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

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15 Summary of Gas Laws Boyle’s Law
16 Charles Law

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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

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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

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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

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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

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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.

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