{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

1A_exercise_key_10

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

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

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 K mol = 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

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

View Full Document