unj fmipa-fisika

1. Sesion#07Gas Ideal Irma Jurusan Kimia FakultasMatematikadanIlmuPengetahuanAlam

3. Indikator Mengkonversi satuan tekanan, volum, suhu dan jumlah gas Menggunakan hukum gas ideal untuk menjawab soal Menghitung P, V, T dan n suatu gas Menghitung tekanan parsial menggunakan hukum Dalton 06/01/2011 © 2010 Universitas Negeri Jakarta | www.unj.ac.id | 3

4. 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 06/01/2011 © 2010 Universitas Negeri Jakarta | www.unj.ac.id | 4

6. Within a few %, ideal gas equation describes most real gases at room temperature and pressures of 1 atm or less

7. In real gases, particles attract each other reducing the pressure

8. Real gases behave more like ideal gases as pressure approaches zero.06/01/2011 © 2010 Universitas Negeri Jakarta | www.unj.ac.id | 5

9. 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 06/01/2011 © 2010 Universitas Negeri Jakarta | www.unj.ac.id | 6

10. Example 1 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 06/01/2011 © 2010 Universitas Negeri Jakarta | www.unj.ac.id | 7

11. Example 2 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? 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 = ? ? 06/01/2011 © 2010 Universitas Negeri Jakarta | www.unj.ac.id | 8

12. 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 06/01/2011 © 2010 Universitas Negeri Jakarta | www.unj.ac.id | 9

13. Example 3 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? 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 06/01/2011 © 2010 Universitas Negeri Jakarta | www.unj.ac.id | 10

14. Example of 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 06/01/2011 © 2010 Universitas Negeri Jakarta | www.unj.ac.id | 11

15. 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 06/01/2011 © 2010 Universitas Negeri Jakarta | www.unj.ac.id | 12

16. 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 06/01/2011 © 2010 Universitas Negeri Jakarta | www.unj.ac.id | 13

17. Example of 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? 06/01/2011 © 2010 Universitas Negeri Jakarta | www.unj.ac.id | 14

18. 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 06/01/2011 © 2010 Universitas Negeri Jakarta | www.unj.ac.id | 15

19. 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 06/01/2011 © 2010 Universitas Negeri Jakarta | www.unj.ac.id | 16

20. 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? 06/01/2011 © 2010 Universitas Negeri Jakarta | www.unj.ac.id | 17

21. 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. 06/01/2011 © 2010 Universitas Negeri Jakarta | www.unj.ac.id | 18

22. Example 4 How many L of O2 are need to react 28.0 g NH3 at24°C and 0.950 atm? 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) 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 06/01/2011 © 2010 Universitas Negeri Jakarta | www.unj.ac.id | 19

23. 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 06/01/2011 © 2010 Universitas Negeri Jakarta | www.unj.ac.id | 20

24. 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 06/01/2011 © 2010 Universitas Negeri Jakarta | www.unj.ac.id | 21

25. Example 5 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 06/01/2011 © 2010 Universitas Negeri Jakarta | www.unj.ac.id | 22

26. 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 + ..... 06/01/2011 © 2010 Universitas Negeri Jakarta | www.unj.ac.id | 23

27. 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 06/01/2011 © 2010 Universitas Negeri Jakarta | www.unj.ac.id | 24

28. 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. 06/01/2011 © 2010 Universitas Negeri Jakarta | www.unj.ac.id | 25

29. Example 6 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? 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 06/01/2011 © 2010 Universitas Negeri Jakarta | www.unj.ac.id | 26