Module 5: C15 - Ideal Gases

Question 1

Equation for Pressure

Answer 1

Pressure = Force/Area

Question 2

How can we express the number of atoms or molecules in a given volume of air

Answer 2

We can express the number of atoms or molecules in a given volume of gas, using moles (mol), the SI unit of measurement for the amount of a substance.

Question 3

What is 1 mol of any substance

Answer 3

1 mol of any substance contains 6.02x10^23 individual atoms or molecules. Therefore, the total number of atoms or molecules in a substance, N, is given by the equation

N = n x Na

(n is the number of moles in the substance)

Question 4

What equation gives the total number of atoms or molecules in a substance

Answer 4

N = n x Na

(n is the number of moles in the substance)

Question 5

Equation for molar mass

Answer 5

m = n x M

Question 6

What is the kinetic theory of matter

Answer 6

The kinetic theory of matter is a model used to describe the behaviour of the atoms or molecules in an ideal gas. Real gases have complex behaviour, so to keep the model simple, a number of assumptions are made about the atoms or molecules in an ideal gas.

Question 7

What are the assumptions made in the kinetic model for an ideal gas

Answer 7

• The gas contains a very large number if atoms or molecules moving in random directions with random speeds
• The atoms or molecules of the gas occupy a negligible volume compared with the volume of the gas.
• The collisions of atoms or molecules with each other and the container walls are perfectly elastic (no kinetic energy is lost)
• The time of collisions between the atoms or molecules is negligible compared to the time between collisions.
• Electrostatic forces between atoms or molecules are negligible except during collisions.

Question 8

6 assumptions made about ideal gases and kinetic theory

Answer 8

1. Molecules are points- the volume of the molecules
   is insignificant compared to the volume of the ideal
   gas.
2. Molecules do not attract each other– if they did
   then the pressure exerted by the gas on its container
   would be reduced.
3. Molecules move in constant random motion.
4. All collisions between gas molecules and their
   container are elastic– there is no loss of kinetic
   energy.
5. The time taken for a collision is much shorter
   than the time between collisions
6. Any sample of an ideal gas contains a very large
   number of molecules.

Question 9

What is the Avogadro Constant

Answer 9

The Avagadro constant NA is equal to the number of atoms in exactly 12g of the isotope carbon 12.

To 3sf. : NA = 6.02 x 10^23

Question 10

How does the pressure and volume relate to each other in ideal gases

Answer 10

If the temperature and mass of a gas remain constant, then the pressure of an ideal gas is inversely proportional to its volume V. This can be expressed as

p ∝ 1/V

or

PVA = constant

Question 11

What is the equation to show When a gas changes
pressure from p1 to p2 while undergoing a volume change from V1 to V2

Answer 11

P1 x V1 = P2 x V2

Question 12

How does the pressure and temperature relate to each other in ideal gases

Answer 12

If the volume and mass of the gas remain constant the pressure p of an ideal gas is directly proportional to its absolute (thermodynamic) temperature T in kelvin. This relationship can be expressed as

p ∝ T
or
p/T = constant

Question 13

What is an ideal gas defined as

Answer 13

An ideal gas is defined as a gas that obeys
Boyle’s law at all pressures.

An ideal gas is defined as a gas that obeys
Boyle’s law at all pressures. Real gases do not obey Boyle’s law at very high pressures or when they are cooled to near their condensation point.

Question 14

What is the equation of state of an ideal gas

Answer 14

pV/T = nR

or

pV = nRT

Question 15

Boyle’s law question:

A gas has an initial volume of 300 m^3 at standard
temperature and pressure (100 kPa). Calculate the final volume of this gas if its pressure is increased by 400 kPa at a constant temperature.

Answer 15

P1 = 100kOa = 100x10^3 Pa
V1 = 300m^3
P2 = 500kPa =500x10^3 Pa
V2 = ?

P1V1 = P2V2

100x10^3 = 300 = 500x10^3 V2
V2 = 60m^3

Question 16

Pressure law question:

A gas has an initial pressure of 100kPa at a temperature of 27°C. Calculate the final pressure if its temperature is increased by 300°C at a constant volume.

Answer 16

P1 = 100kPa = 100x10^3 Pa
T1 = 27°C = 30pK
P2 = ?
T2 = 327°C = 600K

P1/T1 = P2/T2
P2 = P1T2/T1

100x10^3 / 300 = p2 / 600

600 (100x10^3) / 300 = P2

P2 = 200MPa

Question 17

What is Charles’ Law (equation relating volume and temperature in an ideal gas)

Answer 17

For a fixed mass of gas at a constant pressure:

V/T = Constant

When a gas changes volume from V1 to V2 while undergoing a temperature change from T1 to T2 :

V1/T1 = V2/T2

Question 18

Example Question:

A gas has an initial volume of 50 m^3 at a temperature of 127°C. Calculate the final temperature required in 0°C to decrease the volume to 20m^3 at a constant pressure.

Answer 18

V1/T1 = V2/T2
T2 = V2T1/V1

T2 = 20x400 / 50
0K = -273°C
T2 = 160K
160-273 = -113°C

T2 = -113°C

Question 19

What is the Ideal Gas equation and how do you get it

Answer 19

Combining all three gas laws for a constant mass of gas gives:

pV/T = a constant

the constant = nR and so:

Therefore:
pV = nRT

(n = number of moles of the gas
R = molar gas constant = 8.31 J K^-1 mol^-1)

Question 20

Equation for Number of Molecules involving Number of Moles and Avogadros Constant

Answer 20

N = nNa

(Number of molecules = number of moles x avogadros constant)

Question 21

Equation for mass of a gas involving the number of moles and molar mass

Answer 21

m = nM

Mass of Gas = number of moles x Molar Mass

Question 22

Calculate the volume of 1 mole of an ideal gas at 0°C and 101kPa using pV=nRT

Answer 22

pV = nRT
V = nRT/p

V = 1x8.31x273 / 101x10^3
V = 0.0225m^3
V = 2.25x10^-2 m^3

Question 23

Example Question:
A fixed mass of gas has its pressure increased from 101 kPa to 303 kPa, its volume increased by 5 m3 to 1 m3 whilst its temperature is raised from 200C. Calculate its final temperature.

Answer 23

p1V1/T1 = p2V2/T2
T2= P2V2T1/P1V1

T2 = 303x10^3 x 6 x 293 / 101x10^3 x 1

T2 = 5274 K
or
T2 = 5001°C

Question 24

Example Question:

A container of volume 2.0x10^-3 m^3,
temperature 20°C, contains 60g of oxygen of
molar mass 32g. Calculate its pressure.

Answer 24

pV = nRT
becomes: p = nRT/V
But n = m/M
T = 20°C = 293K

n = 60/32
n = 1.875

p = nRT/V
p = 1.875x8.31x293 / 2x10^-3
p = 2.28x10^6 Pa

p = 2.28 MPa

Question 25

What is the Boltzmann Constant, and how it is found

Answer 25

The Boltzmann constant, k = R/Na k = 1.38x10^-23 JK^-1

Question 26

Example Question: Estimate the number of air molecules in this room. [Typical values: room volume = 300m^3] room temperature = 20°C = 293K atmospheric pressure = 101 kPa = 101x10^3 Pa

Answer 26

pV = nRT N = pV/kT N = 101x10^3 x 300 / 1.38x10^-23 x 293 N = 7.49x10^27 N = 7.50x10^27

Question 27

How can pV = nRT become pV = NkT

Answer 27

n = N/Na pV = nRT pV = N/Na x RT pV = R/Ra x NT k = R/Na And so the ideal gas equation can also be stated as: pV = NkT

Question 28

What is the Kinetic Theory of Gases

Answer 28

The kinetic theory of gases states that a gas consists of point molecules moving about in random motion.

Question 29

According to the kinetic theory, what happens when the Volume of a container is decreased

Answer 29

– There will be a greater number of molecules hitting the inside of the container per second – A greater force will be exerted – Pressure will increase

Question 30

According to the kinetic theory, what happens when the Temperature of a container is decreased

Answer 30

If the temperature of a container is increased: – Molecules will be moving at greater speeds. – More molecules will be hitting the inside of the container per second and they will each exert a greater force. – A greater overall force will be exerted – Pressure will increase

Question 31

What is the equation for the Root Mean Square Speed

Answer 31

Crms = √(C1^2 + C2^2 + C3^2 + C4^2 / N) _ C^2 = Crms^2 Where: _ C^2 is the mean square speed

Question 32

Example Question: Calculate the RMS speed of four molecules having speeds 300, 340, 350 and 380 ms^-1

Answer 32

√(300^2 + 340^2 + 350^2 + 380^2 / 4) = 75√21 = 344ms^-1

Question 33

What is the Kinetic Theory Equation

Answer 33

For an ideal gas containing N identical molecules, each of mass, m in a container of volume, V, the pressure, p of the gas is given by: pV = 1/3 Nmc̅^2

Question 34

Example Question: A container of volume 0.05m^3 has 0.4kg of an ideal gas at a pressure of 2.0x10^7 Pa. Calculate the RMS speed of the gas molecules.

Answer 34

pV = 1/3 Nmc̅^2 c̅^2 = 3pV / Nm Nm = mass of the gas = 0.4kg c̅^2 = 3 x 2x10^7 x 0.05 / 0.4 c̅^2 = 7.5x10^6 ms^-1 Crms = √7.5x10^6 ms^-1 Crms = 2740ms^-1

Question 35

How do you work out Average Molecular Kinetic Energy

Answer 35

If 𝒑𝑽 =𝟏/𝟑𝑵𝒎c̅^𝟐 and 𝒑𝑽 = 𝑵𝒌𝑻 then: 𝟏/𝟑𝑵𝒎c̅^𝟐 = 𝑵𝒌𝑻 𝟏/𝟑𝒎c̅^𝟐 = 𝒌𝑻 x 3/2 on both sides 3/2 x 1/3mc̅^2 = 3/2kT 1/2mc̅^2 = 3/2 kT 1/2mc̅^2 is the average translational kinetic energy of a molecule in the gas, so: Ek = 3/2 kT The kinetic energy of a molecule is directly proportional to its absolute temperature

Question 36

How can the equation pV = 1/3Nmc̅^2 be derived

Answer 36

It can be derived by considering how the movement of atoms or molecules of gas inside a sealed box gives rise to pressure. Consider a single gas particle making repeated collisions with a container wall. The container is a cube with sides L. The gas particle has mass m and velocity c. It hits the surface of the wall at right angles. The elastic collision results in a change in momentum of magnitude 2mc. The time t between collisions is the total distance covered by the particle divided by its speed. Therefore t = 2L/c. According to Newton’s second and third laws, the force exerted by the particle on the wall is Force = Δp/Δt = 2mc x c/2L = mc^2/L If there are N particles in the container moving randomly, the average force exerted by each particle must be mc̅^2/L, where c̅^2 is the mean square speed of the particles. On average, because if the random motion of the gas particles, about 1/3 of the particles will be moving between two opposite faces of the container. Consequently, the total force on one container wall of cross-sectional area L^2 due to collisions from all particles must be Force = mc̅^2/L x N x 1/3 = Nmc̅^2/3L Finally, the pressure p exerted by the gas must be equal to the total force exerted by all the particles divided by the cross-sectional area of the wall. Therefore, p = Nmc̅^2/3L x 1/L^2 = Nmc̅^2 /3L^3 = Nmc̅^2/3V where V is the volume of the container. Therefore pV = 1/3Nmc̅^2

Question 37

Example Question: Calculate the mean Ek of air molecules at 0°C

Answer 37

0°C = 273K Ek = 3/2kT Ek = 3/2 x 1.38x10^-23 x 27 Ek = 5.65x10^-21 J

Question 38

Work out the RMS speed of Oxygen-16 where Ek = 5.65x10^-21 J

Answer 38

Ek = 3/2kT Ek = 1/2mc̅^2 3/2kT = 1/2mc̅^2 c̅^2 = 2Ek/m And Ek = 5.65x10^-21 Mass of O2 molecule m = M/Na m = 0.032/6,02x10^23 m = 5.3x10^-26 kg c̅^2 = 2Ek/m c̅^2 = (2 x 5.65x10^-21)/5.3x10^-26 c̅^2 = 2.13x10^5 m^2 s^-2 rms speed = √c̅^2 rms speed = 462ms^-1

Question 39

What is the Nm in the equation Pv = 1/3 Nmc̅^2

Answer 39

The Nm in the equation is the mass of the gas