Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 6, Problem 6.186AP
Interpretation Introduction

Interpretation: The amount of sodium azide needed to inflate a 40×40×20cm bag to a pressure of 1.25 atm at a temperature of 20°C and 10°C is to be calculated.

Concept introduction: The ideal gas law states that “the volume of given amount of gas is directly proportional to the number of moles of gas, temperature and inversely proportional to the pressure”.

To determine: The amount of sodium azide needed to inflate a 40×40×20cm bag to a pressure of 1.25 atm at a temperature of 20°C and 10°C .

Expert Solution & Answer
Check Mark

Answer to Problem 6.186AP

Solution

The amount of sodium azide needed to inflate a 40×40×20cm bag to a pressure of 1.25 atm at a temperature of 20°C and 10°C is 10.8×105g_ and 11.1×105g_ respectively.

Explanation of Solution

Explanation

The ideal gas equation is,

PV=nRT (1)

Where,

  • P is the pressure 1.25atm .
  • V is the volume 320,000L .
  • T is the absolute temperature.
  • n is the numbers of moles.
  • R is the universal gas constant (0.08206Latm/(molK)) .

Rearrange the above expression to calculate the number of moles.

n=PVRT (2)

The conversion of volume of gas from centimeters to liters is done as,

V=(40×40×20)cm=32000cm3=320m3

1cubicmeter=1000L320cubicmeter(V)=320cubicmeter×1000L1cubicmeter=320,000L

The conversion of temperature from °CtoK is done as,

K=°C+273.15

Substitute the given value of temperature in °C in the above expression.

K=°C+273.15=20°C+273.15=293.15K293K

Substitute these values of P , T R and V in equation (1).

PV=nRTn=PVRT=1.25atm×320000L0.08206Latm/(molK)×293K=16,636.45mol

The molar mass of NaN3 is calculated by the formula,

MolarmassofNaN3=Na+3N=(1×22.98)g+(3×14.01)g=65.00mol

The conversion of number of moles into grams is done as,

Numberofmolesintograms=Molarmass×Numberofmoles=65.00mol×16,636.45mol=10.8×105g_

Similarly for T=10°C the amount of azide is calculated by the formula,

PV=nRT (1)

Where,

  • P is the pressure 1.25atm .
  • V is the volume 320,000L .
  • T is absolute temperature 283K .
  • n is the numbers of moles.
  • R is universal gas constant 0.08206 .

Rearrange the above expression,

n=PVRT (2)

The conversion of volume of gas in Centimeters to liters

V=(40×40×20)cm=32000cm3=320m3

For T=10°C

The conversion of temperature from °CtoK is done as,

K=°C+273.15

Substitute the given value of temperature in °C in the above expression.

K=°C+273.15=10°C+273.15=283.15K283K

Substitute these values of P , T R and V in the equation (2).

PV=nRTn=PVRT=1.25atm×320000L0.08206Latm/(molK)×283K=17,224.30mol

Number of moles into grams =Molarmass×Numberofmoles=65.00mol×17,224.30mol=11.1×105g_

Therefore 10.8×105g_ and 11.1×105g_ of sodium azide are needed to inflate 40×40×20cm bag to pressure of 1.25atm at temperature of 20°Cand10°C respectively.

Conclusion

The amount of sodium azide needed to inflate a 40×40×20cm bag to a pressure of 1.25 atm at a temperature of 20°C and 10°C is 10.8×105g_ and 11.1×105g_ respectively

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 6 Solutions

Chemistry

Ch. 6.6 - Prob. 11PECh. 6.7 - Prob. 12PECh. 6.7 - Prob. 13PECh. 6.7 - Prob. 14PECh. 6.8 - Prob. 15PECh. 6.8 - Prob. 16PECh. 6 - Prob. 6.1VPCh. 6 - Prob. 6.2VPCh. 6 - Prob. 6.3VPCh. 6 - Prob. 6.4VPCh. 6 - Prob. 6.5VPCh. 6 - Prob. 6.6VPCh. 6 - Prob. 6.7VPCh. 6 - Prob. 6.8VPCh. 6 - Prob. 6.9VPCh. 6 - Prob. 6.10VPCh. 6 - Prob. 6.11VPCh. 6 - Prob. 6.12VPCh. 6 - Prob. 6.13VPCh. 6 - Prob. 6.14VPCh. 6 - Prob. 6.15VPCh. 6 - Prob. 6.16VPCh. 6 - Prob. 6.17VPCh. 6 - Prob. 6.18VPCh. 6 - Prob. 6.19VPCh. 6 - Prob. 6.20VPCh. 6 - Prob. 6.21VPCh. 6 - Prob. 6.22VPCh. 6 - Prob. 6.23QPCh. 6 - Prob. 6.24QPCh. 6 - Prob. 6.25QPCh. 6 - Prob. 6.26QPCh. 6 - Prob. 6.27QPCh. 6 - Prob. 6.28QPCh. 6 - Prob. 6.29QPCh. 6 - Prob. 6.30QPCh. 6 - Prob. 6.31QPCh. 6 - Prob. 6.32QPCh. 6 - Prob. 6.33QPCh. 6 - Prob. 6.34QPCh. 6 - Prob. 6.35QPCh. 6 - Prob. 6.36QPCh. 6 - Prob. 6.37QPCh. 6 - Prob. 6.38QPCh. 6 - Prob. 6.39QPCh. 6 - Prob. 6.40QPCh. 6 - Prob. 6.41QPCh. 6 - Prob. 6.42QPCh. 6 - Prob. 6.43QPCh. 6 - Prob. 6.44QPCh. 6 - Prob. 6.45QPCh. 6 - Prob. 6.46QPCh. 6 - Prob. 6.47QPCh. 6 - Prob. 6.48QPCh. 6 - Prob. 6.49QPCh. 6 - Prob. 6.50QPCh. 6 - Prob. 6.51QPCh. 6 - Prob. 6.52QPCh. 6 - Prob. 6.53QPCh. 6 - Prob. 6.54QPCh. 6 - Prob. 6.55QPCh. 6 - Prob. 6.56QPCh. 6 - Prob. 6.57QPCh. 6 - Prob. 6.58QPCh. 6 - Prob. 6.59QPCh. 6 - Prob. 6.60QPCh. 6 - Prob. 6.61QPCh. 6 - Prob. 6.62QPCh. 6 - Prob. 6.63QPCh. 6 - Prob. 6.64QPCh. 6 - Prob. 6.65QPCh. 6 - Prob. 6.66QPCh. 6 - Prob. 6.67QPCh. 6 - Prob. 6.68QPCh. 6 - Prob. 6.69QPCh. 6 - Prob. 6.70QPCh. 6 - Prob. 6.71QPCh. 6 - Prob. 6.72QPCh. 6 - Prob. 6.73QPCh. 6 - Prob. 6.74QPCh. 6 - Prob. 6.75QPCh. 6 - Prob. 6.76QPCh. 6 - Prob. 6.77QPCh. 6 - Prob. 6.78QPCh. 6 - Prob. 6.79QPCh. 6 - Prob. 6.80QPCh. 6 - Prob. 6.81QPCh. 6 - Prob. 6.82QPCh. 6 - Prob. 6.83QPCh. 6 - Prob. 6.84QPCh. 6 - Prob. 6.85QPCh. 6 - Prob. 6.86QPCh. 6 - Prob. 6.87QPCh. 6 - Prob. 6.88QPCh. 6 - Prob. 6.89QPCh. 6 - Prob. 6.90QPCh. 6 - Prob. 6.91QPCh. 6 - Prob. 6.92QPCh. 6 - Prob. 6.93QPCh. 6 - Prob. 6.94QPCh. 6 - Prob. 6.95QPCh. 6 - Prob. 6.96QPCh. 6 - Prob. 6.97QPCh. 6 - Prob. 6.98QPCh. 6 - Prob. 6.99QPCh. 6 - Prob. 6.100QPCh. 6 - Prob. 6.101QPCh. 6 - Prob. 6.102QPCh. 6 - Prob. 6.103QPCh. 6 - Prob. 6.104QPCh. 6 - Prob. 6.105QPCh. 6 - Prob. 6.106QPCh. 6 - Prob. 6.107QPCh. 6 - Prob. 6.108QPCh. 6 - Prob. 6.109QPCh. 6 - Prob. 6.110QPCh. 6 - Prob. 6.111QPCh. 6 - Prob. 6.112QPCh. 6 - Prob. 6.113QPCh. 6 - Prob. 6.114QPCh. 6 - Prob. 6.115QPCh. 6 - Prob. 6.116QPCh. 6 - Prob. 6.117QPCh. 6 - Prob. 6.118QPCh. 6 - Prob. 6.119QPCh. 6 - Prob. 6.120QPCh. 6 - Prob. 6.121QPCh. 6 - Prob. 6.122QPCh. 6 - Prob. 6.123QPCh. 6 - Prob. 6.124QPCh. 6 - Prob. 6.125QPCh. 6 - Prob. 6.126QPCh. 6 - Prob. 6.127QPCh. 6 - Prob. 6.128QPCh. 6 - Prob. 6.129QPCh. 6 - Prob. 6.130QPCh. 6 - Prob. 6.131QPCh. 6 - Prob. 6.132QPCh. 6 - Prob. 6.133QPCh. 6 - Prob. 6.134QPCh. 6 - Prob. 6.135QPCh. 6 - Prob. 6.136QPCh. 6 - Prob. 6.137QPCh. 6 - Prob. 6.138QPCh. 6 - Prob. 6.139QPCh. 6 - Prob. 6.140QPCh. 6 - Prob. 6.141QPCh. 6 - Prob. 6.142QPCh. 6 - Prob. 6.143QPCh. 6 - Prob. 6.144QPCh. 6 - Prob. 6.145APCh. 6 - Prob. 6.146APCh. 6 - Prob. 6.147APCh. 6 - Prob. 6.148APCh. 6 - Prob. 6.149APCh. 6 - Prob. 6.150APCh. 6 - Prob. 6.151APCh. 6 - Prob. 6.152APCh. 6 - Prob. 6.153APCh. 6 - Prob. 6.154APCh. 6 - Prob. 6.155APCh. 6 - Prob. 6.156APCh. 6 - Prob. 6.157APCh. 6 - Prob. 6.158APCh. 6 - Prob. 6.159APCh. 6 - Prob. 6.160APCh. 6 - Prob. 6.161APCh. 6 - Prob. 6.162APCh. 6 - Prob. 6.163APCh. 6 - Prob. 6.164APCh. 6 - Prob. 6.165APCh. 6 - Prob. 6.166APCh. 6 - Prob. 6.167APCh. 6 - Prob. 6.168APCh. 6 - Prob. 6.169APCh. 6 - Prob. 6.170APCh. 6 - Prob. 6.171APCh. 6 - Prob. 6.172APCh. 6 - Prob. 6.173APCh. 6 - Prob. 6.174APCh. 6 - Prob. 6.175APCh. 6 - Prob. 6.176APCh. 6 - Prob. 6.177APCh. 6 - Prob. 6.178APCh. 6 - Prob. 6.179APCh. 6 - Prob. 6.180APCh. 6 - Prob. 6.181APCh. 6 - Prob. 6.182APCh. 6 - Prob. 6.183APCh. 6 - Prob. 6.184APCh. 6 - Prob. 6.185APCh. 6 - Prob. 6.186APCh. 6 - Prob. 6.187APCh. 6 - Prob. 6.188AP
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Step by Step Stoichiometry Practice Problems | How to Pass ChemistryMole Conversions Made Easy: How to Convert Between Grams and Moles; Author: Ketzbook;https://www.youtube.com/watch?v=b2raanVWU6c;License: Standard YouTube License, CC-BY