885 kg of O 2 Calculate the pressure of O 2 at 21 o C SOLUTION PLAN We are

# 885 kg of o 2 calculate the pressure of o 2 at 21 o c

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0.885 kg of O 2 . Calculate the pressure of O 2 at 21 o C. SOLUTION: PLAN: We are given V , T and mass, which can be converted to moles ( n ). Use the ideal gas law to find P . V = 438 L T = 21°C = 294 K n = 0.885 kg O 2 (convert to mol) P is unknown = 27.7 mol O 2 P = nRT V = 27.7 mol 294.15 K atm·L mol·K 0.0821 x x 438 L = 1.53 atm 0.885 kg O 2 x 10 3 g 1 kg 1 mol O 2 32.00 g O 2 x
5-25 Sample Problem 5.6 Using Gas Laws to Determine a Balanced Equation PROBLEM: The piston-cylinders is depicted before and after a gaseous reaction that is carried out at constant pressure. The temperature is 150 K before the reaction and 300 K after the reaction. (Assume the cylinder is insulated.) Which of the following balanced equations describes the reaction? (1)A 2 ( g ) + B 2 ( g ) → 2AB( g ) (2) 2AB( g ) + B 2 ( g ) → 2AB 2 ( g ) (3) A( g ) + B 2 ( g ) → AB 2 ( g ) (4) 2AB 2 ( g ) A 2 ( g ) + 2B 2 ( g )
5-26 SOLUTION: PLAN: We are told that P is constant for this system, and the depiction shows that V does not change either. Since T changes, the volume could not remain the same unless the amount of gas in the system also changes. n 1 T 1 = n 2 T 2 Sample Problem 5.6 Since T doubles, the total number of moles of gas must halve – i.e., the moles of product must be half the moles of reactant. This relationship is shown by equation (3). T 1 T 2 = n 2 n 1 = 150 K 300 K = ½ A( g ) + B 2 ( g ) → AB 2 ( g )
5-27 The Ideal Gas Law and Gas Density The density of a gas is - directly proportional to its molar mass and - inversely proportional to its temperature. M x P RT = d = m V m M moles = and m V density = m M RT PV =
5-28 Sample Problem 5.7 Calculating Gas Density (a) At STP, or 273 K and 1.00 atm: PLAN: We can use the molar mass of CO 2 to find its density from the ideal gas equation. PROBLEM: Find the density (in g/L) of CO 2 ( g ) and the number of molecules per liter (a) at STP and (b) at room conditions (20.°C and 1.00 atm). SOLUTION: M x P RT d = = 44.01 g/mol x 1.00 atm 0.0821 x 273 K mol·K atm·L = 1.96 g/L 1.96 g CO 2 1 L 1 mol CO 2 44.01 g CO 2 x x 6.022 x 10 23 molecules 1 mol = 2.68 x 10 22 molecules CO 2 /L
5-29 Sample Problem 5.7 (b) At 20.°C and 1.00 atm: SOLUTION: M x P RT d = = 44.01 g/mol x 1.00 atm 0.0821 x 293 K mol·K atm·L = 1.83 g/L 1.83 g CO 2 1 L 1 mol CO 2 44.01 g CO 2 x x 6.022 x 10 23 molecules 1 mol = 2.50 x 10 22 molecules CO 2 /L T = 20.°C + 273.15 = 293 K
5-30 Molar Mass from the Ideal Gas Law m M n = PV RT = mRT PV M =
5-31 Determining the molar mass of an unknown volatile liquid. based on the method of J.B.A. Dumas (1800-1884)
5-32 Sample Problem 5.8 Finding the Molar Mass of a Volatile Liquid PROBLEM: An organic chemist isolates a colorless liquid from a petroleum sample. She uses the Dumas method and obtains the following data: PLAN: SOLUTION: Use unit conversions, mass of gas, and density- M relationship. Volume of flask = 213 mL Mass of flask + gas = 78.416 g T = 100.0 o C Mass of flask = 77.834 g P = 754 torr Calculate the molar mass of the liquid. m = (78.416 - 77.834) g = 0.582 g M = mRT VP = 0.582 g atm*L mol*K 0.0821 373 K x x 0.213 L x 0.992 atm = 84.4 g/mol
5-33 Gases mix homogeneously in any proportions. Each gas in a mixture behaves as if it were the only gas present. The pressure exerted by each gas in a mixture is called its partial pressure .