Chapter 13 Gas Laws and Kinetic Theoryyslphysics.weebly.com/.../8/2/6/48261583/tutorial_1_c13.pdf · 2019-05-27 · CHAPTER 13 GAS LAWS AND KINETIC THEORY prepared by Yew Sze Ling@Fiona,

CHAPTER 13 GAS LAWS AND KINETIC THEORY prepared by Yew Sze Ling@Fiona, KML

1

Chapter 13 Gas Laws and Kinetic Theory

Curriculum Specification Remarks

Before After Revision

13.1 Ideal Gas Equations

a) Solve problems related to ideal gas equation, 𝑝𝑉 = 𝑛𝑅𝑇

(C3, C4)

b) Discuss the following graphs for an ideal:

i. p-V graph at constant at temperature.

ii. V-T graph at constant at pressure.

iii. p-T graph at constant volume.

(C1, C2)

13.2 Kinetic Theory of Gases

a) State the assumptions of kinetic theory of gases. (C1, C2)

b) Discuss root mean square (rms) speed of gas molecules.

(C1, C2)

c) Solve problems related to root mean square (rms) speed

of gas molecules. (C3, C4)

d) Solve problems related to the equations,

and pressure,

(C3, C4)

13.3 Molecular Kinetic Energy and Internal Energy

a) Discuss translational kinetic energy of a molecule,

(C1, C2)

b) Discuss degrees of freedom, f for monoatomic, diatomic

and polyatomic gas molecules. (C1, C2)

c) State the principle of equipartition of energy. (C1, C2)

d) Discuss internal energy of gas. (C1, C2)

e) Solve problems related to internal energy,

(C3, C4)

𝑝𝑉 =1

3𝑁𝑚𝑣𝑟𝑚𝑠

2 𝑝 =1

3𝜌𝑣𝑟𝑚𝑠

2

𝐾𝑡𝑟 =3

2(𝑅

𝑁𝐴)𝑇 =

3

2𝑘𝑇

𝑈 =1

2𝑓𝑁𝑘𝑇


2

13.1 Ideal Gas Equations

Gas Law Equation p-V Graph

Boyle’s Law

For a fixed amount of gas at a constant

temperature, gas pressure is inversely

proportional to gas volume.

Mathematically,

Thus,

Charles’s Law


pressure, gas volume is directly

proportional to its absolute temperature.

Mathematically,

Thus,

Gay-Lussac (Pressure) Law


volume, gas pressure is directly

proportional to its absolute temperature.

Mathematically,

Thus,

𝑃1𝑉1 = 𝑃2𝑉 2

𝑃𝑉 = constant

𝑃 ∝ 1

𝑉

if T is constant

𝑉 ∝ 𝑇

if P is constant

𝑉

𝑇= constant

𝑉1𝑇1

=𝑉2𝑇2

𝑃 ∝ 𝑇

if V is constant

𝑃

𝑇= constant

𝑃1𝑇1

=𝑃2𝑇2


3

Ideal Gas Law

These 3 laws can be combined into a single more general relation between absolute pressure, volume and absolute temperature of a fixed quantity of gas:

In equation form:

1 mole of any gas at Standard Temperature and Pressure (S.T.P.):

T = 273.15 K, P = 101.3 kPa, Vm = 22.4 liters = 22.4 dm3:

This constant value is denoted by symbol of R and known as molar gas constant. Value of R is

found experimentally same for all gases.

In general, if n moles occupy a volume, V, at pressure, P, and absolute temperature T, we have:

In term of Boltzmann constant, k:

The number of moles of gas, n, can also be expressed as:

𝑃𝑉 ∝ 𝑇

𝑃𝑉

𝑇= constant

𝑃1𝑉1𝑇1

=𝑃2𝑉2𝑇2

Also known as the

combined gas law

𝑃𝑉𝑚𝑇

= 101.3 × 103 22.4 × 10−3

273.

15

𝑃𝑉𝑚𝑇

= 𝟖.𝟑𝟏 𝐉 𝐊−𝟏 𝐦𝐨𝐥−𝟏

𝑃𝑉 = 𝑛𝑅𝑇 where 𝑉 = 𝑛𝑉 𝑚

Molar mass constant

𝑃𝑉 = 𝑁𝑘𝑇 where 𝑘 =𝑅

𝑁𝐴= 1.38 × 10−23 J K− 1

Boltzmann constant

𝑛 =𝑚

𝑀

𝑛 =𝑁

𝑁𝐴

OR

Total mass of gas (in kg)

Molar mass of gas (kg mol-1

)

Number of gas molecules

Avogadro number, 6.02×10

23 mol

-1


4

13.2 Kinetic Theory of Gases

The assumptions of kinetic theory of gases:

1. All gases are made up of atoms or molecules.

2. All atoms or molecules of the gas are in continuous and completely random motion.

3. The volume of the gas molecules is negligible when compared with the volume occupied by the gas.

4. The gas molecules undergo perfectly elastic collisions with each other and with the walls of the

container.

5. The intermolecular forces are negligible except during collisions.

6. The time of collisions between molecules is negligible compared to the time between successive collisions.

7. The mass of each molecule is so small that the effect of gravity on it is negligible.

Additional Knowledge: Force exerted by an ideal gas

According to the kinetic theory of gasses, gas pressure is the result of

the collisions of large number of molecules on the wall of the

container.

Consider a cube of side d, containing N molecules of gas each of mass,

m as shown in the figure.

Each molecule of the gas have the mass, m, and velocity v.

Consider one molecule with component of velocity, vx strikes the wall:

Collision is perfectly elastic, the velocity after collision is –vx.

Thus, change in momentum of the molecule:

Δ𝑝𝑥 = −𝑚𝑣𝑥 − −𝑚𝑣𝑥 = 2𝑚𝑣𝑥

Assume no collision between molecules, the molecule travels a distance of 2d (from A to B

and back to A) before it collides with the same wall.

The time taken for this movement:

If (Fx)1 is the magnitude of the average force exerted by a molecule on the wall in the time Δt, thus by applying Newton’s second law of motion:

Δ𝑡 =2𝑑

𝑣𝑥

𝐹𝑥1 =Δ𝑝𝑥

Δ𝑡=

2𝑚𝑣𝑥

(2𝑑𝑣𝑥

)= (

𝑚

𝑑)𝑣𝑥

2

For one molecule


5

Additional Knowledge: Force exerted by an ideal gas (cont.)

If the cube box contains N molecules moving with velocities of vx1, vx2, vx3, …, vxN then the

force on the wall due to N molecules colliding with it is:

The mean (average) value of the square of the velocity in the x direction for N molecules:

Thus, the x-component for the total force exerted on the wall of the cubical container:

The magnitude of the velocity v:

𝑣2 = 𝑣𝑥2 + 𝑣𝑦

2 + 𝑣𝑧2

⟨𝑣2⟩ = ⟨𝑣𝑥2⟩ + ⟨𝑣𝑦

2⟩ + ⟨𝑣𝑧2⟩

The total force exerted on the wall in all direction, F is given by:

𝐹 =1

3

𝑁𝑚⟨𝑣2⟩

𝑑

Since the N particles move randomly in three dimensions, one-third of them on the average

strike the right wall during the time t.

Additional Knowledge: Pressure of an ideal gas

𝐹𝑥 = (𝑚

𝑑)𝑣𝑥1

2 + (𝑚

𝑑)𝑣𝑥2

2 + ⋯+ (𝑚

𝑑)𝑣𝑥𝑁

2

𝐹𝑥 =𝑚

𝑑 𝑣𝑥1

2 + 𝑣𝑥22 + ⋯+ 𝑣𝑥𝑁

2

⟨𝑣𝑥2⟩ =

𝑣𝑥12 + 𝑣𝑥2

2 + ⋯+ 𝑣𝑥𝑁2

𝑁

𝑣𝑥12 + 𝑣𝑥2

2 + ⋯+ 𝑣𝑥𝑁2 = 𝑁⟨𝑣𝑥

2⟩

Average value

𝐹𝑥 =𝑚

𝑑𝑁⟨𝑣𝑥

2⟩

𝑃 =𝐹

𝐴 where 𝐴 = 𝑑2 and 𝐹 =

1

3

𝑁𝑚⟨𝑣2⟩

𝑑

𝑃 =1

3

𝑁𝑚⟨𝑣2⟩

𝑑3 where 𝑉 = 𝑑3

𝑃 =1

3

𝑁𝑚⟨𝑣2⟩

𝑉

𝑃𝑉 =1

3𝑁𝑚⟨𝑣2⟩


6

Root mean square (rms) speed

The square root of the mean of the square of the speed is called root mean square speed, vrms:

𝑣𝑟𝑚𝑠 = √⟨𝑣2⟩

𝑣𝑟𝑚𝑠2 = ⟨𝑣2⟩

For ideal gas:

A gas which obeys all the assumptions of the kinetic theory of gases is an ideal gas. The ideal gas equations are given by:

13.3 Molecular Kinetic Energy and Internal Energy

Translational Kinetic Energy

From the equation of ideal gas:

Since 𝑃𝑉 = 𝑁𝑘𝑇, average translational kinetic energy for a molecule is

Since , average translational kinetic energy for a molecule can also be written as:

𝑣𝑟𝑚𝑠 = 3𝑘𝑇

𝑚

𝑣𝑟𝑚𝑠 = 3𝑅𝑇

𝑀𝑣𝑟𝑚𝑠 =

3𝑃

𝜌

Mass of a gas

molecule (kg)

Molar mass of

a gas (kg mol-1

)

Density of the

gas molecules

𝑃𝑉 =1

3𝑁𝑚𝑣𝑟𝑚𝑠 2

OR

𝑃 =1

3𝜌𝑣𝑟𝑚𝑠 2 where 𝜌 =

𝑁𝑚

𝑉

Mass of one

molecule, m (kg)

Total volume of

the gas (container)

𝑃𝑉 =1

3𝑁𝑚𝑣𝑟𝑚𝑠 2

𝑃𝑉 =2

3𝑁 (

1

2𝑚𝑣𝑟𝑚𝑠

2 )

𝑃𝑉 =2

3𝑁𝐾 𝑇𝑟

𝐾𝑇𝑟 =3

2 𝑘𝑇

2

3𝐾𝑇𝑟 = 𝑘𝑇

𝑘 =𝑅

𝑁𝐴

𝐾𝑇𝑟 =3

2(𝑅

𝑁𝐴) 𝑇


7

The equation indicates that the Kelvin temperature is directly proportional to the average

translational kinetic energy per molecule in an ideal gas, no matter what the pressure and volume

are. On the average, the molecules have greater kinetic energies when the gas is hotter than when it is cooler.

For N molecules of an ideal gas in the cubical container, the total average (mean) translational

kinetic energy, K is

Since , thus, for n moles of an ideal gas in the cubical container, the total average (mean)

translational kinetic energy, K is

Degrees of freedom, f

The degrees of freedom of a molecule are the number of independent ways in which an atom or molecule can absorb or release or store the energy.

3 type of orientation mode in molecule:

The number of degree of freedom of a molecule depends on whether the molecule is

monoatomic, diatomic or polyatomic.

Monoatomic gas

Example of gases:

Helium, Neon, Argon, Krypton, Xenon and Radon.

3 degrees of freedom which consists of translational (linear) motion along x, y and z.

𝐾 = 𝑁𝐾 𝑇𝑟All (total) molecules One molecule

𝐾 =3

2 𝑁𝑘𝑇 𝐾 =

3

2𝑁 (

𝑅

𝑁𝐴) 𝑇 OR

𝑛 =𝑁

𝑁𝐴

𝐾 = 3

2𝑛𝑅𝑇


8

Diatomic gases/ Linear

polyatomic

Example of gases:

H2, O2, Cl2, N2, CO

5 degrees of freedom

Translational Motion = 3

Rotational Motion = 2

Rotational motion about y-axis is negligible because distance of

the diatom from axis of rotation is very small compare to x and z, thus moment of inertia about y-axis is negligible.

Nonlinear Polyatomic

Example of gases:

H2O, CO2, O3, NH3, N2O4, H2S

6 degrees of freedom

Translational Motion = 3

Rotational Motion = 3

Number of degree of freedom for various types of molecule

discussed applies for gases at normal temperature.

Number of degree of freedom for various types of molecule discussed applies for gases at

normal temperature.

At higher temperature, vibrational motion increases the number of degree of freedom.

Vibrational mode for diatomic at high temperature contributes 2 degrees of

freedom which correspond to the kinetic energy and the potential energy

associated with vibrations along the bond between the atoms.

Principle of Equipartition of Energy

Principle of Equipartition of Energy states that, the mean (average) kinetic energy of each degree

of freedom of a molecule is 𝟏

𝟐𝒌𝑻. Therefore, for a molecule which has f degrees of freedom:

𝐾𝑇𝑟 =𝑓

2 𝑘𝑇 𝐾𝑇𝑟 =

𝑓

2(𝑅

𝑁𝐴) 𝑇 OR


9

Internal Energy, U

The internal energy of a gas is the sum of the total mean kinetic energy (K) and total

potential energy of the gas molecules.

For an ideal gas, the intermolecular forces are assumed to be negligible. Thus, the potential

energy of the molecules can be neglected.

Therefore,

Internal energy, U = Total average kinetic energy, K

For an ideal gas with a total of f degrees of freedom, the total internal energy is

where

U = Internal energy

f = Number of degree of freedom

T = Absolute temperature (Kelvin)

N = Number of molecules

n = Number of moles

k = Boltzmann constant R = Molar gas constant

Summary

𝑈 =𝑓

2𝑛𝑅𝑇𝑈 =

𝑓

2 𝑁𝑘𝑇 OR


10

Exercise

Ideal Gas Equations

1. An ideal gas at 15.5 °C and a pressure of 1.72 ×105 Pa occupies a volume of 2.81 m

3.

a) How many moles of gas are present?

b) If the volume is raised to 4.16 m3 and the temperature raised to 28.2 °C, what will be the

pressure of the gas?

(Answer: 201 mol ; 1.21×105 Pa)

2. A young male adult takes in about 5.0 ×10-4

m3 of fresh air during a normal breath. Fresh air

contains approximately 21% oxygen. Assuming that the pressure in the lungs is 1.0×105 Pa

and that air is an ideal gas at a temperature of 310 K, find the number of oxygen molecules

in a normal breath. (Answer: 2.5×1021

oxygen molecules)

3. The volume of an ideal gas is held constant. Determine the ratio P2/P1 of the final pressure to

the initial pressure when the temperature of the gas rises

a) from 35.0 to 70.0 K b) from 35.0 to 70.0 °C

(Answer: 2.00 ; 1.11)

4. A scuba tank is filled with air to a gauge pressure of 204 atm when the air temperature is

29°C. A diver then jumps into the ocean and, after a short time on the ocean surface, checks

the tank’s gauge pressure and finds that it is only 191 atm. Assuming the diver has inhaled a

negligible amount of air from the tank, what is the temperature of the ocean water?

(Answer: 283 K)

Kinetic Theory of Gases

1. The speeds of some molecules in a volume of hydrogen gas are 3.5×104 m s

-1, 3.0×10

4 m s

-1,

3.6×104 m s

-1, 4.0×10

4 m s

-1. Calculate the rms speed of the hydrogen gas.

(Answer: 3.54×104 m s

-1)

2. Very fine smoke particles are suspended in air. The translational rms speed of a smoke

particle is 2.8×10-3

m s-1

, and the temperature is 301 K. Find the mass of a particle.

(Answer: 1.6×10-15

kg)

3. An oxygen molecule is moving near the earth’s surface. Another oxygen molecule is moving

in the ionosphere (the uppermost part of the earth’s atmosphere) where the Kelvin

temperature is three times greater. Determine the ratio of the translational rms speed in the ionosphere to the translational rms speed near the earth’s surface. (Answer: 1.73)

4. The density of oxygen gas at STP is 1.43 kg m-3

. Estimate the rms speed of oxygen

molecules at STP. Assume that the gas behaves as an ideal gas. (Answer: 461.1 m s-1

)

5. The rms speed of hydrogen molecules at STP is 1.84 km s-1

. Determine the rms speed of the

molecules of oxygen gas at STP. (Molar mass of hydrogen = 2.0 g mol-1

, Molar mass of

oxygen = 32.0 g mol-1

) (Answer: 0.46 km s-1

)

6. The root mean square speed of hydrogen gas is 3×104 m s

-1. If the number of hydrogen

molecules per cm3 is 6.00×10

15, calculate the pressure of the hydrogen gas.

(Answer: 5.98×103 Pa)


11

Molecular Kinetic Energy and Internal Energy

1. What is the average translational kinetic energy of a molecule in an ideal gas at 37°C?

(Answer: 6.24×10-21

J)

2. Determine the average kinetic energy of a gas molecule at 300 K is the gas is

a) He

b) N2

Neglect vibrational mode.

(Answer: 0.62×10-20

J ; 1.04×10-20

J)

3. One mole of oxygen has a mass of 32 g. Assuming oxygen behaves as an ideal gas,

calculate,

a) the volume occupied by one mole of oxygen gas,

b) the density of oxygen gas

c) the r.m.s. speed of its molecules

d) the average translational kinetic energy of a molecule

e) the internal energy of oxygen gas

at 273 K and pressure of 1.01×105 Pa

(Answer: 2.25×10-2

m3 ; 1.42 kg m

-3 ; 461 m s

-1 ; 5.65×10

-21 J ; 5.67×10

3 J)

4. Compressed air can be pumped underground into huge caverns as a form of energy storage.

The volume of a cavern is 5.6×105 m

3, and the pressure of the air in it is 7.7×10

6 Pa. Assume

that air is a diatomic ideal gas.

a) Determine the internal energy produced by the diatomic gas in the cavern.

b) If one home uses 30.0 kW h of energy per day, how many homes could this internal energy serve for one day?

(Answer: 1.1×1013

J ; 1.0×105 homes)

5. Suppose a tank contains 680 m3 of neon (Ne) at an absolute pressure of 1.01×10

5 Pa. The

temperature is changed from 293.2 K to 294.3 K. What is the increase in the internal energy

of the neon? (Answer: 3.9×105 J)

Documents

Chapter 13 Gas Laws and Kinetic Theoryyslphysics.weebly.com/.../8/2/6/48261583/tutorial_1_c13.pdf · 2019-05-27 · CHAPTER 13 GAS LAWS AND KINETIC THEORY prepared by Yew Sze Ling@Fiona,