Chemistry: The Molecular Nature of Matter and Change - Standalone book
Chemistry: The Molecular Nature of Matter and Change - Standalone book
7th Edition
ISBN: 9780073511177
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
bartleby

Videos

Question
Book Icon
Chapter 12, Problem 12.123P

(a)

Interpretation Introduction

Interpretation:

The mass of liquid water that condenses inside the bottle is to be calculated.

Concept introduction:

The ideal gas equation can be expressed as follows,

PV=nRT        (1)

Here,

P is the pressure.

V is the volume.

T is the temperature.

n is the mole of the gas.

R is the gas constant.

The conversion factor to convert °C to Kelvin is as follows:

T(K)=T(°C)+273        (2)

(a)

Expert Solution
Check Mark

Answer to Problem 12.123P

The mass of liquid water that condenses inside the bottle is 0.0027g.

Explanation of Solution

Substitute 22°C in equation (2) to calculate the temperature Tinitial in Kelvin as follows:

Tinitial(K)=22°C+273=295 K

Substitute 0°C in equation (2) to calculate the temperature Tfinal in Kelvin as follows:

Tfinal(K)=0°C+273=273 K

The relative humidity is 44% and therefore the initial pressure is calculated as follows:

Pressure=44%(19.8torr)=8.712torr

Rearrange the equation (1) to calculate the initial number of moles.

ninitial=PVRT        (3)

Substitute the value 8.712torr for P, 295 K for T, 0.75L for V and 0.0821 Latm/Kmol for R in the equation (3).

ninitial=(8.712torr)(0.75L)(0.0821 Latm/Kmol)(295 K)(1atm760torr)=0.000354977mol

The formula to calculate initial mass is as follows:

minitial=(ninitial)(MH2O)        (4)

Substitute the value 0.000354977mol for ninitial and 18.02 g/mol for MH2O in the equation (4).

minitial=(0.000354977mol)(18.02 g/mol)=0.0063966877 g

Rearrange the equation (1) to calculate the final number of moles.

nfinal=PVRT        (5)

Substitute the value 4.6torr for P, 273 K for T, 0.75L for V and 0.0821 Latm/Kmol for R in the equation (5).

ninitial=(4.6torr)(0.75L)(0.0821 Latm/Kmol)(273 K)(1atm760torr)=0.000202534mol

The formula to calculate final mass is as follows:

mfinal=(nfinal)(MH2O)        (6)

Substitute the value 0.000202534mol for nfinal and 18.02 g/mol for MH2O in the equation (6).

mfinal=(0.000202534mol)(18.02 g/mol)=0.0036470057 g 

The formula to calculate the mass of H2O condensed is as follows:

Mass of H2O condensed=minitialmfinal        (7)

Substitute the value 0.0063966877 g for minitial and 0.0036470057 g  for mfinal in the equation (7).

Mass of H2O condensed=0.0063966877 g0.0036470057 g =0.002749682g0.0027g

Conclusion

The mass of liquid water that condenses inside the bottle is 0.0027g.

(b)

Interpretation Introduction

Interpretation:

Whether the liquid water condenses at 10°C is to be determined.

Concept introduction:

The ideal gas equation can be expressed as follows,

PV=nRT        (1)

Here,

P is the pressure.

V is the volume.

T is the temperature.

n is the mole of the gas.

R is the gas constant.

The conversion factor to convert °C to Kelvin is as follows:

T(K)=T(°C)+273        (2)

(b)

Expert Solution
Check Mark

Answer to Problem 12.123P

The liquid water cannot condense at 10°C.

Explanation of Solution

Substitute 10°C in equation (2) to calculate the temperature T in Kelvin as follows:

T(K)=10°C+273=283 K

Rearrange the equation (1) to calculate the number of moles at 10°C.

n=PVRT        (3)

Substitute the value 9.2torr for P, 283 K for T, 0.75L for V and 0.0821 Latm/Kmol for R in the equation (3).

n=(9.2torr)(0.75L)(0.0821 Latm/Kmol)(283 K)(1atm760torr)=0.00040506964mol

The formula to calculate mass at 10°C is as follows:

m=(n)(MH2O)        (4)

Substitute the value 0.00040506964mol for n and 18.02 g/mol for MH2O in the equation (4).

m=(0.00040506964mol)(18.02 g/mol)=0.007041427 g

Mass of water in equilibrium with vapor state is 0.007041427 g and the initial mass is 0.0063 g. Therefore the liquid water cannot condense at 10°C.

Conclusion

The liquid water cannot condense at 10°C.

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 12 Solutions

Chemistry: The Molecular Nature of Matter and Change - Standalone book

Ch. 12.6 - For each of the following crystalline solids,...Ch. 12.6 - Prob. 12.6BFPCh. 12.6 - Prob. 12.7AFPCh. 12.6 - Iron crystallizes in a body-centered cubic...Ch. 12.6 - Prob. 12.8AFPCh. 12.6 - Prob. 12.8BFPCh. 12.6 - Prob. B12.1PCh. 12.6 - Prob. B12.2PCh. 12 - Prob. 12.1PCh. 12 - Prob. 12.2PCh. 12 - Prob. 12.3PCh. 12 - Prob. 12.4PCh. 12 - Prob. 12.5PCh. 12 - Prob. 12.6PCh. 12 - Prob. 12.7PCh. 12 - Name the phase change in each of these events: (a)...Ch. 12 - Prob. 12.9PCh. 12 - Many heat-sensitive and oxygen-sensitive solids,...Ch. 12 - Prob. 12.11PCh. 12 - Prob. 12.12PCh. 12 - Prob. 12.13PCh. 12 - Prob. 12.14PCh. 12 - Prob. 12.15PCh. 12 - Prob. 12.16PCh. 12 - Prob. 12.17PCh. 12 - Prob. 12.18PCh. 12 - From the data below, calculate the total heat (in...Ch. 12 - Prob. 12.20PCh. 12 - Prob. 12.21PCh. 12 - Prob. 12.22PCh. 12 - Prob. 12.23PCh. 12 - Prob. 12.24PCh. 12 - Prob. 12.25PCh. 12 - Prob. 12.26PCh. 12 - Prob. 12.27PCh. 12 - Prob. 12.28PCh. 12 - Prob. 12.29PCh. 12 - Prob. 12.30PCh. 12 - Use Figure 12.10 to answer the following: Carbon...Ch. 12 - Prob. 12.32PCh. 12 - Prob. 12.33PCh. 12 - Prob. 12.34PCh. 12 - Prob. 12.35PCh. 12 - Prob. 12.36PCh. 12 - Distinguish between polarizability and polarity....Ch. 12 - Prob. 12.38PCh. 12 - Prob. 12.39PCh. 12 - Prob. 12.40PCh. 12 - Prob. 12.41PCh. 12 - Prob. 12.42PCh. 12 - Prob. 12.43PCh. 12 - Prob. 12.44PCh. 12 - Prob. 12.45PCh. 12 - Prob. 12.46PCh. 12 - Prob. 12.47PCh. 12 - Prob. 12.48PCh. 12 - Prob. 12.49PCh. 12 - Which liquid in each pair has the lower vapor...Ch. 12 - Which substance has the lower boiling point?...Ch. 12 - Which substance has the higher boiling point?...Ch. 12 - Prob. 12.53PCh. 12 - Prob. 12.54PCh. 12 - Prob. 12.55PCh. 12 - Prob. 12.56PCh. 12 - Why does the antifreeze ingredient ethylene glycol...Ch. 12 - Prob. 12.58PCh. 12 - Prob. 12.59PCh. 12 - Why does an aqueous solution of ethanol (CH3CH2OH)...Ch. 12 - Prob. 12.61PCh. 12 - Prob. 12.62PCh. 12 - Prob. 12.63PCh. 12 - Prob. 12.64PCh. 12 - Prob. 12.65PCh. 12 - Prob. 12.66PCh. 12 - Prob. 12.67PCh. 12 - Prob. 12.68PCh. 12 - Prob. 12.69PCh. 12 - Prob. 12.70PCh. 12 - Prob. 12.71PCh. 12 - Prob. 12.72PCh. 12 - Prob. 12.73PCh. 12 - Prob. 12.74PCh. 12 - Prob. 12.75PCh. 12 - Prob. 12.76PCh. 12 - Prob. 12.77PCh. 12 - Prob. 12.78PCh. 12 - Prob. 12.79PCh. 12 - Prob. 12.80PCh. 12 - Prob. 12.81PCh. 12 - Prob. 12.82PCh. 12 - Prob. 12.83PCh. 12 - Prob. 12.84PCh. 12 - Besides the type of unit cell, what information is...Ch. 12 - What type of unit cell does each metal use in its...Ch. 12 - What is the number of atoms per unit cell for each...Ch. 12 - Calcium crystallizes in a cubic closest packed...Ch. 12 - Chromium adopts the body-centered cubic unit cell...Ch. 12 - Prob. 12.90PCh. 12 - Prob. 12.91PCh. 12 - Prob. 12.92PCh. 12 - Prob. 12.93PCh. 12 - Prob. 12.94PCh. 12 - Prob. 12.95PCh. 12 - Prob. 12.96PCh. 12 - Prob. 12.97PCh. 12 - Prob. 12.98PCh. 12 - Prob. 12.99PCh. 12 - Prob. 12.100PCh. 12 - Prob. 12.101PCh. 12 - Prob. 12.102PCh. 12 - Prob. 12.103PCh. 12 - Polonium, the Period 6 member of Group 6A(16), is...Ch. 12 - Prob. 12.105PCh. 12 - Prob. 12.106PCh. 12 - Prob. 12.107PCh. 12 - Prob. 12.108PCh. 12 - Prob. 12.109PCh. 12 - Prob. 12.110PCh. 12 - Prob. 12.111PCh. 12 - Prob. 12.112PCh. 12 - Prob. 12.113PCh. 12 - Prob. 12.114PCh. 12 - Prob. 12.115PCh. 12 - Prob. 12.116PCh. 12 - Prob. 12.117PCh. 12 - Prob. 12.118PCh. 12 - Prob. 12.119PCh. 12 - Prob. 12.120PCh. 12 - Prob. 12.121PCh. 12 - Prob. 12.122PCh. 12 - Prob. 12.123PCh. 12 - Prob. 12.124PCh. 12 - Prob. 12.125PCh. 12 - Prob. 12.126PCh. 12 - Bismuth is used to calibrate instruments employed...Ch. 12 - Prob. 12.128PCh. 12 - Prob. 12.129PCh. 12 - Prob. 12.130PCh. 12 - Prob. 12.131PCh. 12 - Prob. 12.132PCh. 12 - Prob. 12.133PCh. 12 - Prob. 12.134PCh. 12 - Prob. 12.135PCh. 12 - Prob. 12.136PCh. 12 - Prob. 12.137PCh. 12 - Prob. 12.138PCh. 12 - Prob. 12.139PCh. 12 - Prob. 12.140PCh. 12 - Prob. 12.141PCh. 12 - Prob. 12.142PCh. 12 - Prob. 12.143PCh. 12 - Prob. 12.144PCh. 12 - Prob. 12.145PCh. 12 - The crystal structure of sodium is based on the...Ch. 12 - Prob. 12.147PCh. 12 - One way of purifying gaseous H2 is to pass it...
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
Viscosity, Cohesive and Adhesive Forces, Surface Tension, and Capillary Action; Author: Professor Dave Explains;https://www.youtube.com/watch?v=P_jQ1B9UwpU;License: Standard YouTube License, CC-BY