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 12.137P
Interpretation Introduction

Interpretation:

The concentration of amphetamine when it is in contact with 20°C air is to be calculated.

Concept introduction:

Clausius-Clapeyron equation is used to find the vapor pressure at one temperature when vapor pressure at another temperature and heat of enthalpy is given. It determines the change in vapor pressure with a change in temperature. The expression of the Clausius-Clapeyron equation is as follows:

ln(P2P1)=ΔHvapR(1T21T1)        (1)

Here,

P1 and P2 are the vapor pressure.

T1 and T2 are the temperature.

R is the gas constant.

ΔHvap is the heat of vaporization.

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

T(K)=T(°C)+273.2        (2)

The ideal gas equation can be expressed as follows,

PV=nRT        (3)

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 1L to 1m3 is as follows:

1L=103m3

Expert Solution & Answer
Check Mark

Answer to Problem 12.137P

The concentration of amphetamine when it is in contact with 20°C air is 2.9g/m3.

Explanation of Solution

Substitute 201°C in equation (2) to calculate the temperature T1 in Kelvin as follows:

T1(K)=201°C+273=474 K

Substitute 83°C in equation (2) to calculate the temperature T2 in Kelvin as follows:

T2(K)=83°C+273=356 K

Rearrange the equation (1) to calculate ΔHvap.

ΔHvap=(R)ln(P2P1)(1T21T1)        (4)

Substitute 8.314 J/Kmol for R, 474 K for T1, 356 K for T2 and 760torr for P1 and 13torr for P2 in the equation (4).

ΔHvap=(8.314 J/Kmol)ln(13torr760torr)(1356 K1474 K)=48,370.199 J/mol

Substitute 20°C in equation (2) to calculate the temperature T3 in Kelvin as follows:

T3(K)=20°C+273=293 K

The expression to calculate P3 is as follows:

ln(P3P2)=ΔHvapR(1T31T2)        (5)

Substitute 8.314 J/Kmol for R, 293 K for T3, 356 K for T2 and 48,370.199 J/mol for ΔHvap in the equation (5).

ln(P3P2)=(48,370.199 J/mol)8.314 J/Kmol(1293 K1356 K)=3.5139112

Take the exponential of 3.5139112 and substitute 13torr for P2 as follows:

P313torr=e3.51391124P313torr=0.0297802P3=0.3871426torr(1atm760torr)5.09398×104atm

Rearrange the equation (3) to calculate the number of moles of amphetamine.

n=PVRT        (6)

Substitute the value 5.09398×104atm for P, 293 K for T, 1m3 for V and 0.0821 Latm/Kmol for R in the equation (6).

n=(5.09398×104atm)(1m3)(0.0821 Latm/Kmol)(293 K)(1L103m3)=0.02117612mol

The formula to calculate the mass of amphetamine is as follows:

Mass of amphetamine=(moles ofamphetamine)(molar mass of amphetamine)        (7)

Substitute 0.02117612mol for moles of amphetamine and 135.20g/mol for molar mass of amphetamine in the equation (7).

Mass of amphetamine=(0.02117612mol)(135.20g/mol)=2.8630114g2.9g

Therefore, the mass of amphetamine in 1m3 is 2.9g/m3.

Conclusion

The concentration of amphetamine when it is in contact with 20°C air is 2.9g/m3.

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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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