Fall 07 HW5 solution

Fall 07 HW5 solution - Homework Assignment #5 Name_...

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Homework Assignment #5 Name____________________________ Laboratory Section (circle one) MW1.30 MW6.30 TTh7 1. Chapter 5, Problem 19, 132, 23, 27 (What would the volume be at each temperature if the gas was changed to sulfur trioxide), 39, 34, 43, 49, 52, 54, 56, 60 5.19 a) n fixed b) P halved c) P fixed d) T doubled 5.132 a) cylinder B b) cylinder B c) none d) cylinder C e) cylinder D 5.23 V 2 = V 1 x (P 1 /P 2 ) x (T 2 /T 1 ) x (n 2 /n 1 ) a) Since n 2 = 0.5 n 1 , the volume would be decreased by a factor of 2 . b) This corresponds to both P and T remaining constant, so the volume remains constant . c) Since P 2 = 0.25 P 1 and T 2 = 0.25 T 1 , these two effects offset one another and the volume remains constant . 5.27 V 2 = V 1 (T 2 / T 1 ) (P 1 / P 2 ) = ( ) ( ) 273 14 K 745 torr 3.65 L 298 K 367 torr - = 6.4397 = 6.44 L. No change with change of gas. 5.39 d = M P / RT = ( )( ) ( ) ( ) 137.86 g/mol 1.5 atm L • atm 0.0821 273 120 K mol • K + = 6.40905 = 6.4 g/L 5.34 Air is mostly N 2 (28.02 g/mol), O 2 (32.00 g/mol), and argon (39.95 g/mol). These “heavy” gases dominate the density of dry air. Moist air contains H 2 O (18.02 g/mol). The relatively light water molecules lower the density of the moist air. 5.43 M = mRT / PV = ( ) ( ) ( ) ( )( ) 3 L • atm 0.103 g 0.0821 273 22 K 1 mL 760 mmHg mol • K 747 mmHg 63.8 mL 1 atm 10 L - +   = 39.7809 = 39.8 g/mol The molar masses are N 2 = 28 g/mol, Ne = 20 g/mol, and Ar = 40 g/mol Therefore, the gas is Ar .
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Fall 07 HW5 solution - Homework Assignment #5 Name_...

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