Chapter 11 Molecular Composition of Gases

Avogadro’s Law Equal Volumes of Gases at the Same Temperature & Pressure contain the Same Number of “Particles.”

Balloons Holding 1.0 L of Gas at 25º C and 1 atm. Each balloon contains mole of gas or 2.5 x molecules.

Avogadro’s Law For a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas (at low pressures). V = an a = proportionality constant V = volume of the gas n = number of moles of gas

What are the four quantities needed to describe a gas? Pressure Volume Temperature Moles of gas In Chapter 10, which three of these quantities did we vary? Pressure Volume Temperature Now we will consider the moles of gas.

Moles of Gas The number of moles of gas will always affect at least one of the other three quantities.

Ideal Gas Law 4 Mathematical relationship among 4 pressure, 4 volume, 4 temperature and 4 the number of moles.

Ideal Gas Law 4 Mathematical equation coming from the combination of coming from the combination of - Boyle’s Law - Charles’ Law & - Avogadro’s Law

Ideal Gas Law 4 An equation of state for a gas. 4 “state” is the condition of the gas at a given time. PV = nRT An Ideal Gas is a hypothetical substance. Ideal Gas Law is an empirical equation.

Ideal Gas Law PV = nRT R = proportionality constant R = proportionality constant = ideal gas constant = L atm   mol  = L atm   mol  See Table 11-1 on page 342 for additional values for R.

Ideal Gas Law PV = nRT Because of the units of R, P = pressure in atm P = pressure in atm V = volume in liters V = volume in liters n = moles n = moles T = temperature in Kelvins T = temperature in Kelvins Holds closely at P < 1 atm Holds closely at P < 1 atm

Standard Temperature and Pressure “STP” P = 1 atmosphere T = C

Ideal Gas Law PV = nRT If gas is at STP, R = L atm   mol  R = L atm   mol  P = 1.00 atm P = 1.00 atm T = 273 Kelvin T = 273 Kelvin Then Then V /n = volume in liters/ mole V /n = volume in liters/ mole = RT/ P = RT/ P = L atm x 273 K = L atm x 273 K mole K mole K 1.00 atm 1.00 atm = 22.4 L/ mole = 22.4 L/ mole

Standard Molar Volume of a Gas The volume occupied by one mole of a gas at STP L/ mole

A Mole of Any Gas Occupies a Volume of Approximately L at STP

Problem What is the volume of 77.0 g of nitrogen dioxide at STP?

Solution 77.0 g NO 2 x 1 mole NO 2 x 22.4 L g NO 2 1 mole g NO 2 1 mole = L NO 2

Problem What is the mass of 1.33 x 10 4 mL of oxygen gas at STP?

Solution 1.33 x 10 4 mL x 1 L x 1 mol 1000 mL 22.4 L 1000 mL 22.4 L x g O 2 x g O 2 1 mole O 2 1 mole O 2 = 19.0 g O 2

Problem At STP, 3 L of chlorine is produced during a chemical reaction. What is the mass of this gas?

Solution 3 L x 1 mole x g Cl 2 3 L x 1 mole x g Cl L 1 mole Cl L 1 mole Cl 2 = 9 g Cl 2

Problem What pressure, in atmospheres, is exerted by mole of hydrogen gas in a 4.08 L container at 35  C? Not at STP!!! Use Ideal Gas Equation!!!

Solution: P V = n R T P = nRT/V T (K) = = 308 K P = mole x L atm x 308 K mole K mole K 4.08 L 4.08 L P = 2.01 atm P = 2.01 atm

Problem A sample that contains 4.38 mole of a gas at 250 K has a pressure of atm. What is the volume?

Solution: PV = nRT V = nRT/P V = 4.38 mole x L atm x 250 K mole K mole K atm atm V = 105 L V = 105 L

Molar Mass of a Gas n = # of moles = grams of gas= (m) molar mass molar mass molar mass molar mass

P V = n R T P V = (m/molar mass) R T Therefore, molar mass = m R T P V

Problem At 28  C and atm, 1.00 L of gas has a mass of 5.16 g. What is the molar mass of this gas?

Solution molar mass = m R T P V Molar Mass = 5.16 g x L atm x 301 K mole K atm x 1.00 L Molar Mass = g/mole = 131 g/mole

Also, since density (d) = m V Then, since molar mass = m R T P V molar mass = d R T P

Density = (molar mass) x P R T Temperature in K!! Gas density usually measured in g/L

Problem What is the density of a sample of ammonia gas, NH 3, if the pressure is atm and the temperature is 63.0  C? What is the density of a sample of ammonia gas, NH 3, if the pressure is atm and the temperature is 63.0  C?

Solution Density = (molar mass) x P R T Density = g/mole x atm L atm x 336 K mole K Density = g/ L NH 3

Problem The density of dry air at sea level (1 atm) is g/L at 15  C. (1 atm) is g/L at 15  C. What is the average molar mass of the air?

Solution Density = (molar mass) x P R T Therefore, Molar Mass = Density x R x T P

Solution Molar Mass = Density x R x T P Molar Mass = g x x x x L atm x 288 K L mole K ___ 1 atm Molar Mass = g/mole = 29.0 g/mole

Homework Read pages Complete Worksheet.