7 root mean squareFind (a) the mean (b) mean square, and (c) the root mean square of 1, 3, 5, 7 and 9.

Solution: Mean = (1 + 3 + 5 + 7 + 9) / 5 = 5 Mean square

= (12 + 32 + 52 + 72 + 92)/5

= (1 + 9 + 25 + 49 + 81) / 5 = 33 Root mean square = [(12 + 32 + 52 + 72 + 92)/5] ½ = 5.74

Example 8It is given that the density of hydrogen gas at s.t.p. is 0.09 kg m-3. Find the r.m.s. speed of hydrogen molecules.

Solution: By P = ⅓(cr.m.s.

2)

1.01 x 105 = ⅓(0.09cr.m.s.2)

cr.m.s. = 1835 ms-1

Distribution of molecular speeds A gas contains a large number of

molecules in rapid motions. In each collision between

molecules, some molecules gain energy while others lose energy.

As a result, each molecule has a different speed.

containerParticles move around

freely at high speed

Simulation 1 Simulation 2

Maxwell distribution Note: Maxwell distribution is n

ot symmetric. When temperature incre

ases, the distribution curve flattens out but the area under the curve remains unchanged.

N

cccc N

smr

222

21

...

Most probable speed c0: speed possessed by the most number of

molecules

Root mean square speed cr.m.s.:

It is found that c0 < cm < cr.m.s.

Number of molecules with speed c

0c/m s-1

co cm

N

cccc N

m

21

Mean speed cm: mean speed of molecules

cr.m.s.

Maxwell distribution

Temperature, molar mass and root mean square speed

nRTNmc smr 2..3

1

2..3

1smrNmcPV nRTPV From and

For 1 mole of gas, n = 1, N = NA and the mass of the gas = m

olar mass Mm = mNA

RTmcN smrA 2..3

1

msmr M

RTc

3.. ⇒ ⇒

nRTNmc smr 2..3

1we have

Note: cr.m.s. increases with temperature. cr.m.s. decreases with the mass of the molecule or molar mas

s of the gas.

Tc smr ..

msmr M

RTc

3.. By

m

smrM

c1

..

Example 8Find the root mean square speed of hydrogen molecules at 27oC. It is given that R = 8.31 J K-1 mol-1.

Solution:

Bym

smr M

RTc

3..

cr.m.s. = (3 x 8.31 x 300 / 0.0020.002)½ = 1934 ms-1

Example 9Find the ratio of the root mean square speed of hydrogen to that of oxygen.

Bym

smr M

RTc

3..

For hydrogen: ch = (3RT/0.0020.002) ½

For oxygen: co = (3RT/0.0320.032) ½

ch / co = (0.032/0.002)½ = 4

the ratio of the root mean square speed of hydrogen to that of oxygen is 4.

Translational K.E. and Rotational K.E.

Translational motion

Rotational motion

Translational + rotational motion

Monoatomic and diatomic gases

MonoatomicMonoatomic means having only one atom in the molecule. All of the noble gases are monoatomic.

DiatomicDiatomic means having two atoms in the molecule. Examples are oxygen, nitrogen

kinetic energy of a monoatomic gas Translational K.E. = ½ mc2

Rotational motion = 0 (neglected)

Translational K.E. of a monoatomic gas2

3

1cNmPV nRTPV From and

For 1 mole of gas, n = 1, N = NA and

the mass of the gas = molar mass Mm = mNA

nRTcNm 2

3

1 RTcmN A 2

3

1⇒

Average translational kinetic energy of a molecule

= kTTN

Rcm

A 2

3

2

3

2

1 2 123 JK1038.1 AN

Rk where

which is known as Boltzmann constant.

nRTcNm 2

3

1we have

Note:

All types of gas molecule have the same average translational kinetic energy at the same temperature. ( )

For diatomic gases such as hydrogen, oxygen and nitrogen, the average kinetic energy (translational + rotational) of a molecule = .

(Out of syllabus)

kTcm2

3

2

1 2

kT2

5

Avogadro’s law Avogadro’s law states that under the same temperature

and pressure, equal volumes of different gases contain an equal number of molecules.

Proof

211111 3

1cmNVP

21113

1cmN 2

2223

1cmN

Consider 2 gases under the same temperature, pressure and volume.Let N1 and N2 be the number of molecules in gas 1 and gas 2

respectively.For gas 1:

222222 3

1cmNVP For gas 2:

Same volume and pressure ⇒ V1 = V2 and P1 = P2

∴

Same temperature ⇒ same average kinetic energy of a molecule 2222

12112

1 cmcm ∴

21113

1cmN

22223

1cmN ⇒ N1 = N2

Dalton’s law of partial pressures Dalton's law of partial pressures states that the total

pressure exerted by a mixture of gases is the sum of the pressures exerted by each gas if it were allowed to occupy the whole volume alone.

p1 p2 p

p = p1 + p2

Dalton’s law of partial pressuresPV = nRT

V

RTnP 1

1 V

RTnP 2

2

p1 p2 p

P = P1 + P2

V

RTnnP 21