# Ch. 14 - Gases III. Three More Laws Ideal Gas Law, Daltons Law, & Grahams Law.

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Ch. 14 - Gases III. Three More Laws Ideal Gas Law, Daltons Law, & Grahams Law

1 mol of a gas=22.4 L at STP Molar Volume at STP S tandard T emperature & P ressure 0°C and 1 atm A. Avogadros Principle

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

PV T VnVn PV nT B. Ideal Gas Law = k UNIVERSAL GAS CONSTANT R=0.08206 L atm/mol K R=8.315 dm 3 kPa/mol K = R You dont need to memorize these values! Merge the Combined Gas Law with Avogadros Principle:

B. Ideal Gas Law UNIVERSAL GAS CONSTANT R=0.08206 L atm/mol K R=8.315 dm 3 kPa/mol K PV=nRT You dont need to memorize these values!

GIVEN: P = ? atm n = 0.412 mol T = 16°C = 289 K V = 3.25 L R = 0.0826 L atm/mol K WORK: PV = nRT P(3.25)=(0.412)(0.0821)(289) L mol L atm/mol K K P = 3.01 atm C. Ideal Gas Law Problems b Calculate the pressure in atmospheres of 0.412 mol of He at 16°C & occupying 3.25 L.

GIVEN: V = ?V = ? n = 85 g T = 25°C = 298 K P = 104.5 kPa R = 8.315 dm 3 kPa/mol K C. Ideal Gas Law Problems b Find the volume of 85 g of O 2 at 25°C and 104.5 kPa. = 2.7 mol WORK: 85 g 1 mol O 2 = 2.7 mol 32.00 gO 2 PV = nRT (104.5)V=(2.7) (8.315) (298) kPa mol dm 3 kPa/mol K K V = 64 dm 3

D. Applications of Ideal Gas Law b Can be used to calculate the molar mass of a gas from the density b Substitute this into ideal gas law b And m/V = d in g/L, so

GIVEN: P = 1.50 atm T = 27°C = 300. K d = 1.95 g/L R = 0.08206 L atm/mol K MM = ? WORK: MM = dRT/P MM=(1.95)(0.08206)(300.)/1.50 g/L L atm/mol K K atm MM = 32.0 g/mol D. Applications of Ideal Gas Law b 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 CO 2 T = 25°C = 298 K P = 750. torr R = 0.08206 L atm/mol K MM = 44.01 g/mol MM = dRT/P d = MM P/RT d=(44.01)(.987)/(0.08206)(298) g/mol atm L atm/mol K K d = 1.78 g/L CO 2 D. Applications of Ideal Gas Law b 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. Daltons Law b The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases P total = P 1 + P 2 +...P n

GIVEN: P O2 = ? P total = 1.00 atm P CO2 = 0.12 atm P N2 = 0.70 atm WORK: P total = P CO2 + P N2 + P O2 1.00 atm = 0.12 atm +.70 atm + P O2 P O2 = 0.18 atm D. Daltons Law b A sample of air has a total pressure of 1.00 atm. The mixture contains only CO 2, N 2, and O 2. If P CO2 = 0.12 atm and P N2 = 0.70, what is the partial pressure of O 2 ?

D. Daltons Law When H 2 gas is collected by water displacement, the gas in the collection bottle is actually a mixture of H 2 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: P atm = P gas + P H 2 O

GIVEN: P H2 = ? P total = 94.4 kPa P H2O = 20.4 mm Hg = 2.72 kPa WORK: P total = P H2 + P H2O 94.4 kPa = P H2 + 2.72 kPa P H2 = 91.7 kPa D. Daltons Law b 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 H 2 and water vapor.

GIVEN: P gas = ? P total = 742.0 torr P H2O = 42.2 torr WORK: P total = P gas + P H2O 742.0 torr = P H2 + 42.2 torr P gas = 699.8 torr b 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. Daltons Law The total pressure in the collection bottle is equal to barometric pressure and is a mixture of the gas and water vapor.

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