05_lecture_ppt-Kuila

# 4nh 3 5o 2 4no 6h 2 o 1 mole nh 3 1 mole no at

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4NH 3 + 5O 2 4NO + 6H 2 O 1 mole NH 3 1 mole NO At constant T and P 1 volume NH 3 1 volume NO Subscribe to view the full document.

15 Summary of Gas Laws Boyle’s Law 16 Charles Law Subscribe to view the full document.

17 Avogadro’s Law 18 Ideal Gas Equation Charles’ law: V α T (at constant n and P ) Avogadro’s law: V α n (at constant P and T ) Boyle’s law: P α (at constant n and T ) 1 V V α nT P V = constant x = R nT P nT P R is the gas constant PV = nRT Subscribe to view the full document.

19 The conditions 0 0 C and 1 atm are called standard temperature and pressure (STP). PV = nRT R = PV nT = (1 atm)(22.414L) (1 mol)(273.15 K) R = 0.082057 L • atm / (mol • K) Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L. 20 What is the volume (in liters) occupied by 49.8 g of HCl at STP? PV = nRT V = nRT P T = 0 0 C = 273.15 K P = 1 atm n = 49.8 g x 1 mol HCl 36.45 g HCl = 1.37 mol V = 1 atm 1.37 mol x 0.0821 x 273.15 K L•atm mol•K V = 30.7 L Subscribe to view the full document.

21 Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0 C is heated to 85 0 C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? PV = nRT n, V and R are constant nR V = P T = constant P 1 T 1 P 2 T 2 = P 1 = 1.20 atm T 1 = 291 K P 2 = ? T 2 = 358 K P 2 = P 1 x T 2 T 1 = 1.20 atm x 358 K 291 K = 1.48 atm 22 Density ( d ) Calculations d = m V = P M RT m is the mass of the gas in g M is the molar mass of the gas Molar Mass ( ) of a Gaseous Substance dRT P M = d is the density of the gas in g/L Subscribe to view the full document.

23 A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0 0 C. What is the molar mass of the gas? dRT P M = d = m V 4.65 g 2.10 L = = 2.21 g L M = 2.21 g L 1 atm x 0.0821 x 300.15 K L•atm mol•K M = 54.5 g/mol 24 Gas Stoichiometry What is the volume of CO 2 produced at 37 0 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C 6 H 12 O 6 ( s ) + 6O 2 ( g ) 6CO 2 ( g ) + 6H 2 O ( l ) g C 6 H 12 O 6 mol C 6 H 12 O 6 mol CO 2 V CO 2 5.60 g C 6 H 12 O 6 1 mol C 6 H 12 O 6 180 g C 6 H 12 O 6 x 6 mol CO 2 1 mol C 6 H 12 O 6 x = 0.187 mol CO 2 V = nRT P 0.187 mol x 0.0821 x 310.15 K L•atm mol•K 1.00 atm = = 4.76 L Subscribe to view the full document.

25 Dalton’s Law of Partial Pressures V and T are constant P 1 P 2 P total = P 1 + P 2 26 Consider a case in which two gases, A and B , are in a container of volume V. Subscribe to view the full document. • Fall '07
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