Physical Chemistry 01 2015 1st

Physical Chemistry

Chap 1. The properties of gases

Outline

- Perfect gas

What is its property?

How to describe it? : Perfect gas equation

- Pure gas vs. mixture of gases

- Real gases

How are they different from the perfect gas?

How to describe the behavior of them?

Compression factor

Virial equation of state

Van der Waals equation

The principle of corresponding states

Properties of gas

- collection of molecules in continuous random motions

- average speed increases as the temperature increases.

- molecules in a gas are widely separated from one another.

- molecules in a gas move in paths that are mostly unaffected by

intermolecular forces( most of the time, molecules move without

any interactions)

- intermolecular interactions are negligible

repulsive interaction

attractive interaction

The states of gases

- physical properties: physical state or physical condition of a

substance

- state of pure gas can be specified by

number of moles, n

pressure, p

volume, V

temperature, T

- equation of state: interrelation of variables

p = f(T,V,n) : if we know T, V, n,

then pressure has a fixed value.

for perfect gas p = nRT/V

Pressure

- pressure = force divided by the area to which the force is applied

p = F/A

- [unit] pascal(Pa):

Pa = N/m2 = kg/m∙s2

1 bar: standard pressure for reporting

The measurement of pressure

- barometer: invented by Torricelli

- column of liquid of density ρ and height h

at the surface of the earth

F = mg = (ρV)∙g = (ρAh)∙g ,

p = F/A = ρ∙A∙h∙g/A = ρ∙h∙g

mecahnical equilibrium:

if two regions with high and low pressures is separated

by movable wall ⇒ the wall will be pushed from the

high pressure region until when both pressures

become equal

Temperature

- indicated the direction of the flow of energy

through thermally conducting wall

- boundary diathermic: conducting

adiabatic: insulating

- Energy flows as heat through diathermic wall

from high T to low T until thermal equilibrium

Zeroth law of thermodynamics

: if A is thermal equilibrium with B, and B is in thermal equilibrium with C, then

C is also in thermal equilibrium with A

- measuring the temperature

- θ : Celsius scale, [oC]

- K : Kelvin scale : T = θ + 273.15 [K]

The perfect gas law

-Boyle’s law: pressure volume dependence of perfect gas

pV = constant, at constant n, T

Isotherm: variation of the pressure of a gas a the volume change at constant temperature

Charles’s law

isobar: variation of volume of a gas with temperature changes at constant pressure

isochores: variation of pressure of a gas

with temperature changes at

constant volume

V = constant x T,

at constant n, p

V1/T1 = V2/T2

p= constant x T,

at constant n, V

P1/T1 = p2/T2

- Boyle’s law: pV = constant, at constant n, T

- Charles’s law: V = constant x T, at constant n, p

p= constant x T, at constant n, V

- Avogadro’s principle: V = constant x n, at constant p

⇒ Perfect gas equation

pV = nRT

R = gas constant

*Boyle’s and Charles’s laws are limiting law (strictly try on in a certain limit i.e. low pressures or high temperatures)

Ex 1.2) 100 atm, 300 K ⇒ 500 K, p = ?

R = pV/nT ⇒ p1V1/T1 = p2V2/T2

p2 = T1/T1 x p1 = 500 K/300 K x100 atm = 167 atm

Mixtures of gases

- Partial pressure: pJ = xJp xJ: mole fraction of component J

xJ = nJ/n, n= nA + nB + ∙∙∙

- Total pressure: p = pA + pB + ∙∙∙ = (xA + xB + ∙∙∙ )p

- Dalton’s law for a mixture of gases

: The pressure exerted by a mixture of gases is the sum of the

pressures that each one would exert if it occupied the container

alone

- Partial pressure of perfect gases

Composition of dry air at sea level

Ex) The mass percentage composition of dry air at sea level is approximately

N2:O2:Ar = 75.5:23.2:1.3. What is the partial pressure of each component

when the total pressure is 1.20 atm?

answer: number of mole of each gas for 100 g of air

n(N2) = (75.5 g)/(28.02 g/mol) = 2.69 mol

n(O2) = (23.2 g)/(32.00 g/mol) = 0.725 mol

n(Ar) = ( 1.3 g)/(39.95 g/mol) = 0.033 mol

total = 3.45 mol

mole fraction

x(N2) = 2.69 mol/3.45 mol = 0.780

x(O2) = 0.725 mol/3.45 mol = 0.210

x(Ar) = 0.033mol/3.45 mol = 0.0096

partial pressure: pJ = xjp

p(N2) = 0.780 x 1.2 atm = 0.936 atm

p(O2) = 0.210 x 1.2 atm = 0.252 atm

p(Ar) = 0.0096 x 1.2 atm = 0.012 atm

The kinetic model of gases

- Assumptions for kinetic model of gases

1. The gas consists of molecules of mass m in ceaseless random motion.

2. The size of molecules is negligible, in the sense that their diameters are

much smaller than the average distance travelled between collisions.

3. The molecules interact only through brief, infrequent, and elastic collisions.

- elastic collision: total translational kinetic energy is conserved.

- from kinetic model of gases

pV = 1/3nMvrms2, M = mNA, molar mass

vrms = (ν2)½, ν= speed of the molecules

- pV = nRT

⇒ 𝒗𝐫𝐦𝐬 = 𝟑𝑹𝑻 𝑴 𝟏 𝟐 ∶ molecular speed vs temperature

- Pressure arises from the impact of molecules on the wall

pressure =𝑛𝑀𝑣𝑥

2

𝑉

The Maxwell-Boltzmann distribution of speeds

- 𝒇 𝒗 = 𝟒𝛑 𝑴 𝟐𝝅𝑹𝑻 𝟑 𝟐𝒗𝟐𝒆− 𝑴𝒗𝟐 𝟐𝑹𝑻

- mean values

mean speed of perfect gas: 𝑣mean = 8𝑅𝑇 π𝑀 1 2

most probable speed of perfect gas: 𝑣mp = 2𝑅𝑇 𝑀 1 2

𝒗𝐫𝐦𝐬 = 𝟑𝑹𝑻 𝑴 𝟏 𝟐 : 𝑣mean = 3 8π 1 2𝒗𝐫𝐦𝐬

𝑣mp = 2 3 1 2𝒗𝐫𝐦𝐬

𝑓 𝑣1, 𝑣2 = 𝑣1

𝑣2

𝑓 𝑣 𝑑𝑣

Collisions

collision frequency : 𝒛 = 𝝈𝒗𝒓𝒆𝒍𝒑 𝒌𝑻 ,

collision cross-section: 𝜎 = π𝑑2

mean relative speed: mean speed of one molecule approaches another

: 𝑣rel= 8𝑅𝑇 π𝜇 1 2 , 𝜇 =

𝑀A𝑀B 𝑀A +𝑀B , 𝑀𝐴 𝑜𝑟 𝑀𝐵;𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠

mean free path: average distance a molecule travels between collisions

λ= 𝑣rel 𝑧 ⇒ λ= 𝒌𝑻 𝝈𝒑

Real gases

- do not obey the perfect gas law exactly except in the limit of p → 0

- deviation from the law are particularly important at high pressure and low T.

Molecular interactions

- repulsive forces(short range)

- attractive forces: important when the molecules

are fairly close

- repulsive forces assist expansions

attractive forces assist compressions

Variation of the potential energy of two molecules on their separation

The compression factor

- Z: the ratio of its measured molar volume(Vm) to the molar volume of a

perfect gases(Vmo) at the same pressure and temp

, Vmo = RT/p for perfect gas

∴ Z = pVm/RT ⇒ pVm = RTZ

- deviation of Z from 1

⇒ departure from perfect behavior

- Z = 1 : perfect gas,

~ 1 : low pressure

> 1 : high pressure, Vm > Vm0

repulsive interaction

< 1 : intermediate pressure, Vm < Vm0

attractive interaction

Virial coefficients

- at large molar volume and high temperature

⇒ real gases do not obey the perfect gas law ⇒ power of variables

- Virial equation of state(p or V)

⇒ pVm = RT(1 + B’p + C’p2 + ∙∙∙ )

pVm = RT(1 + B/Vm + C/Vm2 + ∙∙∙ )

B, C, ∙∙∙ : virial coefficient depend on T

B/Vm ≫ C/Vm2

- Virial equation can be written with Z

pVm = RTZ ⇒ Z = 1 + B’p + C’p2 + ∙∙∙

= 1 + B/Vm + C/Vm2 + ∙∙∙

- for perfect gas: Z = 1, dZ/dp = 0

- for real gas: Z → 1 as p → 0

dZ/dp → B’ as Vm → ∞

dZ/d(1/Vm) → B as Vm → ∞

- Boyle temperature TB, B = 0

: At TB, : properties of the real gas do coincide with those of perfect gas

C/Vm2 and higher terms are negligibly small

- Temperature dependence of B

at high T, dZ/dp → B(T) > 0 as p → 0

at low T, dZ/dp → B(T) < 0 as p → 0

at TB dZ/dp = 0 as p → 0

Second order viral coefficients. B/(cm3/mol)

Condensation

- Condensation: At low temperature a gas changes to a liquid.

- Vapor pressure: pressure of the gas in equilibrium with the liquid at a certain T

- Isotherm at 20oC A → B → C → D → E → F

A → B: V decreases when p increases

at C: V decreases without p increases

liquid formation started

C → D → E: liquid increases

p = vapor pressure of liquid

E → F: liquid compression

small volume decreases

with large p increases

Critical constant

- critical point: the point where a gas phase and a liquid phase are not

separated

- critical constant: pc, Vc, Tc

- below Tc: gas condensed to liquid when it pressurized

- above Tc: no phase separation between liquid and gas

- supercritical fluid: T > Tc single phase

Critical constants

The van der Waals equations

- for real gases,

repulsive and attractive interactions between molecules

molecules have volume ⇒ reduce the total volume available

attractive interaction ⇒ reduce the pressure

Formulation of the equation

- van der Waals equation:

⇒ Vm = V/n

- van der Waals coefficients

a: represents the strength of attractive interaction

b: repulsive interaction, related with the volume of molecules

- a, b correlated with critical temperature, vapor pressure, and enthalpy of

vaporization reflect the strength of intermolecular interaction

Repulsive interaction

- consider the volume of molecules

⇒ V - nb, nb: approximately the total volume o taken by the molecules

themselves

p = nRT/V ⇒ p = nRT/(V-nb), when repulsion is significant

- Hard-sphere molecule with radius r

Vmolecule = 4/3πr3 ⇒ volume excluded = 4/3π(2r)3

⇒ volume excluded per molelcule = 4Vmolecule

⇒ b ~ 4Vmolecule∙NA

- Intermolecular interaction↑⇒ b↑ van der Waals coefficients

Attractive interaction

- pressure depends on collision frequency and force of each collision

⇒ both are reduced by the attractive forces

strength ∝the molar concentrations, n/V

⇒ reduce the pressure = -a(n/V)2

- intermolecular interaction↑⇒ a↑

- small a ⇒ weak interaction

The features of the equation

- virial equation(experimental coefficients) should use for numerical analysis.

- van der Waals equation: it is not perfect but analytical

provide some general conclusions about he real gases

Equation of states

Features of van der Waals equation

- perfect gas isotherms are obtained at high T and large molar volumes

⇒ (Vm≫b)

- liquids and coexist when cohesive and dispersing effects are in balance

- the critical constants are related to the van der Waals coefficients.

⇒ at critical point(T = Tc)

dp/dVm = 0 = - RT/(Vm-b)2 + 2a/Vm3

d2p/d2Vm = 0 = 2RT/(Vm-b)3 – 6a/Vm4

⇒ Vc = 3b, pc = a/27b2, Tc = 8a/27Rb

⇒ critical compression factor

Zc = pcVc/RTc = 3/8

van der Waals isotherms with different T/Tc

⇒

The principle of corresponding states

- Reduced variables(dimensionless)

Vr = Vm/Vc, pr = p/pc, Tr = T/Tc ⇒ p = prpc, T = TrTc, Vm = VrVc

- van der Waal equation by reduced variables

p = RT/(Vm-b) – a/Vm2 ⇒ prpc = RTrTc/(VrVc-b) – a/Vr

2Vc2

⇒ Vc = 3b, pc = a/27b2, Tc = 8a/27Rb

∴

⇒ isotherms based on reduced variables

are the same curves for all gases

T/Tc

Exercises: 1A.7(b), 1B.3(b), 1C.6(a)

Problems: 1A.2, 1B.3,1C.9