Ch. 14 - Gases

III. Three III. Three More LawsMore LawsIII. Three III. Three More LawsMore Laws

Ideal Gas Law, Ideal Gas Law, Dalton’s Dalton’s Law, & Graham’s LawLaw, & Graham’s Law

1 mol of a gas=22.4 Lat STP

Molar Volume at Molar Volume at STPSTPMolar Volume at Molar Volume at STPSTP

Standard Temperature & Pressure0°C and 1 atm

A. Avogadro’s PrincipleA. Avogadro’s PrincipleA. Avogadro’s PrincipleA. Avogadro’s Principle

kn

VV

n

A. Avogadro’s PrincipleA. Avogadro’s PrincipleA. Avogadro’s PrincipleA. Avogadro’s Principle

Equal volumes of gases contain equal numbers of moles• at constant temp & pressure• true for any gas• n = number of moles

PV

TVn

PVnT

B. Ideal Gas LawB. Ideal Gas LawB. Ideal Gas LawB. Ideal Gas Law

= kUNIVERSAL GAS

CONSTANTR=0.08206

Latm/molKR=8.315

dm3kPa/molK

= R

You don’t need to memorize these values!

Merge the Combined Gas Law with Avogadro’s Principle:

B. Ideal Gas LawB. Ideal Gas LawB. Ideal Gas LawB. Ideal Gas Law

UNIVERSAL GAS CONSTANTR=0.08206

Latm/molKR=8.315

dm3kPa/molK

PV=nRT

You don’t need to memorize these values!

GIVEN:

P = ? atm

n = 0.412 mol

T = 16°C = 289 K

V = 3.25 LR = 0.0826Latm/molK

WORK:

PV = nRT

P(3.25)=(0.412)(0.0821)(289) L mol Latm/molK K

P = 3.01 atm

C. Ideal Gas Law ProblemsC. Ideal Gas Law ProblemsC. Ideal Gas Law ProblemsC. Ideal Gas Law Problems Calculate the pressure in atmospheres of

0.412 mol of He at 16°C & occupying 3.25 L.

GIVEN:

V = ?

n = 85 g

T = 25°C = 298 K

P = 104.5 kPaR = 8.315 dm3kPa/molK

C. Ideal Gas Law ProblemsC. Ideal Gas Law ProblemsC. Ideal Gas Law ProblemsC. Ideal Gas Law Problems

Find the volume of 85 g of O2 at 25°C and 104.5 kPa.

= 2.7 mol

WORK:

85 g 1 mol O2 = 2.7 mol

32.00 g O2

PV = nRT(104.5)V=(2.7) (8.315) (298) kPa mol dm3kPa/molK K

V = 64 dm3

D. Applications of Ideal Gas D. Applications of Ideal Gas LawLawD. Applications of Ideal Gas D. Applications of Ideal Gas LawLaw

Can be used to calculate the molar mass of a gas from the density

Substitute this into ideal gas law

And m/V = d in g/L, so

MMmassmolar

mass

massmolar

gas a of grams gas a of moles

mn

)(

)(

VMM

RTm

V

nRTP

P

dRTMM

MM

dRTP or

GIVEN:

P = 1.50 atm

T = 27°C = 300. K

d = 1.95 g/LR = 0.08206 Latm/molK

MM = ?

WORK:

MM = dRT/P

MM=(1.95)(0.08206)(300.)/1.50 g/L Latm/molK K atm

MM = 32.0 g/mol

D. Applications of Ideal Gas D. Applications of Ideal Gas LawLawD. Applications of Ideal Gas D. Applications of Ideal Gas LawLaw

The density of a gas was measured at 1.50 atm and 27°C and found to be 1.95 g/L. Calculate the molar mass of the gas.

GIVEN:

d = ? g/L CO2

T = 25°C = 298 KP = 750. torr

R = 0.08206 Latm/molK

MM = 44.01 g/mol

MM = dRT/P →d = MM P/RT

d=(44.01)(.987)/(0.08206)(298) g/mol atm Latm/molK K

d = 1.78 g/L CO2

D. Applications of Ideal Gas D. Applications of Ideal Gas LawLawD. Applications of Ideal Gas D. Applications of Ideal Gas LawLaw

Calculate the density of carbon dioxide gas at 25°C and 750. torr.

WORK:

750 torr 1 atm = .987 atm 760 torr

= .987 atm

D. Dalton’s LawD. Dalton’s LawD. Dalton’s LawD. Dalton’s Law

The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases

Ptotal = P1 + P2 + ...Pn

GIVEN:

PO2 = ?

Ptotal = 1.00 atm

PCO2 = 0.12 atm

PN2 = 0.70 atm

WORK:Ptotal = PCO2 + PN2 + PO2

1.00 atm = 0.12 atm + .70 atm + PO2

PO2 = 0.18 atm

D. Dalton’s LawD. Dalton’s LawD. Dalton’s LawD. Dalton’s Law A sample of air has a total pressure of

1.00 atm. The mixture contains only CO2, N2, and O2. If PCO2 = 0.12 atm and PN2 = 0.70, what is the partial pressure of O2?

D. Dalton’s LawD. Dalton’s LawD. Dalton’s LawD. Dalton’s Law

•When H2 gas is collected by water displacement, the gas in the collection bottle is actually a mixture of H2 and water vapor.

•Water exerts a pressure known as water-vapor pressure•So, to determine total pressure of gas and water vapor inside a container, make total pressure inside bottle = atmosphere and use this formula:

Patm = Pgas + PH2O

GIVEN:

PH2 = ?

Ptotal = 94.4 kPa

PH2O = 20.4 mm Hg

= 2.72 kPa

WORK:

Ptotal = PH2 + PH2O

94.4 kPa = PH2 + 2.72 kPa

PH2 = 91.7 kPa

D. Dalton’s LawD. Dalton’s LawD. Dalton’s LawD. Dalton’s Law

Hydrogen gas is collected over water at 22.5°C. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa.

Look up water-vapor pressure for 22.5°C, convert to kPa

Sig Figs: Round to lowest number of decimal places.

The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H2 and water vapor.

GIVEN:

Pgas = ?

Ptotal = 742.0 torr

PH2O = 42.2 torr

WORK:

Ptotal = Pgas + PH2O

742.0 torr = PH2 + 42.2 torr

Pgas = 699.8 torr

A gas is collected over water at a temp of 35.0°C when the barometric pressure is 742.0 torr. What is the partial pressure of the dry gas?

Look up water-vapor pressure for 35.0°C.

Sig Figs: Round to least number of decimal places.

D. Dalton’s LawD. Dalton’s LawD. Dalton’s LawD. Dalton’s Law

The total pressure in the collection bottle is equal to barometric pressure and is a mixture of the “gas” and water vapor.