20090124_HW1%20answer - Homework 1 Answer (Due 1/21/2009...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
1 Homework 1 Answer (Due 1/21/2009 Wednesday) P1.2) Consider a gas mixture in a 2.00-dm 3 flask at 27.0°C. For each of the following mixtures, calculate the partial pressure of each gas, the total pressure, and the composition of the mixture in mole percent: a. 1.00 g H 2 and 1.00 g O 2 b. 1.00 g N 2 and 1.00 g O 2 c. 1.00 g CH 4 and 1.00 g NH 3 To calculate the partial pressures we use the ideal gas law: a) () ( ) ( ) ( ) Pa 10 6.18 mol kg 10 2.02 m 10 2.00 K 300.15 mol K J 8.314472 kg 0.001 V M T R m V T R n p 5 1 3 - 3 3 - 1 1 2 2 2 2 × = × × × × × = = = H H H H ( ) ( ) ( ) Pa 10 3.90 mol kg 10 32.0 m 10 2.00 K 300.15 mol K J 8.314472 kg 0.001 V M T R m V T R n p 4 1 3 - 3 3 - 1 1 2 2 2 2 × = × × × × × = = = O O O O Pa 10 57 . 6 p p p 5 total 2 2 × = + = O H % 4.1 9 mol kg 10 32.0 kg 0.001 mol kg 10 2.02 kg 0.001 mol kg 10 2.02 kg 0.001 100 O mol H mol H mol 100 H % mol 1 3 - 1 3 - 1 3 - 2 2 2 2 = × + × × × = + × = % .9 5 mol kg 10 32.0 kg 0.001 mol kg 10 2.02 kg 0.001 mol kg 10 32.0 kg 0.001 100 O mol H mol O mol 100 O % mol 1 3 - 1 3 - 1 3 - 2 2 2 2 = × + × × × = + × =
Background image of page 1

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

View Full DocumentRight Arrow Icon
2 b) () ( ) ( ) ( ) Pa 10 4.45 mol kg 10 28.02 m 10 2.00 K 300.15 mol K J 8.314472 kg 0.001 V M T R m V T R n p 4 1 3 - 3 3 - 1 1 2 2 2 2 × = × × × × × = = = N N N N ( ) ( ) ( ) Pa 10 3.90 mol kg 10 32.0 m 10 2.00 K 300.15 mol K J 8.314472 kg 0.001 V M T R m V T R n p 4 1 3 - 3 3 - 1 1 2 2 2 2 × = × × × × × = = = O O O O Pa 10 35 . 8 p p p 4 total 2 2 × = + = O H % 3.3 5 mol kg 10 32.0 kg 0.001 mol kg 10 28.02 kg 0.001 mol kg 10 28.02 kg 0.001 100 O mol N mol N mol 100 N % mol 1 3 - 1 3 - 1 3 - 2 2 2 2 = × + × × × = + × = % 6.7 4 mol kg 10 32.0 kg 0.001 mol kg 10 28.02 kg 0.001 mol kg 10 32.0 kg 0.001 100 O mol N mol O mol 100 O % mol 1 3 - 1 3 - 1 3 - 2 2 2 2 = × + × × × = + × = c) ( ) ( ) ( ) Pa 10 7.32 mol kg 10 17.03 m 10 2.00 kg 300.15 mol K J 8.314472 kg 0.001 V M T R m V T R n p 4 1 3 - 3 3 - 1 1 3 3 3 3 × = × × × × × = = = NH NH NH NH ( ) ( ) ( ) Pa 10 7.77 mol kg 10 16.04 m 10 2.00 kg 300.15 mol K J 8.314472 kg 0.001 V M T R m V T R n p 4 1 3 - 3 3 - 1 1 4 4 4 4 × = × × × × × = = = CH CH CH CH Pa 10 51 . 1 p p p 5 total 2 2 × = + = O H % 8.5 4 mol kg 10 17.03 kg 0.001 mol kg 10 16.04 kg 0.001 mol kg 10 17.03 kg 0.001 100 CH mol NH mol NH mol 100 NH % mol 1 3 - 1 3 - 1 3 - 4 3 3 3 = × + × × × = + × =
Background image of page 2
3 () % 1.5 5 mol kg 10 17.03 kg 0.001 mol kg 10 16.04 kg 0.001 mol kg 10 16.04 kg 0.001 100 CH mol NH mol CH mol 100 CH % mol 1 3 - 1 3 - 1 3 - 4 3 4 4 = × + × × × = + × = P1.4) In a normal breath, about 0.5 L of air at 1.0 atm and 293 K is inhaled. About 25.0% of the oxygen in air is absorbed by the lungs and passes into the bloodstream. For a respiration rate of 18
Background image of page 3

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

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 12

20090124_HW1%20answer - Homework 1 Answer (Due 1/21/2009...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online