Gases

Question 1

Ideal gas

Answer 1

1. No interparticle forces
2. 0 volume
   Though these are false assumptions, if a gas has a low enough density, the forces and volumes become negligible due to high particle separation.

Question 2

Ideal gas equation

Answer 2

pV = nRT
Pa, m^3, K

Question 3

What is R equivalent to?

Answer 3

KbNa where Kb is the Boltzmann constant and Na is Avogadro’s constant

Question 4

Ideal gas equation of state

Answer 4

The equation can also be used to calculate the effect of changing conditions on other properties. E.g., if a gas is cooled at a fixed volume, pressure will change. P changes proportionally to T.

Question 5

Mixtures of ideal gases

Answer 5

It is assumed that all ideal gases behave in the same way; as a result, properties are determined by the number of moles rather than the identity of the particles.

Question 6

Dalton’s Law

Answer 6

The pressure of a mixture of gases is equal to the sum of the partial pressures of each gas.

Question 7

Mixture of ideal gases equation

Answer 7

P(tot) = n(a)RT/V + n(b)RT/V

Question 8

Elastic collisions

Answer 8

Collisions that conserve total kinetic energy. Conserve momentum.

Question 9

Inelastic collisions

Answer 9

Collisions that involve a change in kinetic energy. Conserve momentum.

Question 10

Collisions in atomic gases

Answer 10

All collisions are elastic; atoms bounce off of each other, not converting any energy.

Question 11

Collisions in molecular gases

Answer 11

Most collisions are inelastic, as kinetic energy can be translated into rotational or vibrational energy and vice versa. Because both options are possible, the collisions are seen as elastic on average.

Question 12

Real gases

Answer 12

Refers to gases that deviate from ideal behaviour, especially under high pressure and low temp conditions. They have intermolecular forces and finite volumes.

Question 13

Compression factor

Answer 13

When the ideal gas equation is rearranged to: pV/nRT = 1, it can be used for real gases as well; however, because they deviate from ideal behaviour, the 1 is replaced by Z, the compression factor.

Question 14

Z values

Answer 14

If Z > 1, the real gas is harder to compress than an ideal gas.
If Z < 1, the real gas is easier to compress than an ideal gas.

Question 15

Van der Waal’s equation

Answer 15

(p + a/v^2)(V - b) = nRT
1: Corrects for the attractive forces between gas molecules. It becomes most significant when the gas molecules are close together, where intermolecular attractions (a) cause the gas to exert a lower pressure than expected.
2: Accounts for the finite volume of real gas molecules. The volume available for the molecules to move around in is reduced by the volume taken up by the molecule (b).

Question 16

Berthelot equation

Answer 16

Derived in the same way as the VdW equation but with temperature dependence for internal pressure.

Question 17

Virial equation of state

Answer 17

pV(m) = RT(1 + B/V(m) + C/V(m^2) + …)
The terms in the bracket are equal to Z. By measuring Z at different conditions, the values of the coefficients can be determined.

Question 18

Liquefaction

Answer 18

Liquefaction isn’t possible for ideal gases, as there are no interparticle interactions. Real gases can be liquefied.

Question 19

Supercritical fluid formation

Answer 19

Forms through sealing a liquid inside a vessel, causing a vapour to form above. As temperature increases, the vapour pressure increases until the liquid and gas have the same density, until finally a single phase exists at a critical temperature.

Question 20

Supercritical fluid

Answer 20

On the phase diagram, there is a line representing the end of the coexistence of liquid and gas. After this point, only supercritical fluid exists. As it cannot be compressed, it is considered a gas; however, it can have a much higher density than subcritical gases given the pressure is high enough.

Question 21

Uses of supercritical fluids

Answer 21

Supercritical CO2 can be used to extract caffeine from coffee. They can be used to aid chemical reactions without relying on the use of traditional, environmentally harmful solvents.

Question 22

Measuring speed of gas particles

Answer 22

Experimental: Uses spinning discs with slits that are offset from the adjacent discs’. The different rates of spin allow for different particle speeds to be selected. Measuring the number of particles passing through gives a normal distribution of speeds (Maxwell distribution).

Question 23

Maxwell averages

Answer 23

The peak indicates the most probable speed: V(mp)
Calculated: (2KbT/m)^1/2

Mean: V(m) = (8KbT/πm)^1/2

Root-mean-square: V(rms) = (3KbT/m)^1/2
Equal to the root of the integral v^2f(v) dv

When expressing the averages with molar mass, Kb is replaced by the gas constant R (kgmol^-1)

Question 24

How factors affect maxwell

Answer 24

Temperature and particle mass changes impact the height and width of the graph, but not the shape. Higher temps and lower masses lead to wider distributions.

Question 25

Distribution for particle kinetic energies

Answer 25

Temperature has a greater impact on kinetic energy distributions as the dependence is linear, whereas with speed distributions is is square rooted.

Question 26

Particle kinetic energy averages

Answer 26

Most probable: E(mp) = KbT/2
Mean: E(m) = 3KbT/2
E(m) = 1/2mV(rms)^2

Question 27

Collision frequency

Answer 27

Number of collisions a particle experiences each second.

Question 28

Collision density

Answer 28

Number of collisions every particle in 1m^3 experiences each second.

Question 29

Mean free path

Answer 29

How far, on average, each particle travels between collisions.

Question 30

Collision theory

Answer 30

All three measures depend on their mean relative speeds and sizes.

Question 31

Mean relative speeds - collision theory

Answer 31

Related to mean speed (maxwell), however, as it is relative between two particles, m is replaced with reduced mass: μ
V(rel) = (8KbT/μπ)^1/2

Question 32

Collision cross section

Answer 32

Calculated using the equation for area of a circle except you add the radii of the two colliding molecules to form an “effective radius.”

Question 33

Effusion

Answer 33

The process that occurs when a small hole is opened in a vessel and vacuum is on the other side.

Question 34

Graham’s law of effusion

Answer 34

States that lighter gases are able to effuse more quickly than heavier gases due to their molecules having higher velocity and reaching the hole quicker.

Question 35

Rate of effusion

Answer 35

dN/dt = –A(0)Z(w)
Where A0 is the area of the hole and Zw relates to how many molecules are approaching the hole per unit time.