Ideal Gas Law The equality for the four variables involved in Boyle’s Law, Charles’ Law, Gay-Lussac’s Law and Avogadro’s law can be written PV = nRT R = ideal gas constant

Ideal Gases Behave as described by the ideal gas equation; no real gas is actually ideal Within a few %, ideal gas equation describes most real gases at room temperature and pressures of 1 atm or less In real gases, particles attract each other reducing the pressure Real gases behave more like ideal gases as pressure approaches zero.

PV = nRT R is known as the universal gas constant Using STP conditions R = PV = (1.00 atm)(22.4 L) nT (1mol) (273K) n T = 0.0821 L-atm mol-K

Learning Check G15 What is the value of R when the STP value for P is 760 mmHg of 1 mole of gas?

Solution G15 What is the value of R when the STP value for P is 760 mmHg? R = PV = (760 mm Hg) (22.4 L) nT (1mol) (273K) = 62.4 L-mm Hg mol-K

Learning Check G16 Dinitrogen monoxide (N2O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23°C, what is the pressure (mmHg) in the tank in the dentist office?

Solution G16 Adjust to match the units of R Set up data for 3 of the 4 gas variables Adjust to match the units of R V = 20.0 L 20.0 L T = 23°C + 273 296 K n = 2.86 mol 2.86 mol P = ? ?

Rearrange ideal gas law for unknown P P = nRT V Substitute values of n, R, T and V and solve for P P = (2.86 mol)(62.4L-mmHg)(296 K) (20.0 L) (K-mol) = 2.64 x 103 mm Hg

Learning Check G17 A 5.0 L cylinder contains oxygen gas at 20.0°C and 735 mm Hg. How many grams of oxygen are in the cylinder?

Solution G17 Solve ideal gas equation for n (moles) n = PV RT = (735 mmHg)(5.0 L)(mol K) (62.4 mmHg L)(293 K) = 0. 20 mol O2 x 32.0 g O2 = 6.4 g O2 1 mol O2

Molar Mass of a gas n = PV = (0.813 atm) (0.215 L) = 0.00703 mol What is the molar mass of a gas if 0.250 g of the gas occupy 215 mL at 0.813 atm and 30.0°C? n = PV = (0.813 atm) (0.215 L) = 0.00703 mol RT (0.0821 L-atm/molK) (303K) Molar mass = g = 0.250 g = 35.6 g/mol mol 0.00703 mol

Density of a Gas Calculate the density in g/L of O2 gas at STP. From STP, we know the P and T. P = 1.00 atm T = 273 K Rearrange the ideal gas equation for moles/L PV = nRT PV = nRT P = n RTV RTV RT V

Substitute (1.00 atm ) mol-K = 0.0446 mol O2/L (0.0821 L-atm) (273 K) Change moles/L to g/L 0.0446 mol O2 x 32.0 g O2 = 1.43 g/L 1 L 1 mol O2 Therefore the density of O2 gas at STP is 1.43 grams per liter

Formulas of Gases A gas has a % composition by mass of 85.7% carbon and 14.3% hydrogen. At STP the density of the gas is 2.50 g/L. What is the molecular formula of the gas?

Formulas of Gases Calculate Empirical formula 85.7 g C x 1 mol C = 7.14 mol C/7.14 = 1 C 12.0 g C 14.3 g H x 1 mol H = 14.3 mol H/ 7.14 = 2 H 1.0 g H Empirical formula = CH2 EF mass = 12.0 + 2(1.0) = 14.0 g/EF

Using STP and density ( 1 L = 2.50 g) 2.50 g x 22.4 L = 56.0 g/mol 1 L 1 mol n = EF/ mol = 56.0 g/mol = 4 14.0 g/EF molecular formula CH2 x 4 = C4H8

Gases in Chemical Equations On December 1, 1783, Charles used 1.00 x 103 lb of iron filings to make the first ascent in a balloon filled with hydrogen Fe(s) + H2SO4(aq)  FeSO4(aq) + H2(g) At STP, how many liters of hydrogen gas were generated?

Solution lb Fe  g Fe  mol Fe  mol H2  L H2 1.00 x 103 lb x 453.6 g x 1 mol Fe x 1 mol H2 1 lb 55.9 g 1 mol Fe x 22.4 L H2 = 1.82 x 105 L H2 1 mol H2 Charles generated 182,000 L of hydrogen to fill his air balloon.

Learning Check G18 How many L of O2 are need to react 28.0 g NH3 at 24°C and 0.950 atm? 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)

Solution G18 Find mole of O2 28.0 g NH3 x 1 mol NH3 x 5 mol O2 17.0 g NH3 4 mol NH3 = 2.06 mol O2 V = nRT = (2.06 mol)(0.0821)(297K) = 52.9 L P 0.950 atm

Summary of Conversions with Gases Volume A Volume B Grams A Moles A Moles B Grams B Atoms or Atoms or molecules A molecules B

Daltons’ Law of Partial Pressures The % of gases in air Partial pressure (STP) 78.08% N2 593.4 mmHg 20.95% O2 159.2 mmHg 0.94% Ar 7.1 mmHg 0.03% CO2 0.2 mmHg PAIR = PN + PO + PAr + PCO = 760 mmHg 2 2 2 Total Pressure 760 mm Hg

Learning Check G19 A.If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O2 in the air? 1) 35.6 2) 156 3) 760 B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N2 in the air? 1) 557 2) 9.14 3) 0.109

Solution G19 A.If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O2 in the air? 2) 156 B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N2 in the air? 1) 557

Partial Pressure Partial Pressure Pressure each gas in a mixture would exert if it were the only gas in the container Dalton's Law of Partial Pressures The total pressure exerted by a gas mixture is the sum of the partial pressures of the gases in that mixture. PT = P1 + P2 + P3 + .....

Partial Pressures The total pressure of a gas mixture depends on the total number of gas particles, not on the types of particles. STP P = 1.00 atm P = 1.00 atm 0.50 mol O2 + 0.20 mol He + 0.30 mol N2 1.0 mol He

Health Note When a scuba diver is several hundred feet under water, the high pressures cause N2 from the tank air to dissolve in the blood. If the diver rises too fast, the dissolved N2 will form bubbles in the blood, a dangerous and painful condition called "the bends". Helium, which is inert, less dense, and does not dissolve in the blood, is mixed with O2 in scuba tanks used for deep descents.

Learning Check G20 A 5.00 L scuba tank contains 1.05 mole of O2 and 0.418 mole He at 25°C. What is the partial pressure of each gas, and what is the total pressure in the tank?

Solution G20 P = nRT PT = PO + PHe V 2 PT = 1.47 mol x 0.0821 L-atm x 298 K 5.00 L (K mol) = 7.19 atm