1A_exercise_key_10 - 1 Chapter 10 Exercise Key Chapter 10...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 10 Exercise Key 1 Chapter 10 Exercise Key Exercise 10.1 – Using the Ideal Gas Equation: Krypton gas does a better job than argon of slowing the evaporation of the tungsten filament in an incandescent light bulb. Because of its higher cost, however, krypton is used only when longer life is considered to be worth the extra expense. a. How many moles of krypton gas must be added to a 175-mL incandescent light bulb to yield a gas pressure of 117 kPa at 21.6 ° C? ( ) 3 117 kPa 175 mL PV 1 L PV = nRT n = = 8.3145 L kPa RT 10 mL 294.8 K Km o l ⎛⎞ ⎜⎟ ⎝⎠ = 8.35 × 10 3 mol Kr b. What is the volume of an incandescent light bulb that contains 1.196 g of Kr at a pressure of 1.70 atm and a temperature of 97 ° C? V = ? g = 1.196 g T = 97 ° C + 273.15 = 370 K P = 1.70 atm g PV = RT M 0.082058 L • atm 1.196 g Kr 370 K gRT K•mole V = = g PM 1.70 atm 83.80 mole = 0.255 L Kr c. What is the density of krypton gas at 18.2 ° C and 762 mmHg? g V = ? P = 762 mmHg T = 18.2 ° C + 273.15 = 291.4 K g PV = RT M () g 762 mmHg 83.80 g PM 1 atm mole = = L atm V RT 760 mmHg 0.082058 291.4 K K mole = 3.51 g/L Copyright 2004 Mark Bishop
Background image of page 1

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

View Full DocumentRight Arrow Icon
Chapter 10 Exercise Key 2 Exercise 10.2 – Using the Combined Gas Law Equation: A helium weather
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 3

1A_exercise_key_10 - 1 Chapter 10 Exercise Key Chapter 10...

This preview shows document pages 1 - 3. Sign up to view the full document.

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