Presentation on theme: "Ideal Gas Law The equality for the four variables involved in Boyles Law, Charles Law, Gay-Lussacs Law and Avogadros law can be written PV = nRT R = ideal."— Presentation transcript:
Ideal Gas Law The equality for the four variables involved in Boyles Law, Charles Law, Gay-Lussacs Law and Avogadros 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 P V 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 (N 2 O), 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 Set up data for 3 of the 4 gas variables Adjust to match the units of R V = 20.0 L20.0 L T = 23°C + 273 296 K n = 2.86 mol2.86 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 10 3 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 O 2 x 32.0 g O 2 = 6.4 g O 2 1 mol O 2
Molar Mass of a gas 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 O 2 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 = nRTPV = nRT P = n RTV RTV RT V
Substitute (1.00 atm ) mol-K = 0.0446 mol O 2 /L (0.0821 L-atm) (273 K) Change moles/L to g/L 0.0446 mol O 2 x 32.0 g O 2 = 1.43 g/L 1 L 1 mol O 2 Therefore the density of O 2 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 = CH 2 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 CH 2 x 4 = C 4 H 8
Gases in Chemical Equations On December 1, 1783, Charles used 1.00 x 10 3 lb of iron filings to make the first ascent in a balloon filled with hydrogen Fe(s) + H 2 SO 4 (aq) FeSO 4 (aq) + H 2 (g) At STP, how many liters of hydrogen gas were generated?
Solution lb Fe g Fe mol Fe mol H 2 L H 2 1.00 x 10 3 lb x 453.6 g x 1 mol Fe x 1 mol H 2 1 lb 55.9 g 1 mol Fe x 22.4 L H 2 = 1.82 x 10 5 L H 2 1 mol H 2 Charles generated 182,000 L of hydrogen to fill his air balloon.
Learning Check G18 How many L of O 2 are need to react 28.0 g NH 3 at 24°C and 0.950 atm? 4 NH 3 (g) + 5 O 2 (g) 4 NO(g) + 6 H 2 O(g)
Solution G18 Find mole of O 2 28.0 g NH 3 x 1 mol NH 3 x 5 mol O 2 17.0 g NH 3 4 mol NH 3 = 2.06 mol O 2 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 molecules A molecules B
Daltons Law of Partial Pressures The % of gases in air Partial pressure (STP) 78.08% N 2 593.4 mmHg 20.95% O 2 159.2 mmHg 0.94% Ar 7.1 mmHg 0.03% CO 2 0.2 mmHg P AIR = P N + P O + P Ar + P CO = 760 mmHg 2 2 2 Total Pressure760 mm Hg
Learning Check G19 A. If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O 2 in the air? 1) 35.6 2) 156 3) 760 B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N 2 in the air? 1) 557 2) 9.143) 0.109
Solution G19 A. If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O 2 in the air? 2) 156 B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N 2 in the air? 1) 557
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. P T = P 1 + P 2 + P 3 +.....
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 1.0 mol He 0.50 mol O 2 + 0.20 mol He + 0.30 mol N 2
Health Note When a scuba diver is several hundred feet under water, the high pressures cause N 2 from the tank air to dissolve in the blood. If the diver rises too fast, the dissolved N 2 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 O 2 in scuba tanks used for deep descents.
Learning Check G20 A 5.00 L scuba tank contains 1.05 mole of O 2 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 P T = P O + P He V 2 P T = 1.47 mol x 0.0821 L-atm x 298 K 5.00 L(K mol) =7.19 atm