Presentation on theme: "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."— Presentation transcript:
1Ideal Gas LawThe equality for the four variables involved in Boyle’s Law, Charles’ Law, Gay-Lussac’s Law and Avogadro’s law can be writtenPV = nRTR = ideal gas constant
2Ideal GasesBehave as described by the ideal gas equation; no real gas is actually idealWithin a few %, ideal gas equation describes most real gases at room temperature and pressures of 1 atm or lessIn real gases, particles attract each other reducing the pressureReal gases behave more like ideal gases as pressure approaches zero.
3PV = 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= L-atmmol-K
4Learning Check G15What is the value of R when the STP value for P is 760 mmHg of 1 mole of gas?
5Solution G15What 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)= L-mm Hg mol-K
6Learning Check G16Dinitrogen 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?
7Solution G16 Adjust to match the units of R Set up data for 3 of the 4 gas variablesAdjust to match the units of RV = L LT = 23°C Kn = mol 2.86 molP = ? ?
8Rearrange ideal gas law for unknown P P = nRTVSubstitute values of n, R, T and V and solve for PP = (2.86 mol)(62.4L-mmHg)(296 K)(20.0 L) (K-mol)= x 103 mm Hg
9Learning Check G17A 5.0 L cylinder contains oxygen gas at 20.0°C and 735 mm Hg. How many grams of oxygen are in the cylinder?
10Solution 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 O2 x g O2 = 6.4 g O21 mol O2
11Molar Mass of a gas n = PV = (0.813 atm) (0.215 L) = 0.00703 mol 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) = molRT ( L-atm/molK) (303K)Molar mass = g = g = g/molmol mol
12Density of a GasCalculate the density in g/L of O2 gas at STP. From STP, we know the P and T.P = atm T = 273 KRearrange the ideal gas equation for moles/LPV = nRT PV = nRT P = nRTV RTV RT V
13Substitute(1.00 atm ) mol-K = mol O2/L( L-atm) (273 K)Change moles/L to g/Lmol O2 x g O2 = g/L1 L mol O2Therefore the density of O2 gas at STP is1.43 grams per liter
14Formulas of GasesA 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?
15Formulas of Gases Calculate Empirical formula 85.7 g C x 1 mol C = mol C/7.14 = 1 C12.0 g C14.3 g H x 1 mol H = 14.3 mol H/ 7.14 = 2 H1.0 g HEmpirical formula = CH2EF mass = (1.0) = 14.0 g/EF
16Using STP and density ( 1 L = 2.50 g) 2.50 g x L = g/mol1 L moln = EF/ mol = g/mol = 414.0 g/EFmolecular formulaCH2 x = C4H8
17Gases 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 hydrogenFe(s) + H2SO4(aq) FeSO4(aq) + H2(g)At STP, how many liters of hydrogengas were generated?
18Solution lb Fe g Fe mol Fe mol H2 L H2 1.00 x 103 lb x g x 1 mol Fe x 1 mol H21 lb g mol Fex L H2 = x 105 L H21 mol H2Charles generated 182,000 L of hydrogen to fill his air balloon.
19Learning Check G18How many L of O2 are need to react 28.0 g NH3 at 24°C and atm?4 NH3(g) + 5 O2(g) NO(g) H2O(g)
20Solution G18 Find mole of O2 28.0 g NH3 x 1 mol NH3 x 5 mol O2 17.0 g NH mol NH3= 2.06 mol O2V = nRT = (2.06 mol)(0.0821)(297K) = LP atm
21Summary of Conversions with Gases Volume A Volume BGrams A Moles A Moles B Grams BAtoms or Atoms ormolecules A molecules B
22Daltons’ Law of Partial Pressures The % of gases in air Partial pressure (STP)78.08% N mmHg20.95% O mmHg0.94% Ar mmHg0.03% CO mmHgPAIR = PN + PO + PAr + PCO = 760 mmHgTotal Pressure mm Hg
23Learning Check G19A.If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O2 in the air?1) ) ) 760B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N2 in the air?1) ) )
24Solution G19A.If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O2 in the air?2) 156B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N2 in the air?1) 557
25Partial Pressure Partial Pressure Pressure each gas in a mixture would exert if it were the only gas in the containerDalton's Law of Partial PressuresThe total pressure exerted by a gas mixture is the sum of the partial pressures of the gases in that mixture.PT = P P P
26Partial Pressures The total pressure of a gas mixture depends on the total number of gas particles, not onthe types of particles.STPP = atm P = atm0.50 mol O2mol Hemol N21.0 mol He
27Health NoteWhen a scuba diver is several hundred feetunder 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.
28Learning Check G20A 5.00 L scuba tank contains 1.05 mole of O2 and mole He at 25°C. What is the partial pressure of each gas, and what is the total pressure in the tank?
29Solution G20 P = nRT PT = PO + PHe V 2 PT = mol x L-atm x 298 K5.00 L (K mol)= 7.19 atm