General Chemistry I 1 KINETIC MOLECULAR DESCRIPTION OF THE STATES OF MATTER CHAPTER 9 The Gaseous State General Chemistry I U N I T III CHAPTER 10 Solids,

General Chemistry I 1

KINETIC MOLECULARDESCRIPTION OF THESTATES OF MATTER

CHAPTER 9The Gaseous State

General Chemistry I

U N I T III

CHAPTER 10Solids, Liquids, and Phase Transitions

CHAPTER 11Solutions


392

Gas

Liquid

Solid


THE GASEOUS STATE

9.1 The Chemistry of Gases

9.2 Pressure and Temperature of Gases

9.3 The Ideal Gas Law

9.4 Mixtures of Gases

9.5 The Kinetic Theory of Gases

9.6 Real Gases: Intermolecular Forces

9CHAPTER

General Chemistry I


395

Li + H2O → LiOH + H2


9.1 THE CHEMISTRY OF GASES396


397

CaCO3(s) + 2HCl(g) →

CaCl2(s) + CO2(g) + H2O(l)

K2SO3(s) + 2HCl(g) →

2KCl(s) + SO2(g) + H2O(l)


9.2 PRESSURE AND TEMPERATURE OF GASES

398

Pressure

Torricelli’s barometerEvangelista Torricelli (ITA, 1608-1647)


Force exerted by the mercury column at its base F = mg

Pressure : /

F mg mg mP gh gh

A A V h V

= 13.5951 g cm–3 → density of Hg(l) at 0 oC

g = 9.80665 m s–2 → gravitational acceleration

h = 76 cm → height of mercury column

399

1 atmosphere pressure (1 atm) is

1.01325 x 105 kg/ms2 or Pa; the SI unit= 101.325 kPa

1.01325 bar (1 bar = 105 Pa760 mm Hg (at 273 K)760 torr (at any temperature)


Pressure and Boyle’s Law

~ Experiments on the compression and expansion of air. “The spring of the Air and Its Effects” (1661).

Boyle’s J-tube experiment

Trapped air at the closed end of the J-

tube:

Add Hg and measure the volume of air (V):

1

(in mm)1 atm

760 mm atm

hP

1CP C

V V

Robert Boyle (UK, 1627-1691) PV Cor

C: a constant at constant T and fixed amount of gas

400


Fig. 9.3 Boyle’s J-tube.

(a)Same Hg height on two sides

Pconfined = Pair

(b) Hg added

Difference in Hg heights, h

Vconfined compressed

400


(a) P vs. V hyperbola

(b) P vs. 1/V straight line passing through the origin (slope: C)

(c) PV vs. P straight line independent of P

(parallel to P-axis, intercept C on PV-axis)

The value of C at 0oC and for 1 mol of gas,

~ good for all gases at very low pressure

C = PV = 22.414 L atm

401


Temperature and Charles’s law

Jacques Charles (France,1746-1823)

constant (at constant and ) V T n P

o

0

273.15 C 1V

tV

0 0o1

273.15 C 273.15

t TV V V

(Kelvin) 273.15 (Celsius)T t

402


Fig. 9.5 Volume of a gas confined at constant P increases as T increases.

403


Fig. 9.6 The volume of a sample of a gas is a function of temperature at constant pressure.

404


◈ Absolute Temperature Scale, K Kelvin, Lord William Thomson (UK, 1824-1907) 0 K : ‘absolute zero’ temperature 273.15 K : triple point of water 0 K = - 273.15 oC 0 oC = 273.15 K

Kelvin scale: absolute, thermodynamic temperature

scale - ‘absolute zero’ is the temperature at which all ther

mal motion ceases in the classical description of thermo

dynamics.

405


9.3 THE IDEAL GAS LAW405

√ Equation of state

√ Limiting law for real gases

as P 0

√ Universal gas constant, R

R = 8.314 J·K–1·mol–1

= 8.206 x 10–2 L·atm·K–1

PV = nRT

Boyle’s law: V 1/P (at constant T and n)Charles’ law: V T (at constant P and n)Avogadro’s hypothesis: V n (at constant T and P)

nT

VP


408

EXAMPLE 9.5 Concentrated nitric acid acts on copper

to give nitrogen dioxide and dissolved copper ions according

to the balanced chemical equation

Cu(s) + 4H+(aq) + 2NO3-(aq) → 2NO2(g) + Cu2+(aq) + 2H2O(l)

Suppose that 6.80 g copper is consumed in this reaction, and

that the NO2 is collected at a pressure of 0.970 atm and a

temperature of 45oC. Calculate the volume of NO2 produced.


9.4 MIXTURES OF GASES408

▶ Partial pressure (Pi) of the ith gas in a mixture of gases →

pressure that the ith gas would exert if it occupied the container

alone


◆ Dalton’s Law of Partial Pressures

The total pressure of a mixture of gases is the

sum of the partial pressures of its component.

A B ii

P P P P

▶ Mole fraction of the component A is xA

AA A B

A B

, 1n

x x xn n

AA ,

n RTP

V A B

nRT RTP n n

V V A

A AA B

n P

P x Pn n

409

PA = xAP


409

EXAMPLE 9.6 When NO2 is cooled to room temperature, some of it reacts

to form a dimer, N2O4, through the reaction

2NO2(g) → N2O4(g)

Suppose 15.2 g of NO2 is placed in a 10.0 L flask at high temperature and the

Flask is cooled to 25oC. The total pressure is measured to be 0.500 atm.

What partial pressures and mole fractions of NO2 and N2O4 are present?

(a)

(b)

(Dalton’s Law)


9.5 THE KINETIC THEORY OF GASES410

1. A gas consists of a collection of molecules in

continuous random motion.

2. Gas molecules are infinitesimally small (mass) points.

3. The molecules move in straight lines until they collide.

4. The molecules do not influence one another except

during collisions.


- Collision with walls: consider molecules traveling only in one dimension, x, with a velocity of vx.

All the molecules within a distance vxt of the walland traveling toward it will strike the wall during theinterval t.

If the wall has area A, all the particles in a volumeAvxt will reach the wall if they are traveling towardit.

411

The change in momentum (final – initial)of one molecule: -2mvx = 2mvx momentumchange for the wall


The number of molecules in the volume Avxt is that fraction of the total volume V, multiplied by the total number of molecules:

The average number of collisions with the wall during the interval t is half the number in the volume Avxt:

The total momentum change = number of collisions × individual wall momentum change

411


Force = rate of change of momentum = (total momentum change)/t

mean-square speed

412


412

- Kinetic energy of NA molecules,

- average kinetic energy per molecule, kB = R/NA

- root-mean-square speed

M = molar mass = NAm

Root mean square speeds of some gases at 25 oC


Maxwell-Boltzmann distribution of speed

23/2

2 /2 ( ) with ( ) 4 2

M RTMf f e

RT

vN

= v v v vN

James Clerk Maxwell Ludwig Eduard Boltzmann (Scotland, 1831-1879) (Austria, 1844-1906)

2B

3/2

/22

B

or ( ) 4 2

m k Tmf e

k T

vv v

23 1A/ 1.38066 10 J KBk R N Boltzmann constant:

414

Fraction ofmolecules withspeeds between v and v + v

NN


Fig. 9.13 A device for measuring the distribution of molecular speeds.

414


415

Fig. 9.14 Maxwell-Boltzmann distribution of molecularspeeds in N2 at three different temperatures.


mpvrmsv

v

416Different speeds associated with a collectionof gas molecules or atoms

1. Most probable speed (vmp)

df(v)dv

= 0v = vmp

vmp = 2kBTm

or 2RTM

2. Average speed (v)_

v = vf(v)dv0

oo_ = 8kBTm

or 8RTM

3. Mean square speed (v2)_

v2_

v2f(v)dv0

oo

= = 3kBTm

or 3RTM

4. Root mean square speed (vrms)

vrms = 3kBTm

or 3RT

Mv2_

=

vmp < v < vrms

_= 1.000:1.128:1.225


9.6 REAL GASES: INTERMOLECULAR FORCES

417

▶ Compression (or Compressibility) factor, Z z is a measure of deviation from ideality

- For an ideal gas, z = 1

- For real gases, z deviates from

1 as P increases:

z < 1 when attractive forces dominate

z > 1 for repulsive forces dominate

z =Vm

Vmideal =

Vm

RT/P=

PVm

RT=

nRT

Vm = molar volume = V/n

PV


z = PV/nRT

418


◈ The Van der Waals Equation of State

▶ Corrections to the ideal equation of state

- Attraction at long distance:

Reduction in collision frequency

Reduction in intensity of collision

/n V/n V

2

ideal 2

nP P a

V

- Repulsion at short distance:

No overlap of molecules → excluded volume effect

Reduction in free volume n

idealV V bn

418


2

2

nP a V nb nRT

V

▶ Van der Waals equation:

a: atm L2 mol-2

b: L mol-1

R: L atm mol-1 K-1

419

Rearranging this equation to solve for P gives the form of the equation that is most often used in calculations:

See Slide 37

P =nRT

V nb

_ an2

V2_


Repulsive forces (through b) increase z above 1.

Attractive forces (through a) reduce z.

The Boyle temperature TB

Influence of van der Waals parameters a and b onthe compressibility factor z

Substituting the approximation1

1 nb/V~ 1 +

nb

V_

into the equation above gives,

z ~ 1 + baRT

n

V_

+...

+....

(at low n/V)


-Note that the Constant b is the volume excluded by 1 mol of molecules

and should be close to Vm, the volume per mole in the liquid state.

The temperature at which the coefficient in brackets is 0 (so z = 1)is the Boyle temperature (TB)

TB =a

Rb

For T > TB repulsive forces dominate and z > 1

For T < TB attractive forces dominate and z < 1


419


Example of the use of the van der Waalsequation for the calculation of pressure

Solution

A 10.0-L tank containing 25 mol of O2 is stored in a diving supplyshop at 25 oC. Use the data in table 4.5 and the van der Waalsequation to estimate the pressure in the tank.

P =(25 mol) x (0.08206 L atm K-1 mol-1) x (298 K)

10.0 L (25 mol) x (3.19 x 10-2 L mol-1)_

_ (1.364 L2 atm mol-2) x(25 mol)2

(10.0 L)2

= 58 atm

Rewriting the van der Waals equation so as to solve for P,

P + a n2

V2(V nb) = nRT becomes P = nRT

(V nb)a

n2

V2__ _


Intermolecular Forces

12 6

LJ ( ) 4 V RR R

Lennard-Jones Potential:

where is the depth and is the distance at which V(R) passesthrough zero.

421

(Ar)

(He) (Ar)

(He)


421


10 Problem Sets

For Chapter 9,

6, 18, 30, 40, 42, 50, 64, 72, 76, 92

Documents

General Chemistry I 1 KINETIC MOLECULAR DESCRIPTION OF THE STATES OF MATTER CHAPTER 9 The Gaseous State General Chemistry I U N I T III CHAPTER 10 Solids,