Equal volumes of gases contain equal numbers of

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Equal volumes of gases contain equal numbers of molecules. The gas doesn’t matter. 2 2 1 1 n V n V
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Tro's Introductory Chemistry, Chapter 11 85 Avogadro’s Law, Continued
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mol added = n 2 n 1 , 2 2 1 1 n V n V Example 11.5—A 0.22 Mol Sample of He Has a Volume of 4.8 L. How Many Moles Must Be Added to Give 6.4 L? Since n and V are directly proportional, when the volume increases, the moles should increase, and it does. V 1 =4.8 L, V 2 = 6.4 L, n 1 = 0.22 mol n 2 , and added moles Check: Solution: Solution Map: Relationships: Given: Find: 2 1 2 1 n V V n V 1 , V 2 , n 1 n 2 mol 29 . 0 L 4.8 L 6.4 mol 0.22 1 2 1 2 V V n n mol 07 . 0 added moles 22 . 0 29 . 0 added moles
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Tro's Introductory Chemistry, Chapter 11 95 Practice—If 1.00 Mole of a Gas Occupies 22.4 L at STP, What Volume Would 0.750 Moles Occupy?
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2 2 1 1 n V n V Practice—If 1.00 Mole of a Gas Occupies 22.4 L at STP, What Volume Would 0.750 Moles Occupy?, Continued Since n and V are directly proportional, when the moles decreases, the volume should decrease, and it does. V 1 =22.4 L, n 1 = 1.00 mol, n 2 = 0.750 mol V 2 Check: Solution: Solution Map: Relationships: Given: Find: 2 1 2 1 V n n V V 1 , n 1 , n 2 V 2 L 8 . 16 mol 1.00 mol 0.750 L 22.4 1 2 1 2 n n V V
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Tro's Introductory Chemistry, Chapter 11 97 Ideal Gas Law By combining the gas laws, we can write a general equation. R is called the Gas Constant . The value of R depends on the units of P and V . We will use 0.0821 and convert P to atm and V to L . Use the ideal gas law when you have a gas at one condition, use the combined gas law when you have a gas whose condition is changing. K mol L atm nRT PV R or T n V P
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1 atm = 14.7 psi T (K) = t (°C) + 273 K mol L atm 0.0821 nRT, R PV Example 11.7—How Many Moles of Gas Are in a Basketball with Total Pressure 24.2 Psi, Volume of 3.2 L at 25 °C? 1 mole at STP occupies 22.4 L at STP; since there is a much smaller volume than 22.4 L, we expect less than 1 mole of gas. V = 3.2 L, P = 24.2 psi, t = 25 °C, n , mol Check: Solution: Solution Map: Relationships: Given: Find: RT PV n P, V, T, R n mol 22 . 0 K 98 2 0.0821 L 2 . 3 atm 62 4 1.6 K mol L atm T R V P n atm 62 4 1.6 psi 14.7 atm 1 psi 24.2 K 298 273 C 25 (K) T T
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Practice—Calculate the Volume Occupied by 637 g of SO 2 (MM 64.07) at 6.08 x 10 3 mmHg and –23 °C. Tro's Introductory Chemistry, Chapter 11 107
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1 atm = 760 mmHg T (K) = t (°C) + 273, 1 mol SO 2 = 64.07 g K mol L atm 0.0821 nRT, R PV Practice—Calculate the Volume Occupied by 637 g of SO 2 (MM 64.07) at 6.08 x 10 3 mmHg and –23 °C, Continued . m SO2 = 637 g, P = 6.08 x 10 3 mmHg, t = −23 °C, V , L Solution: Solution Map: Relationships: Given: Find: P nRT V P, n, T, R V L 55 . 2 atm 0 . 80 K 0 5 2 0.0821 mol 2 4 9.9 K mol L atm P T R n V atm 0 . 0 8 mmHg 760 atm 1 mmHg 10 6.08 3 K 0 25 273 C 23 - (K) T T g n g 64.07 mol 1 2 2 2 SO mol 2 4 9.9 g 64.07 SO mol 1 SO g 637
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Practice—Calculate the Density of a Gas at 775 torr and 27 °C if 0.250 moles Weighs 9.988 g.
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