Gases & Liquids

Ch.12

(12-1) Properties of Gases

• Fluids

• Low density

• Compressible

• Fill a container & exert P equally in all directions

• Influenced by T

Kinetic-Molecular Theory

• Explains behavior of gases

• 2 major assumptions:1. Collisions are elastic

2. V of individual gas molecules is negligible

KE is lost KE is maintained

Ideal Gas

• Describes the behavior of gases under most conditions

• High T & low P gases act more ideal

Kinetic Energy

• T of a gas determines the avg. KE of its particles

• KE = ½ mv2 – Where, m = mass, v = speed

Pressure

• Pressure (P) = force (F)

area (A)

• SI units: – F: newtons (N)– P: pascal (Pa) = 1 N/m2

Standard Temp. & Pressure

• STP: std. conditions for a gas

• Temp. (T): 0 °C (273 K)

• Pressure (P): 1 atm (101.325 kPa)– See Table 12-1, p.428 for more P units

Greenhouse Effect

• Inc. in the T of Earth caused by reflected solar radiation that’s trapped in the atmosphere

• Inc. in greenhouse gases such as CO2 & CFCs

Greenhouse Effect

Free Radical

• Atom or molecule that has 1 or more unpaired e- & is very reactive– UV radiation breaks apart CFCs, making Cl•

• Chain rxn: self-sustaining rxn in which the product from 1 step acts as a reactant for the next step– Cl• + O3 ClO + O2 + O– ClO + O Cl• + O2

(12-2) The Gas Laws

• Symbols:– P = pressure– T = temp in K– V = volume– n = # of moles

Boyle’s Law

• At constant T:– Inc. P, dec. V– Dec. P, inc. V

• P1V1 = P2V2

Boyle’s Law Practice

If the P exerted on a 300 mL sample of H2 gas at constant T is inc. from 0.500 atm to 0.750 atm, what will be the final V of the sample?

1. List known V1 = 300 mL P1 = 0.500 atm

V2 = ? mL P2 = 0.750 atm

Boyle’s Law Practice

2. Write eq.P1V1 = P2V2 V2 = P1V1

P2

3. Substitute & solveV2 = (0.500 atm)(300 mL) = 200 mL

0.750 atm

Dalton’s Law of Partial P’s

• Total P in a gas mixture is the sum of the partial P’s of the individual components

• Ptotal = PA + PB + PC…

– Where, Ptotal = total P, PA = partial P of A

Dalton’s Law Practice

A mixture of O2, N2, & H2 gases exerts a total P of 278 kPa. If the partial P’s of O2 & H2 are 112 kPa & 101 kPa respectively, what would be the partial P of the N2?

1. List the knownPtotal = 278 kPa PN2

= ? kPa

PO2 = 112 kPa PH2

= 101 kPa

Dalton’s Law Practice

2. List the eq. & rearrangePtotal = PO2

+ PH2 + PN2

PN2 = Ptotal - PO2

- PH2

3. Substitute & solvePN2

= 278 kPa – 112 kPa – 101 kPa = 65 kPa

Mole Fraction

• # of moles of 1 component compared w/ the total # of moles in the mixture

• Mole fraction (X) = ____mol A___

total mols

• To calc. partial P:

• PA = PTXA

Mole Fraction Example

The total P of a mixture of gases is 0.97 atm. The mole fraction for N2 is 0.78. What’s the partial P of N2?

1. List the knownPtotal = 0.97 atm XN2

= 0.78

Mole Fraction Example

2. Write the eq.PN2

= Ptotal XN2

3. Substitute & solve

PN2 = (0.97 atm)(0.78)

= 0.76 atm

Charles’ Law

• At constant P:– V inc., T inc.– V dec., T dec.

• V1 = V2

T1 T2

Charles’ Law Practice

Gas in a balloon occupies 2.5 L at 300 K. The balloon is dipped into liquid N2 at 80 K. What V will the gas in the balloon occupy at this T?

1. List knownV1 = 2.5 L T1 = 300 K

V2 = ? L T2 = 80 K

Charles’ Law Practice

2. Write eq.V1 = V2 V2 = V1T2

T1 T2 T1

3. Substitute & solveV2 = (2.5 L)(80 K) = 0.67 L

(300 K)

Pressure & Temp.

• P inc. w/ inc. in T at constant V

• P1 = P2

T1 T2

P & T Practice

Gas in a sealed can has a P of 3.00 atm at 25°C. A warning says not to store the can in a place where the T will exceed 52°C. What would the gas P in the can be at 52°C?

1. List knownP1 = 3.00 atm T1 = 25°C = 298 K

P2 = ? atm T2 = 52°C = 325 K

P & T Practice

2. Write eq.P1 = P2 P2 = P1T2

T1 T2 T1

3. Substitute & solveP2 = (3.00 atm)(325 K) = 3.27 atm

(298 K)

Avogadro’s Law

• V’s of different gases under the same T & P’s have the same # of molecules

• V1 = V2

n1 n2

Gay-Lussac’s Law

• Law of Combining Volumes: at constant T & P, gases react in whole # V proportions

• H2 + Cl2 2 HCl

• 1 V + 1 V 2 V

Effusion

• Motion of a gas through a small opening

• Diffusion: gas particles disperse from areas of high to low conc.

Graham’s Law of Effusion

• At the same T & P, 2 gases rates of effusion can be measured by:

• ½ MAvA2 = ½ MBvB

2 or vA = MB

vB MA

• Where:– v = speed of effusion (2 gases, A & B)– M = molar mass

Graham’s Law Practice

O2 has an avg. speed of 480 m/s at room T. On avg., how fast is SO3 traveling at the same T?

1. List knownvO2

= 480 m/s MO2 = 32 g/mol

vSO3 = ? m/s MSO3

= 80.07 g/mol

Graham’s Law Practice

2. Write eq.

vSO3 = MO2

vSO3 = vO2

MO2

vO2 MSO3

MSO3

3. Substitute & solvevSO3

= (480 m/s) (32 g/mol)

(80.07 g/mol) = 300 m/s

More Graham’s Practice

Compare the rate of effusion of H2O vapor w/ O2 gas at the same T & P.

1. List known

MH2O = 18.02 g/mol

MO2 = 32 g/mol

More Graham’s Practice

2. Write eq.

vH2O = MO2

vO2 MH2O

3. Substitute & solvevH2O = 32 g/mol = 1.33

vO2 18.02g/mol

H2O effuses 1.33X faster

than O2

(12-3) Ideal Gas Law

• PV = nRT

• Where:– R = 8.314 L•kPa / mol•K or– R = 0.0821 L•atm / mol•K

Ideal Gas Law Practice

Calculate the V of 1.00 mol of CO2 gas at STP.

1. List knownP = 1.00 atm V = ? Ln = 1.00 mol R = 0.0821 L•atm/mol•K T = 273 K

Ideal Gas Law Practice

2. Write eq. PV = nRT V = nRT

P

3. Substitute & solveV = (1.00 mol)(0.0821 L•atm/mol•K)(273 K)

(1.00 atm) = 22.4 L

Combined Gas Law

• Moles remain constant, but other conditions change

• P1V1 = P2V2

T1 T2

Combined Gas Law Practice

A sample of CO2 gas occupies 45 L at 750 K & 500 kPa. What’s the V of this gas at STP?

1. List knownP1 = 500 kPa P2 = 101.325 kPa

V1 = 45 L V2 = ? L

T1 = 750 K T2 = 273 K

Combined Gas Law Practice

2. Write eq. P1V1 = P2V2 V2 = P1V1T2

T1 T2 T1P2

3. Substitute & solveV2 = (500 kPa)(45 L)(273 K)

(750 K)(101.325 kPa) = 81 L

Gas Stoichiometry

• Gas V’s can be determined from mole ratios in bal. eqs.

• 3H2 + N2 2NH3

3 L 1 L 2 L

22 L N2 x 3 L H2 = 66 L H2

1 L N2

(12-4) Changes of State

• Evaporation: l g

• Condensation: g l

• Sublimation: s g

Vapor Pressure

• P exerted by a vapor in equilibrium w/ its liquid state at a given T– H2O(l) H2O(g)

Phase Diagrams

• Shows T’s & P’s at which a substance exists in different phases

• Phases are at equilibrium along the lines

• Phase: substance has uniform composition & properties

Phase Diagrams (cont.)

• Normal bp: boiling T at 1 atm

• Critical point: T & P above which the properties of vapor can’t be distinguished from a liquid– Supercritical fluids

• Triple point: T & P where 3 phases exist in equil.

Phase Diagram: Water

C.P.

N.B.P.N.M.P.

Phase Diagram: Carbon Dioxide