Lesson_7_-_Example_Solutions-2009

Lesson_7_-_Example_Solutions-2009 - Lesson 7 Example 7-1...

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Unformatted text preview: Lesson 7 Example 7-1 Two grams of He and 2.00 g of H 2 are placed in a 15 L container at 30 o C. What are their partial pressures and the total pressure? Basis: 2.00 g H 2 and 2.00 g of He Data: m He = 2.0 g m H2 = 2.0 g V = 15 L T = 30 o C = 303 K M He = 4 g/mol M H2 = 2 g/mol 1 Lesson 7 ( 29 ( 29 atm 65 . 1 atm 83 . 48 . 2 P P P atm 83 . atm 48 . 2 333 . P y P atm 48 . 2 L 15 K 303 K . mol atm . L 082 . mol 5 . 1 V nRT P 667 . 333 . 1 y 1 y 333 . 5 . 1 5 . n n y mol 5 . 1 . 1 5 . n n n mol 1 g 2 mol 1 g 2 n mol 5 . g 4 mol 1 g 2 n He 2 H He He 2 H 2 H He He 2 H He 2 H He =- =- = = = = = = = =- =- = = = = = + = + = = = = = 2 Lesson 7 Example 7-2 A 2.35 L container of H 2 at 762 mmHg and 24 o C is connected to a 3.17 L container of He at 728 mmHg and 24 o C. After mixing, what is the total pressure, in mmHg, with the temperature remaining at 24 o C? Basis: not needed Data: V H2 = 2.35 L P H2 = 762 mmHg V He = 3.17 L P He = 728 mmHg T = 24 o C 3 Lesson 7 mmHg 47 . 742 L ) 35 . 2 17 . 3 ( ) L 35 . 2 )( mmHg 762 ( ) L 17 . 3 )( mmHg 728 ( P V V P V P V RT RT V P V P V RT ) n n ( V RT n P RT V P n RT V P n T T 2 H 2 H He He T 2 H 2 H He He T 2 H He T T T 2 H 2 H 2 H He He He = + + = + = + = + = = = = 4 Lesson 7 Example 7-3 The chemical composition of air that is exhaled is different from ordinary air. Is the density of expired air greater or less than that of ordinary air at the same temperature and pressure? (Calculate the density for both at 37 o C and 1.0 atm.) Volumetric Analysis (V%) Basis: 1 mol air Data: M N 2 = 28 g/mol M O 2 = 32 g/mol M CO 2 = 44 g/mol 5 4 0 0 .9 0 .9 3 A r 1 8 5 .9 H 2 O 4 4 3 .8 0 .0 3 6 C O 2 3 2 1 5 .2 2 0 .9 5 O 2 2 8 7 4 .2 7 8 .0 8 N 2 M ( g /m o l) E x h a le d a ir A ir 4 0 0 .9 0 .9 3 A r 1 8 5 .9 H 2 O 4 4 3 .83 ....
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Lesson_7_-_Example_Solutions-2009 - Lesson 7 Example 7-1...

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