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

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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 .
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern