Chapter 11 Gases 11.6 The Combined Gas Law Under water, the pressure on a diver is greater than the atmospheric pressure.

Summary of Gas Laws

Combined Gas Law The combined gas law uses Boyle’s law, Charles’s law, and Gay-Lussac’s law (n is constant). P1 V1 = P2 V2 T1 T2

Example of Using the Combined Gas Law A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm, and a temperature of 29 °C. At what temperature (°C) will the helium have a volume of 90.0 mL and a pressure of 3.20 atm (n constant)? STEP 1 Organize the data in a table of initial and final conditions. Conditions 1 Conditions 2 P1 = 0.800 atm P2 = 3.20 atm V1 = 0.180 L (180. mL) V2 = 90.0 mL T1 = 29 °C + 273 = 302 K T2 = ??

Example of Using the Combined Gas Law (continued) STEP 2 Rearrange the gas law for the unknown. Solve the combined gas law for T2. P1 V1 = P2 V2 T1 T2 T2 = T1 P2V2 P1V1 STEP 3 Substitute values into the gas law to solve for the unknown. T2 = 302 K x 3.20 atm x 90.0 mL = 604 K 0.800 atm 180. mL = 604 K - 273 = 331 °C

Learning Check A gas has a volume of 675 mL at 35 °C and 0.850 atm pressure. What is the volume (mL) of the gas at -95 °C and a pressure of 802 mmHg (n constant)?

Solution STEP 1 Organize the data in a table of initial and final conditions. Conditions 1 Conditions 2 T1 = 35 °C + 273 = 178K T2 = -95 °C + 273 = 178K V1 = 675 mL V2 = ??? P1 = 646 mmHg P2 = 802 mmHg STEP 2 Rearrange the gas law for the unknown. P1 V1 = P2V2 T1 T2 V2 = V1 P1T2 P2T1

Solution (continued) STEP 3 Substitute values into the gas law to solve for the unknown. V2 = 675 mL x 646 mmHg x 178K = 314 mL 802 mmHg x 308K

Chapter 11 Gases 11.7 Volume and Moles (Avogadro’s Law) Balloons rise in the air because helium is less dense than air.

Avogadro's Law: Volume and Moles Avogadro’s law states that the volume of a gas is directly related to the number of moles(n) of gas T and P are constant V1 = V2 n1 n2

Learning Check If 0.75 mol of helium gas occupies a volume of 1.5 L, what volume will 1.2 mol of helium occupy at the same temperature and pressure? 1) 0.94 L 2) 1.8 L 3) 2.4 L Balloons rise in the air because helium is less dense than air.

Solution STEP 1 Organize the data in a table of initial and final conditions. Conditions 1 Conditions 2 Know Predict V1 = 1.5 L V2 = ? V increases n1 = 0.75 mole n2 = 1.2 moles n increases

Solution (continued) STEP 2 Rearrange the gas law for the unknown. V1 = V2 n1 n2 V2 = V1n2 n1 STEP 3 Substitute values into the gas law to solve for the unknown. V2 = 1.5 L x 1.2 mol He = 2.4 L 0.75 mol He

STP The volumes of gases can be compared at STP (standard temperature and pressure) when they have the same temperature Standard temperature (T) = 0 °C or 273 K the same pressure Standard pressure (P) = 1 atm (760 mmHg)

Molar Volume The molar volume of a gas is measured at STP (standard temperature and pressure) is 22.4 L for exactly1 mol of a gas

Molar Volume as a Conversion Factor The molar volume of a gas is about the same as the volume of three basketballs. The molar volume at STP has about the same volume as three basketballs can be used to write conversion factors 22.4 L and 1 mol 1 mol 22.4 L

Using Molar Volume

Example of Using Molar Volume What is the volume occupied by 2.75 mol of N2 gas at STP? STEP 1 Identify given and needed. Given 2.75 mol of N2 Need liters of N2 STEP 2 Write a plan. moles of N2 liters of N2

Example of Using Molar Volume (continued) STEP 3 Write conversion factors. 1 mol of gas = 22.4 L 1 mol gas and 22.4 L 22.4 L 1 mol gas STEP 4 Set up problem with factors to cancel units. 2.75 mol N2 x 22.4 L = 61.6 L of N2 1 mol N2

Learning Check How many grams of He gas are present in 8.00 L of He at STP? 1) 25.6 g 2) 0.357 g 3) 1.43 g

Solution STEP 1 Identify given and needed. Given 8.00 L of He Need grams of He STEP 2 Write a plan. liters of He moles of He grams of He

Solution (continued) STEP 3 Write conversion factors. 1 mol of gas = 22.4 L 1 mol gas and 22.4 L 22.4 L 1 mol gas 1 mol of He = 32.00 g 4.003 g He and 1 mol He 1 mol He 4.003 g He STEP 4 Set up problem with factors to cancel units. 8.00 L x 1 mol He x 4.003 g He = 1.43 g of He (3) 22.4 L 1 mol He

Chapter 11 Gases 11.8 The Ideal Gas Law Dinitrogen oxide is used as an anesthetic in dentistry.

Ideal Gas Law The relationship between the four properties (P, V, n, and T) of a gas can be written equal to a constant R. PV = R nT Rearranging this relationship gives the ideal gas law. PV = nRT

Universal Gas Constant, R The universal gas constant, R, can be calculated at STP using a temperature of 273 K, a pressure of 1.00 atm, a quantity of 1 mol of a gas, and a molar volume of 22.4 L. P V R = PV = (1.00 atm)(22.4 L) nT (1 mol) (273K) n T = 0.0821 L • atm mol • K

Learning Check Another value for the universal gas constant is obtained using mmHg for the pressure at STP. What is the value of R when a pressure of 760 mmHg is placed in the R value expression?

Solution What is the value of R when a pressure of 760 mmHg (at STP) is placed in the R value expression? R = PV = (760 mmHg) (22.4 L) nT (1 mol) (273 K) = 62.4 L • mmHg mol • K

Summary of Units for the Universal Gas Constant

Using the Ideal Gas Law

Learning Check Dinitrogen oxide (N2O), laughing gas, is used by dentists as an anesthetic. If a 20.0-L tank of laughing gas contains 2.86 mol of N2O at 23 °C, what is the pressure (mmHg) in the tank? Dinitrogen oxide is used as an anesthetic in dentistry.

Solution STEP 1 Organize the data given for the gas. R = 62.4 L • mmHg mol • K V = 20.0 L T = 23°C + 273 = 296 K n = 2.86 mol P = ?

Solution (continued) STEP 2 Solve the ideal gas law for the unknown. Rearrange the ideal gas law for P. P = nRT V STEP 3 Substitute gas data and calculate the unknown quantity. P = (2.86 mol)(62.4 L • mmHg)(296 K) (20.0 L) (mol • K) = 2.64 x 103 mmHg

Learning Check A cylinder contains 5.0 L of O2 at 20 °C and 0.85 atm. How many grams of oxygen are in the cylinder?

Solution STEP 1 Organize the data given for the gas. P = 0.85 atm, V = 5.0 L, T = 293 K, R = 0.0821 L • atm , n ( or g =?) mol • K STEP 2 Solve the ideal gas law for the unknown. Rearrange the ideal gas law for n (moles). n = PV RT

Solution (continued) STEP 3 Substitute gas data and calculate the unknown quantity. = (0.85 atm)(5.0 L)(mol • K) = 0.18 mol of O2 (0.0821atm • L)(293 K) Convert the moles of gas to the grams of gas using its molar mass. = 0. 18 mol O2 x 32.00 g O2 = 5.8 g of O2 1 mol O2

Calculating the Molar Mass of a Gas What is the molar mass of a gas if 0.250 g of the gas occupies 215 mL at 0.813 atm and 30.0 °C? STEP 1 Organize the data given for the gas. R = 0.0821 L atm/mol K P = 0.813 atm V = 0.215 L n = ? mol T = 30.0 °C + 273 = 303 K

Calculating the Molar Mass of a Gas (continued) STEP 2 Solve the ideal gas law for the unknown. Solve for the moles (n) of gas. n = PV = (0.813 atm) (0.215 L) mol•K = 0.00703 mol RT (0.0821 L • atm)(303 K) STEP 3 Substitute gas data and calculate the unknown quantity. Set up the molar mass relationship. Molar mass = g = 0.250 g = 35.6 g/mol mol 0.00703 mol