Chemistry: The Molecular Nature of Matter and Change
Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781259631757
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 12, Problem 12.136P

(a)

Interpretation Introduction

Interpretation:

The mass of water that must be removed every time the inside air is completely replaced with outside air 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)

The conversion factor to convert m3 to L is as follows:

  1m3=1000L

(a)

Expert Solution
Check Mark

Answer to Problem 12.136P

The mass of water that must be removed is 49.3tons.

Explanation of Solution

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

  T(K)=22°C+273.2=295.2 K

Rearrange the equation (1) to calculate the number of moles at 31torr.

  n1=PVRT        (3)

Substitute the value 31torr for P, 295.2 K for T, 2.4×106m3 for V and 0.0821 Latm/Kmol for R in the equation (3).

  n1=(31torr)(1atm760torr)(2.4×106m3)(1L103m3)(0.0821 Latm/Kmol)(295.2 K)=4,039,244.229mol

The formula to calculate mass at 31torr is as follows:

  m1=(n1)(MH2O)        (4)

Substitute the value 4,039,244.229mol for n1 and 18.02 g/mol for MH2O in the equation (4).

  m1=(4,039,244.229mol)(18.02 g/mol)=72787181 g(1kg1000g)(1ton1000kg)=72.7872tons

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

  n2=PVRT        (5)

Substitute the value 10torr for P, 295.2 K for T, 2.4×106m3 for V and 0.0821 Latm/Kmol for R in the equation (5).

  n2=(10torr)(1atm760torr)(2.4×106m3)(1L103m3)(0.0821 Latm/Kmol)(295.2 K)=1,302,980.7mol

The formula to calculate mass at 10torr is as follows:

  m2=(n2)(MH2O)        (6)

Substitute the value 1,302,980.7mol for n2 and 18.02 g/mol for MH2O in the equation (6).

  m2=(1,302,980.7mol)(18.02 g/mol)=23479712.2 g(1kg1000g)(1ton1000kg)=23.4797tons

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

  Mass of H2O removed=m1m2        (7)

Substitute the value 72.7872tons for m1 and 23.4797tons for m2 in the equation (7).

  Mass of H2O  removed=72.7872tons23.4797tons=49.3075tons49.3tons.

Conclusion

49 metric tons of water must be removed every time the inside air is completely replaced with outside air.

(b)

Interpretation Introduction

Interpretation:

The heat released when 49.3tons of water condenses is to be calculated.

Concept introduction:

The expression to calculate the moles of solute when given mass and molecular mass of compound are given is as follows:

  Moles of compound(mol)=[given massof compound(g)(1moleof compound(mol)molecular mass of compound(g))]

(b)

Expert Solution
Check Mark

Answer to Problem 12.136P

The heat released when 49.3tons of water condenses is 1.11×108kJ.

Explanation of Solution

The heat of condensation for water is 40.7kJ/mol. The formula to calculate the heat released when 49.3tons mass of water condenses is as follows:

  Heat released=(given mass of watermolar mass of water)(heat of condensation for water)        (8)

Substitute the value 49.3tons for given mass of water, 18.02 g/mol for molar mass of water and 40.7kJ/mol for heat of condensation for water in the equation (8).

  Heat released=(49.3tons18.02 g/mol)(40.7kJ/mol)=1.113660×108kJ1.11×108kJ.

Conclusion

Condensation is an exothermic reaction that is heat is released when water is condensed. The heat released when 49.3tons of water condenses is 1.11×108kJ.

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Chapter 12 Solutions

Chemistry: The Molecular Nature of Matter and Change

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...
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