Lecture 25 © slgCHM 151

Topics:

1. Gas State Introduction2. P,V,T,n Relationships

The Gas State of Matter Chapter 12

Let us turn to the gas state, and learn howto measure and calculate amounts of matter whenfound in this high energy situation.

SOLID

(heat)

LIQUID

(heat)

GAS

Consider:

H2O(s)H2O(l)

H2O(g)

MP, 0oC BP, 100oC

CH3OH(s) CH3OH(l)CH3OH(g)

MP, -94oCBP, 65oC

NaCl(s)NaCl(l)

NaCl(g)MP, 801oC BP, 1413oC

CH4(s) CH4(l)CH4 (g)

MP, -182oC BP, -164oC

SOLID LIQUID GAS

Also, consider three lidded containers, holdingrespectively a solid, liquid and gas sample of material:

Retains shape, volume

Retains volume,takes shape ofcontainer

Takes shape,volume ofcontainer

Differences in MP’s and BP’s, and retention or loss of shape and volume, all are due to attractions between the molecules or ions or atoms which make up the sample.

Let us look more closely at each state in these terms...

Solids are made up of particles (molecules or ions or atoms) which are highly attracted to each other and packed as closely together as possible in some sort of “crystal lattice” arrangement which keeps themrigidly in place.

Solid particles can vibrate in position but not flow past each other. They retain their shape and volume due to this internal attraction between particles.

• solid state volume doesn’t change with pressure due to close packing of particles, and

• volume can be calculated from mass and a reference density

Liquid samples are made up of particles which arestill strongly attracted to each other, but possess the energy to flow. They are still packed as closetogether as possible, but are not confined to a“crystal lattice”.

Accordingly, liquids retain their volume when poured from one container to another, but assume the shape of the container.

•Like solids, liquids resist volume change under pressure due to “close packing”and

•volume can be calculated from mass and a reference density.

Gases are made up of independent particles with the necessary energy to escape the attractions of neighboring particles. They move freely within the volume in which they are confined: they fill volume completely, assuming both the shape and the volume of the container.

Pressure deeply affects gas volumes, as the particles are as far apart as the container allows: gases are readily compressed into smaller volumes.

To correctly assess the amount of a gas sample, itis necessary to consider four factors: temperature,pressure, number of particles (“moles”), and volume.

We will return to the nature of the attractionsbetween particles which make some materials solidsat room temperature, and others liquids or gases.

We should have gathered from what we had just examined that high attraction leads to solids andlow attraction to gases....

First however, we need to consider the inter-relationship of the four factors needed to define anysample in the gas state: P, pressure; T, temperature;V, volume; and n, number of moles.

Solids: Definite shape, volume: know mass, calculate volume know volume, calculate mass

Liquids: Definite volume, shape determined by container: know mass, calculate volume know volume, calculate mass

D= mass/ volume

Gases: Volume, “shape” determined by container

To determine volume, mass, molar mass, P and T required.....

GAS PRESSURE

Gas samples are composed of independent particlesmoving in a straight line path until collision with thewalls of the container (or another particle).

Collision with the walls of the container results in apressure, defined as a force per unit area.

The pressure created depends on how frequently thecollisions occur and how forceful they happen to be.

Rapid random movement of particles results in frequentcollisions with the walls of the container, creating pressure:

Lidded container

collisions

Pressure Dependence on T, V, n

Using the container as our model, let’s considerhow P relates to the other variables, consideringeach as we hold the other two constant:

We’ll hold number of moles (n) and T constant first,and decide how P changes with changing V.

Pressure and Volume, constant n, T

If the volume of the container is decreased, the particles hit the wall more frequently (they haven’t as far to travel!). The P goes up as the number ofcollisions increases.

If the volume of the container is increased, collision will occur less frequently as the particles have furtherto travel and the pressure will go down.

In a nutshell, n,T constant:

as V increases, P decreases as V decreases, P increases

In mathematical terms, P is “inversely proportional”

to V at constant n, T.

This relationship was studied first by Robert Boyle in the late 1600’s, and the law defining this relationshipis termed “Boyle’s Law”.

P 1 V

CB

P =

V

PV = CB

ProportionalityConstant

CB is a “proportionality constant” obtained bygraphing P vs. 1 / V. This constant, CB, is the slope of the straight line obtained.

Boyle’s Law states that for a given sample of gas at aconstant temperature, any product of the P times theV equal a constant:

P1V1 = CB = P2V2 (n, T constant)

Pressure and Temperature, constant n, V

If the temperature of the sample is increased, the particles hit the wall harder and more frequently (they will be moving faster, with the added energy heat provides).

The P goes up as the number and intensity of collisions increases.

If the temperature of the sample is decreased, collision will occur less frequently and with less force as theless energetic particles move at a slower pace. The pressure will go down.

So, in a nutshell, at constant V, n:

as T increases, P increases as T decreases, P decreases

P T P = Ck T P1 P2

= Ck =T1 T2

Pressure is directly proportional to T (in the absolute orKelvin scale) as can be shown graphically. The straightline obtained plotting P vs. T yields the constant “Ck”.

Summary, to date:

Boyle’s Law: given sample of gas, constant n, T:

P1V1 = P2V2

Charles’s Law: given sample of gas, constant n, V:

P1 P2

= T1 T2

Combined Gas Law constant n

We can combine the three factors which describe a “confined gas” (constant n, number of particles)as follows:

P 1 / V, P T :

P T P =Cc T V V

PV = Cc P1V1 = Cc =P2V2

T T1 T2

If we hold any factor constant for this “confinedgas”, then :

P1V1 = P2V2

T1 T2

V1 = V2

T1 T2

Constant T

P1V1 = P2V2

Constant P

Constant V

P1 = P2 T1 T2

Units Utilized in Gas Law Problems:

Pressure: generally done in reference to a column of mercury immersed in a dish of mercury open to atmospheric pressure. (CD ROM)

760 mm Hg = 1 atmosphere (atm) = 760 Torr

= 14.7 lbs/ft2(psi)

= 101.325 X 103 Pascals (Pa) (Newton/m2)= 1.013 bar

Volume: Liters, L

Temperature: Absolute or Kelvin Scale; k= oC +273.15

Since gas law problems always include lots of data, itis a standard practice to set up a table of all given databefore proceeding: For the combined gas law, anappropriate table is below:

V1 =

P1 =

T1 =

n is consta nt

INITIAL

V2 =

P2 =

T2 =

FINAL

A gas exerts a pressure of 735 mm Hg in volume of 2.50 L at a temperature of 73o C. What pressure would itexert if the temperature were increased to 110o C and the volume increased to 3.00 L?

V1 = 2.50 L

P1 = 735 mm Hg

T1 = 73 oC +273

= 346 k

n is consta nt

V2 = 3.00L

P2 = ?

T2 = 110oC + 273 = 383 k

INITIAL FINAL

P1V1

T1

P2V2

T2

=P2 = P1V1T2

T1 V2

P2 =

V1 = 2.50 L

P1 = 735 mm Hg

T1 = 73 oC +273

= 346 k

V2 = 3.00L

P2 = ?

T2 = 110oC + 273 = 383 k

INITIALFINAL

735 mm Hg X 2.50 L X 383 k

346 k X 3.00 L

= 678 mm Hg

data

formula

solve

A sample of CO2 gas has a pressure of 56.5 mm Hg in a 125 mL flask. The gas is transferred to a new flask where it has a pressure of 62.3 mm Hg at the sametemperature. What is the volume of the new flask?

V1 =125 ml = .125 L

P1 = 56.5 mm Hg

T1 = same

V2 = ?

P2 = 62.3 mm Hg

T2 = same

INITIALFINAL

P1V1

T1

P2V2

T2

=V2 = P1V1

P2

V2 =

V1 =125 ml = .125 L

P1 = 56.5 mm Hg

T1 = s ame

V2 = ?

P2 = 62.3 mm Hg

T2 = s ame

INITIALFINAL

56.5 mm Hg X .125 L

62.3 mm Hg

= .113 L = 113 mL