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Lecture PLUS Timberlake 20001 Ideal Gas Law The equality for the four variables involved in Boyles Law, Charles Law, Gay-Lussacs Law and Avogadros law.

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Presentation on theme: "Lecture PLUS Timberlake 20001 Ideal Gas Law The equality for the four variables involved in Boyles Law, Charles Law, Gay-Lussacs Law and Avogadros law."— Presentation transcript:

1 Lecture PLUS Timberlake 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

2 Lecture PLUS Timberlake 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.

3 Lecture PLUS Timberlake 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 = L-atm mol-K

4 Lecture PLUS Timberlake Learning Check G15 What is the value of R when the STP value for P is 760 mmHg?

5 Lecture PLUS Timberlake 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

6 Lecture PLUS Timberlake 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?

7 Lecture PLUS Timberlake 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 K n = 2.86 mol2.86 mol P = ? ?

8 Lecture PLUS Timberlake 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

9 Lecture PLUS Timberlake 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?

10 Lecture PLUS Timberlake 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) = mol O 2 x 32.0 g O 2 = 6.4 g O 2 1 mol O 2

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

12 Lecture PLUS Timberlake 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

13 Lecture PLUS Timberlake Substitute (1.00 atm ) mol-K = mol O 2 /L ( L-atm) (273 K) Change moles/L to g/L 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

14 Lecture PLUS Timberlake 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?

15 Lecture PLUS Timberlake 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 = (1.0) = 14.0 g/EF

16 Lecture PLUS Timberlake 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 = g/EF molecular formula CH 2 x 4 = C 4 H 8

17 Lecture PLUS Timberlake 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?

18 Lecture PLUS Timberlake Solution lb Fe g Fe mol Fe mol H 2 L H x 10 3 lb x 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.

19 Lecture PLUS Timberlake Learning Check G18 How many L of O 2 are need to react 28.0 g NH 3 at 24°C and atm? 4 NH 3 (g) + 5 O 2 (g) 4 NO(g) + 6 H 2 O(g)

20 Lecture PLUS Timberlake Solution G18 Find mole of O g NH 3 x 1 mol NH 3 x 5 mol O g NH 3 4 mol NH 3 = 2.06 mol O 2 V = nRT = (2.06 mol)(0.0821)(297K) = 52.9 L P atm

21 Lecture PLUS Timberlake Summary of Conversions with Gases Volume A Volume B Grams A Moles A Moles B Grams B Atoms or molecules A molecules B

22 Lecture PLUS Timberlake Daltons Law of Partial Pressures The % of gases in air Partial pressure (STP) 78.08% N mmHg 20.95% O mmHg 0.94% Ar 7.1 mmHg 0.03% CO mmHg P AIR = P N + P O + P Ar + P CO = 760 mmHg Total Pressure760 mm Hg

23 Lecture PLUS Timberlake 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) ) 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

24 Lecture PLUS Timberlake 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

25 Lecture PLUS Timberlake 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

26 Lecture PLUS Timberlake 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 mol He mol N 2

27 Lecture PLUS Timberlake 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.

28 Lecture PLUS Timberlake Learning Check G20 A 5.00 L scuba tank contains 1.05 mole of O 2 and mole He at 25°C. What is the partial pressure of each gas, and what is the total pressure in the tank?

29 Lecture PLUS Timberlake Solution G20 P = nRT P T = P O + P He V 2 P T = 1.47 mol x L-atm x 298 K 5.00 L(K mol) =7.19 atm


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