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Notes on Gases - Survey of Chemistry I | CHEM 1151K, Study notes of Chemistry

Material Type: Notes; Professor: Osborne; Class: Survey of Chemistry I; Subject: Chemistry; University: Gordon College; Term: Spring 2010;

Typology: Study notes

2009/2010

Uploaded on 05/03/2010

ewhaley
ewhaley 🇺🇸

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Chapter 7 Gases
7.6
The Combined Gas Law
General, Organic, and Biological Chemistry
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Chapter 7 Gases

The Combined Gas Law

General, Organic, and Biological Chemistry

Summary of Gas Laws The gas laws can be summarized as follows: General, Organic, and Biological Chemistry 2

A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29 °C. At what temperature (°C) will the helium have a volume of 90.0 mL and a pressure of 3.20 atm ( n constant)? Step 1 Set up data table: Conditions 1 Conditions 2 P 1 = 0.800 atm P 2 = 3.20 atm V 1 = 0.180 L (180 mL) V 2 = 90.0 mL T 1 = 29 °C + 273 = 302 K T 2 =?

Combined Gas Law Calculation

General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.

STEP 2 Solve for T 2 P 1 V 1 = P 2 V 2 T 1 T 2 T 2 = T 1 x P 2 x V 2 P 1 V 1 STEP 3 Substitute values to solve for unknown. T 2 = 302 K x 3.20 atm x 90.0 mL = 604 K 0.800 atm 180. mL T 2 = 604 K  273 = 331 °C

Combined Gas Law Calculation

(continued)

General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.

STEP 1 Set up data table Conditions 1 Conditions 2 T 1 = 308 K T 2 = -95 °C + 273 = 178 K V 1 = 675 mL V 2 =? P 1 = 646 mmHg P 2 = 802 mmHg STEP 2 Solve for V 2 V 2 = V 1 x P 1 x T 2 P 2 T 1 STEP 3 Substitute values to solve for unknown. V 2 = 675 mL x 646 mmHg x 178 K = 314 mL 802 mmHg x 308 K

Solution

General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.