A. Ideal and Real Gases

Question 1

According to Boltzmann distribution,

(a) at low temperatures ____ molecules are in states of ______

(b) at high temperatures, _____ molecules are in ____

Answer 1

(a) At low temperatures, most molecules are in states of low energy.

(b) At high temperatures, some molecules can populate states of high energy

Question 2

Formula for pressure

Answer 2

P = F/A

→ the ratio of force (F) to the area (A) which the force is exerted

Question 3

Pressure units conversion:

1N = ?
1 Pa = ?
1 bar = ?
1 atm = ?

Answer 3

1N = kg m / s^2
1 Pa = 1 N / m^2 = 1 kg / m s^2
1 bar = 10^5 Pa
1 atm = 101325 Pa

Question 4

All molar properties [Xm = X/n] are ___ (extensive/intensive) property

Answer 4

intensive

NOTE: it is intensive despite both X and n being extensive

Question 5

The state of any sample of substance can be specified by giving the values of what properties?

Answer 5

1. volume
2. pressure
3. temperature
4. amount of substance

Question 6

The four
quantities V, p, T, and n are ___ (dependent/independent) of one
another.

Hence, substance obeys a/n ____, an equation of the form of ____

Answer 6

dependent

equation of state in the form:
p = f (n, V, T)

Question 7

All gases obey the perfect/ideal gas
equation of state ever more closely as _____. Hence, it is an example of a ____.

Answer 7

the pressure is reduced towards zero

limiting law - law that becomes increasingly valid as a variable approaches a certain limit
(e.g., p→0)

Question 8

Differentiate perfect/ideal gas and real gas

Answer 8

Perfect gas - obeys the ideal gas law (PV=nRT) at all pressures

Real gas - behaves more and more like a perfect gas as its pressure is reduced towards zero.

Question 9

Why real gas behaves differently from a perfect gas?

Answer 9

attractions and repulsions that exist between actual molecules of real gas

Question 10

In a Boyle’s law plot (p vs 1/V), the observed relationship of real gas deviates from perfect gas as ____

Answer 10

pressure increases and volume decreases (1/V increases) at constant temperature

Question 11

At constant pressure, the observed V vs T plot deviates from perfect gas isotherm as ____

Answer 11

volume and temperature decreases (↑P)

Question 12

Why Avogadro’s principle remain as principle and not a law?

Answer 12

it started as a theoretical insight rather than an observed experimental law unlike Charle’s, Boyle’s and Gay-Lussac’s.

Question 13

The molar volumes of gases at SATP (298.15 K and 1 bar) are ____ whatever its chemical identity. Why?

Answer 13

the same because 1 bar is low enough for gases to behave perfectly

→ given the same T and P, the molar volume stays the same

NOTE: Based on IUPAC, STP = 1 bar and 273.15 K (0°C)

Question 14

1 L = 1 __^3
1 mL = 1 __^3

Answer 14

1 L = 1 dm^3
1 mL = 1 cm^3

Question 15

Define Dalton’s law

Answer 15

The pressure exerted by a mixture of perfect gases is the sum of the pressures that each gas would
exert if it were alone in the container at the same temperature:

p = pA + pB + … + pJ

Question 16

Formula for partial pressure

Answer 16

pJ = xJ p

xJ = mole fraction
xJ = n of substance J n total

Question 17

How to compute for the partial pressure of a perfect gas?

Answer 17

pJ = xJ p ; xJ = nJ/ntot and pV = nRT

pJ = nJ RT/V

Question 18

linear momentum formula

Answer 18

momentum = mv

Question 19

Kinetic model of gases assumptions:

Answer 19

1. A gas consists of molecules in ceaseless random motion
2. The size of the molecules is negligible (they are ‘point-like’) in the sense that their diameters are much smaller than the average distance travelled between collisions.
3. The molecules do not interact, except during elastic collisions.

Question 20

Define elastic collision

Answer 20

The translational kinetic energy of the colliding molecules is the same before and after the collision (so no translational energy is lost by exciting vibrations or rotations).

Question 21

Based on kinetic model of gases, the potential energy of molecules is _____ and the total energy of a sample of gas is therefor ____

Answer 21

PE = 0 (independent of their separation)
Total E = sum of all kinetic energies of all the molecules present

Question 22

faster the gas molecules travel:
_____ kinetic energy
_____ total energy

Answer 22

↑ KE
↑ Total E

Question 23

Pressure of gas according to the kinetic model

Answer 23

p = n M (rms)^2 / 3V

rms = root-mean-square speed

Question 24

Definition of root-mean-square speed

Answer 24

basically, the mean translational kinetic energy of gas molecules

the square root of the average of the square of each gas molecules

1. square each speed
2. get the average
3. get the square-root

Question 25

Mean (average) speed and rms relationship

Answer 25

ave speed = 0.921 (rms) = (8/3π)^1/2 (rms)

Question 26

For samples consisting of large numbers of molecules, the mean speed is ____ than the r.m.s. speed.

Answer 26

slightly smaller ave speed = 0.921 (rms)

Question 27

rms and ave speed of gas molecules

Answer 27

rms = (3RT/M)^1/2 ave speed = (8RT/πM)^1/2

Question 28

1 J = 1 ____?

Answer 28

1 J = 1 kg m^2 / s^2

Question 29

Doubling the absolute temperature ____ the r.m.s. and the mean speed of molecules by ___

Answer 29

increases by a factor of √2 rms ∝ √T

Question 30

What is Maxwell distribution of speed?

Answer 30

Molecules in a sample of gas have a speed that lies in a particular range at any instant given in an expression: P (v, v + Δv) ≈ ρ(v)Δv whereas ρ(v) = 4π [(M/2πRT)^3/2] (v^2) [(e^-(Mv^2/2RT)] P - probability ρ(v) - probability density ρ - rho symbol

Question 31

Why Maxwell distribution of speed is strictly only valid when Δv is narrow?

Answer 31

ρ(v) can be regarded as a constant throughout the narrow range. Hence, formally written as P (v, v + dv) ≈ ρ(v)dv

Question 32

If at a given speed, the range of interest is doubled (but still kept narrow), then the probability of molecules lying in that range _____

Answer 32

doubles also P (v, v + Δv) ≈ ρ(v)Δv ↑Δv, ↑P (v, v + Δv)

Question 33

the probability that molecules will be found with very high speeds is ___

Answer 33

very small e^(-ax^2) becomes very small when ax^2 is large.

Question 34

heavy molecules (large M) are unlikely to be found with very ____ speeds

Answer 34

high speeds (e^-(Mv^2/2RT) → the e approaches zero very rapidly with large M

Question 35

at ____ temperatures, there is a greater probability of the molecules having high speeds.

Answer 35

high temperature = high probability of faster gas molecules (e^-(Mv^2/2RT) → the e approaches zero very slowly with large T

Question 36

the probability of finding molecules with VERT low speeds is ____

Answer 36

very small (v^2) [(e^-(Mv^2/2RT)] → this factor goes to zero as v goes to zero

Question 36

Differentiate effusion and diffusion

Answer 36

Diffusion is the spreading of the molecules of one substance into a region initially occupied by another species. (both substances move) Effusion is the escape of molecules through a small hole in a confining wall.

Question 37

↑ Molar Mass ___ Rate of Effusion/Diffusion This observation (rate of effusion and molar mass) is summarized by ____

Answer 37

↑ Molar Mass ↓Rate of Effusion/Diffusion Graham's law of effusion

Question 38

↑ Temperature ___ Rate of Effusion/Diffusion

Answer 38

↑ Temperature ↑ Rate of Effusion/Diffusion

Question 39

Define mean free path λ and its formula for a perfect gas

Answer 39

average distance that a molecule travels between collisions λ = kT/[(√2) σp] σ - collision cross-section = πd^2 k - Boltzmann's constant = 1.38×10^−23 J/K

Question 40

Define collision frequency (z) and time of flight

Answer 40

collision frequency - average number of collisions one molecule makes in a given time interval divided by the duration of the interval time of flight (1/z) - average time that a molecule spends between two collisions

Question 41

Formula for relative mean speed and collision frequency

Answer 41

relative mean speed = λz z = (2σp / kT) (8RT/πM)^1/2 λ - mean free path z - collision frequency σ - collision cross-section k - Boltzmann's constant = 1.38×10^−23 J/K (8RT/πM)^1/2 = average speed

Question 42

The mean free path ____ as the pressure increases. Why?

Answer 42

decreases λ = kT/[(√2) σp] → λ ∝ 1/p in a fixed volume: ↑ p = ↑ gas molecules = shorter distance to travel before collision

Question 43

The mean free path is ____ for large molecules

Answer 43

shorter λ = kT/[(√2) σp] → λ ∝ 1/σ large molecule = ↑ d = ↑σ = ↓λ

Question 44

the collision frequency ____ with the increasing pressure of the gas

Answer 44

increases z = (2σp / kT) (8RT/πM)^1/2 → z ∝ p

Question 45

Given the same σ, heavy molecules have ____ collision frequencies than light molecules

Answer 45

lower z = (2σp / kT) (8RT/πM)^1/2 → z ∝ average speed average speed = (8RT/πM)^1/2 → average speed ∝ 1/M^1/2 ↑ M = ↓ ave speed = ↑ collisions (z)

Question 46

Condition for perfect gas behavior

Answer 46

λ >>>> d

Question 47

Describe the molecular interactions of gases

Answer 47

At large nuclear separation, gas molecules attract each other and lower the potential energy As soon as they come into contact, they repel each other (attractive force increases, increases the potential energy as they are squash together)

Question 48

Explain the plateau in the isotherm (Vm vs P) of a real gas at 20°C

Answer 48

1. An ideal gas follows the trend of Boyle's law at low pressure, high volume 2. It experiences no change in volume at some point as pressure increases → point where condensation begins → this plateau corresponds to molecules being close to each other and increases the attractive forces → transform to liquid 3. After condensation finished, a very large pressure is required to further compress a liquid (decrease volume) → force molecules that are already in contact even closer together (overcome strong repulsive interaction)

Question 49

Define critical temperature

Answer 49

The maximum temperature for a gas to be condensed to a liquid by applying pressure (a gas cannot be condensed to a liquid by the application of pressure unless the temperature is below the critical temperature.)

Question 50

Differentiate vapor and gas in terms of critical temperature

Answer 50

1. vapor - gaseous phase of a substance below its critical temperature (and which can therefore be liquefied by compression alone) 2. gas - gaseous phase of a substance above its critical temperature (and which cannot therefore be liquefied by compression alone)

Question 51

Define supercritical fluid

Answer 51

Dense fluid that resulted from compressing a gas above its critical temperature 1. Not a liquid: it fills any container and never possesses a surface to separate from vapor 2. Not a gas: very dense

Question 52

Define compression factor (Z)

Answer 52

measures how far a real gas departs from ideal gas Z = Vm of real gas / Vm of perfect gas Z = 1 if Vm of real gas = Vm of perfect gas => Z = p Vm(real) / RT

Question 53

Differentiate the forces present when: 1. Z<1 2. Z>1

Answer 53

1. at Z<1: normally occurs at low pressure, attractive forces are dominant 2. at Z>1: happens as pressure increases. repulsive forces are dominant NOTE: hydrogen gas has a very weak attractive forces so that the Z>1 even at low pressures

Question 54

Define the virial equation of state

Answer 54

This is to account the intermolecular forces acting on real gases p = nRT/V [1 + nB/V + n^2C/V^2 + ...] For a fixed amount of gas (n) and very large volume (V), only the term 1 remains in the bracket and real gas behave as perfect gas

Question 55

Define Boyle temperature

Answer 55

temperature at which B at virial equation of state is equal to 0 → Z = 1 and the gas behaves almost perfectly over a limited range of molar volumes.

Question 56

Define van der Waals equation of state

Answer 56

To better visualize and account the repulsive and attractive forces of real gas (p + an^2/V^2)(V - nb) = nRT (p + an^2/V^2) → attractive forces; slow the gas molecules, strike walls less frequently and weaker; large a when attractive forces are large (V - nb) → repulsive forces; volume is reduced to an extent; large b if molecules are large

Question 57

A gas may be liquefied by cooling it at which temperature?

Answer 57

Cooling below its boiling point

Question 58

How to condense a gas if the boiling point is too low? What is this process called?

Answer 58

allow gas to expand (increase separation) = kinetic energy transform to potential energy = ↓ average speed of gas molecules = ↓ temperature (cooling) If it can be reduced to the point that neighbors can capture each other by their intermolecular attractions, then the cooled gas will condense to a liquid. => Joule-Thomson effect

Question 59

Limitation of Joule-Thomson effect

Answer 59

For gases with Z<1 or the attractive forces are dominant ONLY. Molecules need to overcome the attractive forces for it to slow down and cools For Z>1 (repulsion forces are dominant), gases will become warmer