Chapter 4 Ideal Gases

Question 1

Amount of Substance

Answer 1

-In thermodynamics, the amount of substance is measured in the SI unit ‘mole’
-This has the symbol mol
-The mole is a unit of substance, not a unit of mass
-The mole is defined as:
The SI base unit of an ‘amount of substance’. It is the amount containing as many particles (e.g. atoms or molecules) as there are atoms in 12 g of carbon-12

Question 2

The Avogadro Constant

Answer 2

• A carbon-12 atom has a mass of:

12 u = 12 × 1.66 × 10^-27 = 1.99 × 10^-26 kg

• The exact number for a mole is defined as the number of molecules in exactly 12 g of carbon:

1 mole= 0.012/1.99x10^-26= 6.02 x 10²³ molecules

• Avogadro’s constant (N_A) is defined as:

The number of atoms of carbon-12 in 12 g of carbon-12; equal to 6.02 × 10²³ mol^-1

• For example, 1 mole of sodium (Na) contains 6.02 × 10²³ atoms of sodium
• The number of atoms can be determined if the number of moles is known by multiplying by N_A, for example:

2.0 mol of nitrogen contains: 2.0 × N_A = 2.0 × 6.02 × 10²³ = 1.20 × 10²⁴ atoms

Question 3

Mole and the Atomic Mass

Answer 3

• One mole of any element is equal to the relative atomic mass of that element in grams
  
  • E.g. Helium has an atomic mass of 4 – this means 1 mole of helium has a mass of 4 g
• If the substance is a compound, add up the relative atomic masses, for example, water (H₂O) is made up of
  
  • 2 hydrogen atoms (each with atomic mass of 1) and 1 oxygen atom (atomic mass of 16)
  • So, 1 mole of water would have a mass of (2 × 1) + 16 = 18 g

Question 4

Molar Mass

Answer 4

• The molar mass of a substance is the mass, in grams, in one mole
  
  • Its unit is g mol^-1
• The number of moles from this can be calculated using the equation:

no moles= mass(g)/molar mass (g mol^-1)

Question 5

Ideal Gases

Answer 5

• An ideal gas is one which obeys the relation:

pV ∝ T

• Where:
  
  • p = pressure of the gas (Pa)
  • V = volume of the gas (m³)
  • T = thermodynamic temperature (K)

Question 6

Gas molecules move about randomly. what other conditions affect

Answer 6

• temperature
• momentum
• pressure
• volume

Question 7

Gas molecules move about randomly. what other conditions affect: temperature

Answer 7

• The temperature of a gas is related to the average speed of the molecules:
  
  • The hotter the gas, the faster the molecules move
  • Hence the molecules collide with the surface of the walls more frequently

Question 8

Gas molecules move about randomly. what other conditions affect: momentum

Answer 8

• Each collision applies a force across the surface area of the walls
• The faster the molecules hit the walls, the greater the force on them

Question 9

Gas molecules move about randomly. what other conditions affect: pressure

Answer 9

• Since pressure is the force per unit area
  
  • Higher temperature leads to higher pressure

Question 10

Gas molecules move about randomly. what other conditions affect: volume

Answer 10

• If the volume V of the box decreases, and the temperature T stays constant:
  
  • There will be a smaller surface area of the walls and hence more collisions
  • This also creates more pressure

Question 11

greater force per unit area, pressure in an ideal gas is therefore defined by:

Answer 11

The frequency of collisions of the gas molecules per unit area of a container

Question 12

Boyle’s Law

Answer 12

• If the temperature T is constant, then Boyle’s Law is given by:

p ∝ 1/V

• This leads to the relationship between the pressure and volume for a fixed mass of gas at constant temperature:

P₁V₁ = P₂V₂

Question 13

Charles’s Law

Answer 13

• If the pressure P is constant, then Charles’s law is given by:

V ∝ T

• This leads to the relationship between the volume and thermodynamic temperature for a fixed mass of gas at constant pressure:

V₁/T₁ = V₂/T₂

Question 14

Pressure Law

Answer 14

• If the volume V is constant, the the Pressure law is given by:

P ∝ T

• This leads to the relationship between the pressure and thermodynamic temperature for a fixed mass of gas at constant volume:
• P₁/T₁ = P₂/T₂

Question 15

Ideal Gas Equation

Answer 15

pV = nRT

or

pV= NkT

• An ideal gas is therefore defined as:

A gas which obeys the equation of state pV = nRT at all pressures, volumes and temperatures

Question 16

The Boltzmann Constant

Answer 16

• The Boltzmann constant k is used in the ideal gas equation and is defined by the equation:

K= R/N_A

• Where:
  
  • R = molar gas constant
  • N_A = Avogadro’s constant
• Boltzmann’s constant therefore has a value of

K=8.13/6.02x10²³=1,38x10^-23J K^-1

Question 17

• The Boltzmann constant relates the properties of ..

Answer 17

• microscopic particles (e.g. kinetic energy of gas molecules) to their macroscopic properties (e.g. temperature)
  
  • This is why the units are J K^-1
• Its value is very small because the increase in kinetic energy of a molecule is very small for every incremental increase in temperature

Question 18

Kinetic Theory of Gases

Answer 18

• Gases consist of atoms or molecules randomly moving around at high speeds
• The kinetic theory of gases models the thermodynamic behaviour of gases by linking the microscopic properties of particles (mass and speed) to macroscopic properties of particles (pressure and volume)

Question 19

Assumptions of the Kinetic Theory of Gases

Answer 19

• Molecules of gas behave as identical, hard, perfectly elastic spheres
• The volume of the molecules is negligible compared to the volume of the container
• The time of a collision is negligible compared to the time between collisions
• There are no forces of attraction or repulsion between the molecules
• The molecules are in continuous random motion
• The number of molecules of gas in a container is very large, therefore the average behaviour (eg. speed) is usually considered

Question 20

Root-Mean-Square Speed

Answer 20

• The pressure of an ideal gas equation includes the mean square speed of the particles:

2>

• Where
  
  • c = average speed of the gas particles
  • 2> has the units m² s^-2

Question 21

In order to find the pressure of the gas, the

Answer 21

• velocities must be squared
  
  • This is a more useful method, since a negative or positive number squared is always positive

Question 22

• To calculate the average speed of the particles in a gas, take the square root of the mean square speed:

Answer 22

• c_r.m.s is known as the root-mean-square speed and still has the units of m s^-1
• The mean square speed is not the same as the mean speed

Question 23

5 Step Derivation of the Kinetic Theory of Gases Equation

Answer 23

1.Find the change in momentum as a single molecule hits a wall perpendicularly

2. Calculate the number of collisions per second by the molecule on a wall

3.Find the change in momentum per second

4. Calculate the total pressure from N molecules

5. Consider the effect of the molecule moving in 3D space

Question 24

5 Step Derivation of kinetic theory of gases: No. 1

Find the change in momentum as a single molecule hits a wall perpendicularly

Answer 24

• One assumption of the kinetic theory is that molecules rebound elastically
• This means there is no kinetic energy lost in the collision
• If they rebound in the opposite direction to their initial velocity, their final velocity is -c
• The change in momentum is therefore:

Δp = −mc − (+mc) = −mc − mc = −2mc

Question 25

5 Step Derivation of kinetic theory of gases: No. 2

Calculate the number of collisions per second by the molecule on a wall

Answer 25

• The time between collisions of the molecule travelling to one wall and back is calculated by travelling a distance of 2l with speed c:
• time between collisions = distance/speed = 2l/c
• Note: c is not taken as the speed of light in this scenario

Question 26

5 Step Derivation of kinetic theory of gases: No. 3

Find the change in momentum per second

Answer 26

• The force the molecule exerts on one wall is found using Newton’s second law of motion:

Force = rate of change of momentum = ∇p/∇t = 2mc/(2l/c) = mc²/l

• The change in momentum is +2mc since the force on the molecule from the wall is in the opposite direction to its change in momentum

Question 27

5 Step Derivation of kinetic theory of gases: No. 4

Calculate the total pressure from N molecules

Answer 27

• The area of one wall is l²
• The pressure is defined using the force and area:

Pressure p = Force/area =(mc²/l) / l² = mc² / l³

• This is the pressure exerted from one molecule
• To account for the large number of N molecules, the pressure can now be written as:

p= Nmc²/l³

• Each molecule has a different velocity and they all contribute to the pressure
• The mean squared speed of c² is written with left and right-angled brackets 2>
• The pressure is now defined as:
• p= Nm2> / l³

Question 28

5 Step Derivation of kinetic theory of gases: No. 5

Consider the effect of the molecule moving in 3D space

Answer 28

• The pressure equation still assumes all the molecules are travelling in the same direction and colliding with the same pair of opposite faces of the cube
• In reality, all molecules will be moving in three dimensions equally
• Splitting the velocity into its components c_x, c_y and c_z to denote the amount in the x, y and z directions, c² can be defined using pythagoras’ theorem in 3D:

c² = c_x² + c_y² + c_z²

• Since there is nothing special about any particular direction, it can be determined that:

x²> = y²> = z²>

• Therefore, x²> can be defined as:

x²> = ⅓ x²>

• The box is a cube and all the sides are of length l
  
  • This means l³ is equal to the volume of the cube, V
• Substituting the new values for 2> and l³ back into the pressure equation obtains the final equation:

pV=⅓ Nmx²>

Question 29

Kinetic Theory of Gases equation

Answer 29

• This can also be written using the density ρ of the gas:

p= mass/volume = Nm/V

• Rearranging the pressure equation for p and substituting the density ρ:

p = ⅓ p²>

Question 30

Average Kinetic Energy of a Molecule

Answer 30

• An important property of molecules in a gas is their average kinetic energy
• This can be deduced from the ideal gas equations relating pressure, volume, temperature and speed

Question 31

Average Kinetic Energy of a Molecule equation

Answer 31

• Recall the ideal gas equation:

pV = NkT

• Also recall the equation linking pressure and mean square speed of the molecules:

pV=⅓ Nm²>

• The left hand side of both equations are equal (pV)
• This means the right hand sides are also equal:

⅓ Nm²≥ NkT

• N will cancel out on both sides and multiplying by 3 obtains the equation:

m2> = 3kT

• Recall the familiar kinetic energy equation from mechanics:

kinetic energy = ½mv²

• Instead of v² for the velocity of one particle, 2> is the average speed of all molecules
• Multiplying both sides of the equation by ½ obtains the average translational kinetic energy of the molecules of an ideal gas:

E_k = ½ m²≥3/2 kT

• Where:
  
  • E_K = kinetic energy of a molecule (J)
  • m = mass of one molecule (kg)
  • 2> = mean square speed of a molecule (m² s^-2)
  • k = Boltzmann constant
  • T = temperature of the gas (K)

Question 32

• A key feature of this equation is that the mean kinetic energy of an ideal gas molecule is proportional to its thermodynamic temperature

Answer 32

E_k ∝ T

Question 33

Translational kinetic energy is defined as

Answer 33

The energy a molecule has as it moves from one point to another

• A monatomic (one atom) molecule only has translational energy, whilst a diatomic (two-atom) molecule has both translational and rotational energy