Solution manual for physical chemistry for the life sciences 1st edition by engel

Is the system open or closed if you place the boundary just outside the liquid water?. sys-If the system boundaries are just outside of the liquid water, the system is open because water

Chapter 1: Fundamental Concepts of Thermodynamics

Questions on Concepts

Q1.1) The location of the boundary between the system and the surroundings is a choice that

must be made by the thermodynamicist Consider a beaker of boiling water in an airtight room

Is the system open or closed if you place the boundary just outside the liquid water? Is the tem open or closed if you place the boundary just inside the walls of the room?

sys-If the system boundaries are just outside of the liquid water, the system is open because water can escape from the top surface The system is closed if the boundary is just inside the walls, be-cause the room is airtight

Q1.2) Real walls are never totally adiabatic Order the following walls in increasing order with

respect to their being diathermal: 1-cm-thick concrete, 1-cm-thick vacuum, 1-cm-thick copper, 1-cm-thick cork

1-cm-thick vacuum < 1-cm-thick cork < 1-cm-thick concrete < 1-cm-thick copper

Q1.3) Why is the possibility of exchange of matter or energy appropriate to the variable of

in-terest a necessary condition for equilibrium between two systems?

Equilibrium is a dynamic process in which the rates of two opposing processes are equal ever, if the rate in each direction is zero, no exchange is possible, and the system cannot reach equilibrium

How-Q1.4) At sufficiently high temperatures, the van der Waals equation has the form P ≈ RT/(V m –

b) Note that the attractive part of the potential has no influence in this expression Justify this

behavior using the potential energy diagram of Figure 1.7

In this case, the energy well depth is small compared to the total energy of the particle fore, the particle is unaffected by the attractive part of the potential

There-Q1.5) Parameter a in the van der Waals equation is greater for H2O than for He What does this say about the form of the potential function in Figure 1.7 for the two gases?

This means that the depth of the attractive potential is greater for H2O than for He

( ) ( ) ( )

(0.005kg) (8.314472JK mol ) 723.6K

molkg10 0.083m0.001Pa

101.00R

m

MVp

P1.2) Consider a gas mixture in a 2.00-dm3 flask at 27.0°C For each of the following mixtures,

calculate the partial pressure of each gas, the total pressure, and the composition of the mixture

KJ8.314472kg

0.001V

M

TRmV

TRn

1 3

3

-3 -

1 1

2

2 2

H

(2.00 10 m ) (32.0 10 kgmol ) 3.90 10 Pa

kg300.15mol

KJ8.314472kg

0.001V

M

TRmV

TRn

1 3

3

-3 -

1 1

2

2 2

O

Pa10 57.6pp

molkg10 32.0

kg0.001mol

kg10 2.02

kg0.001

molkg10 2.02

kg0.001100

OmolH

mol

Hmol100

H

%mol

1 3

1

-3 -

1 3

2

kg 0.001 mol

kg 10 2.02

kg 0.001

mol kg 10 32.0

kg 0.001 100

O mol H

mol

O mol 100

O

% mol

1 3

1

-3 -

1 3

-2 2

( ) ( ) ( )

molkg10 28.02m

10 2.00

kg300.15mol

KJ8.314472kg

0.001V

M

TRmV

TRn

1 3

3

-3 -

1 1

2

2 2

N

(2.00 10 m ) (32.0 10 kgmol ) 3.90 10 Pa

kg300.15mol

KJ8.314472kg

0.001V

M

TRmV

TRn

1 3

3

-3 -

1 1

2

2 2

O

Pa10 35.8pp

molkg10 32.0

kg0.001mol

kg10 28.02

kg0.001

molkg10 28.02

kg0.001100

OmolN

mol

Nmol100

N

%mol

1 3

1

-3 -

1 3

2

kg 0.001 mol

kg 10 28.02

kg 0.001

mol kg 10 32.0

kg 0.001 100

O mol N

mol

O mol 100

O

% mol

1 3

1

-3 -

1 3

-2 2

KJ8.314472kg

0.001V

M

TRmV

TRn

1 3

3

-3 -

1 1

3

3 3

NH

(2.00 10 m ) (16.04 10 kgmol ) 7.77 10 Pa

kg300.15mol

KJ8.314472kg

0.001V

M

TRmV

TRn

1 3

3

-3 -

1 1

4

4 4

CH

Pa10 51.1pp

( )

%8.54mol

kg10 17.03

kg0.001mol

kg10 16.04

kg0.001

molkg10 17.03

kg0.001100

CHmolNH

mol

NHmol100

NH

%mol

1 3

1

-3 -

1 3

-4 3

3 3

kg 10 17.03

kg 0.001 mol

kg 10 16.04

kg 0.001

mol kg 10 17.03

kg 0.001 100

CH mol NH

mol

CH mol 100

CH

% mol

1 3

1

-3 -

1 3

-4 3

4 4

P1.3) Approximately how many oxygen molecules arrive each second at the mitochondrion of

an active person? The following data are available: oxygen consumption is about 40 mL of O2

per minute per kilogram of body weight, measured at T = 300 K and P = 1.0 atm An adult with

a body weight of 64 kg has about 1 × 1012 cells Each cell contains about 800 mitochondria With nRTpV= the number of moles per minute and per kg of body weight is:

3 -5

minmol101.6249min

1molKJ8.314472K

300

m104.0Pa101325T

R

Vp

min1kg64min

mol101.6249

P1.4) In a normal breath, about 0.5 L of air at 1.0 atm and 293 K is inhaled About 25.0% of the

oxygen in air is absorbed by the lungs and passes into the bloodstream For a respiration rate of

18 breaths per minute, how many moles of oxygen per minute are absorbed by the body? sume the mole fraction of oxygen in air is 0.21 Compare this result with Example Problem 1.1

As-We use the ideal gas law to calculate the number of moles of air inhaled every minute:

(2930K) (8.314472JK mol ) 0.0208mol

m105.0Pa101325T

R

Vp

P1.5) Suppose that you measured the product PV of 1 mol of a dilute gas and found that PV =

22.98 L atm at 0.00°C and 31.18 L atm at 100°C Assume that the ideal gas law is valid, with T =

t(°C) + a, and that the value of R is not known Determine R and a from the measurements

n

Vp

2

2 2 1

1 1

+

=+

C2.280 1

atmL31.18

atmL22.98

atmL31.18

atmL22.98C

100-C0

1Vp

Vp

Vp

VpT-Ta

2 2

1 1

2 2

1 1 2 1

atmL22.98a

T

n

Vp

So that on the Kelvin scale:

(8.314472Jmol-1K-1) (293K) 2436.14Jmol-1T

P1.7) A rigid vessel of volume 0.500 m3 containing H2 at 20.5°C and a pressure of 611 × 103 Pa

is connected to a second rigid vessel of volume 0.750 m3 containing Ar at 31.2°C at a pressure of

433 × 103 Pa A valve separating the two vessels is opened and both are cooled to a temperature

of 14.5°C What is the final pressure in the vessels?

We need to first calculate the number of moles of H2 and Ar using the ideal gas law:

(273.15K 20.5K) (8.314472JK mol ) 125mol

m.5000Pa10611T

R

Vp

×+

R

Vp

×+

×

×

=

Using the final number of moles, the final pressure at 14.5ºC is then:

molKJ8.314472K

5.14K273.15mol

125mol281V

TRn

3 3

1 1 total

+

×+

×+

=

P1.8) In normal respiration, an adult exhales about 500 L of air per hour The exhaled air is

saturated with water vapor at body temperature T = 310 K At this temperature water vapor in

equilibrium with liquid water has a pressure of P = 0.062 atm Assume water vapor behaves

ide-ally under these conditions What mass of water vapor is exhaled in an hour?

Using the ideal gas law, and solving for m yields:

TRM

mTR

n V

3 -1

1 -3

hkg.02270 h

1molKJ8.314472K

300

m0.5atm

Pa101325atm

0.062mol

kg1018.02T

R

VpM

P1.9) At T = 293 K and at 50.% relative humidity, the pressure of water vapor in equilibrium

with liquid water is 0.0115 atm Using the information in Problem P1.8, determine what mass of

water is inhaled per hour and the net loss of water through respiration per hour

Using the ideal gas law, and solving for m, with the volume from P1.8 yields:

TRM

mTR

n V

3 -1

1 -3

hkg10.1520.5h

1molKJ8.314472K

293

m0.5atm

Pa101325atm

0.0115mol

kg1018.02T

R

VpM

With the results from P1.8, the net loss of water is then:

-1 -1

-3 -1 -2.15 10 kgh 0.0205kghh

kg0.0227loss

P1.10) A compressed cylinder of gas contains 1.50 × 103 g of N2 gas at a pressure of 2.00 × 107

Pa and a temperature of 17.1°C What volume of gas has been released into the atmosphere if the final pressure in the cylinder is 1.80 × 105 Pa? Assume ideal behavior and that the gas tempera-ture is unchanged

Let ni and nf be the initial and final number of moles of N2 in the cylinder, respectively:

f

f i

i

p

TRnp

TRn

kg1028.01

kg1.50p

pn

5 1

3 - i

f i

Pa101325

molKJ8.314472K

290.25mol

0.482 -mol53.55p

RTnn

P1.11) As a result of photosynthesis, 1.0 kg of carbon is fixed per square meter of forest

As-suming air is 0.046% CO2 by weight, what volume of air is required to provide 1.0 kg of fixed

carbon? Assume T = 298 K and P = 1.00 atm Also assume that air is approximately 20.%

oxy-gen and 80.% nitrooxy-gen by weight

We first calculate the number of moles necessary to provide one 1 kg of CO2:

( ) ( ( ) ) (44 10( )kgmol ) 22.722mol

kg1CO

M

COmCO

00046.0

2.0kg

1OM

OmO

8.0kg

1NM

NmN

2 2

2 tot

air

m6.8511

Pa013251

K298mol

KJ8.314472mol

62111.8mol

13586.962mol

22.722

p

TRNnOnCOnp

TRnV

=

×

×+

+

=

++

=

=

P1.12) A balloon filled with 10.50 L of Ar at 18.0°C and 1 atm rises to a height in the

atmos-phere where the pressure is 248 Torr and the temperature is –30.5°C What is the final volume of the balloon?

We first calculate the number of moles of Ar at 1 atm using the ideal gas law:

( ) ( ) ( )

(248Torr) (133.32PaTorr ) 0.0268m 26.8L

molKJ8.314472K

30.5K291.150.4395mol

p

TR

n

1 -

1 1

P1.13) One liter of fully oxygenated blood can carry 0.20 L of O2 measured at T = 273 K and P

= 1.00 atm Calculate the number of moles of O2 carried per liter of blood Hemoglobin, the gen transport protein in blood, has four oxygen-binding sites How many hemoglobin molecules are required to transport the O2 in 1.0 L of fully oxygenated blood?

oxy-With nRTpV= the number of moles of O2 in one liter of fully oxygenated blood is:

m 0.0002 Pa

01325 1 T

R

V p O

1 1

Fi-nally, four binding sites per Hemoglobin molecule have to be considered, so that the number of

O2 molecules required is:

molecules10

1.344 / molecules10

5.377required

O

P1.14) Myoglobin is a protein that stores oxygen in the tissues Unlike hemoglobin, which has

four oxygen-binding sites, myoglobin has only a single oxygen-binding site How many oglobin molecules are required to transport the oxygen absorbed by the blood in Problem 1.13?

my-Since myoglobin can only bind ¼ of the amount of O2 that hemoglobin can bind the number of myoglobin molecules is:

molecules10

5.36 4molecules10

1.344myoglobinmolecules

P1.15) Consider a 20.0-L sample of moist air at 60.°C and 1 atm in which the partial pressure of

water vapor is 0.120 atm Assume that dry air has the composition 78.0 mol % N2, 21.0 mol %

O2, and 1.00 mol % Ar

a What are the mole percentages of each of the gases in the sample?

b The percent relative humidity is defined as %RH = P H

2O P H

2O

* where

P H2O is the partial pressure of water in the sample and P H

2O

* = 0.197 atm is the equilibrium vapor pressure of

water at 60.°C The gas is compressed at 60.°C until the relative humidity is 100% What volume does the mixture contain now?

c What fraction of the water will be condensed if the total pressure of the mixture is

isother-mally increased to 200 atm?

atm 0.88 0.78 100 p

p 100 N

% mol

total

N 2

p 100 O

% mol

total

O 2

p 100 Ar

% mol

p 100 O H

% mol

total

O H 2

V p V

p ′H2O ′ = H2O , where the prime refers to 100% RH

(0.197 atm) 12.2L

L 20.0 atm

0.12 p

V p V

O H

O H

atm 24.0 0.12 atm 200 O H fraction mol

p

24.0 condensed

P1.16) A mixture of 2.50 × 10–3 g of O2, 3.51 × 10–3 mol of N2, and 4.67 × 1020 molecules of

CO is placed into a vessel of volume 3.50 L at 5.20°C

a Calculate the total pressure in the vessel

b Calculate the mole fractions and partial pressures of each gas

a) The pressure in the vessel can be calculated by using the total number of moles:

mol g 32.0

g 10 2.5 O

1 -

-3

mol molecules 10

6.02214

molecules 10

4.67 CO

1 - 23

2885.375

m 0.0035

K 278.35 mol

K J 8.314472 10

3.51 mol 10 7.7547 mol

10 7.8125

V

T R n p

3

1 1 3

4 5

× +

mol 10 7.8125 O

mol 10 3.51 N

mol 10 7.7547 CO

parts per million (ppm) When the CO level increases to 800 ppm, dizziness, nausea, and sciousness occur, followed by death Assuming the partial pressure of oxygen in air at sea level

uncon-is 0.20 atm, what ratio of O2 to CO is fatal?

Converting the partial pressure of O2 in the atmosphere to ppm using

atm 0.2 O

P1.18) A normal adult inhales 0.500 L of air at T = 293 K and 1.00 atm To explore the surface

of the moon, an astronaut requires a 25.0-L breathing tank containing air at a pressure of 200 atm How many breaths can the astronaut take from this tank?

We first need to calculate the number of moles inhaled at 1 atm using the ideal gas law:

m 0.0005 Pa

101325 T

R

V p

n breaths of

R

pV

3 -3

g180.182

nMn

M

glucose glucose

P1.21) A sample of propane (C3H8) is placed in a closed vessel together with an amount of O2

that is 3.00 times the amount needed to completely oxidize the propane to CO2 and H2O at

con-stant temperature Calculate the mole fraction of each component in the resulting mixture after oxidation assuming that the H2O is present as a gas

The reaction we have to consider is:

( )g 5O ( )g 3CO ( )g 4H O( )gH

If m moles of propane are present initially, there must be 15 m moles of O2 After the reaction is complete, there are 3 m moles of CO2, 4 m moles of H2O, and 10 m moles of O2 Therefore:

.1760 m17

m3

m17

m4

m17

m10

xO2 = =

P1.22) Calculate the volume of all gases evolved by the complete oxidation of 0.25 g of the

amino acid alanine (NH2CHCH3COOH) if the products are liquid water, nitrogen gas, and

car-bon dioxide gas and the total pressure is 1.00 atm and T = 310 K

The reaction equation is:

( )s 9O ( )g N ( )g 6H O( )l 6CO ( )gCOOH

CHCHNH

g0.25M

m

1 alanine

alanine

From the reaction equation we know that:

mol101.4194n

( ) ( ) ( )

molKJ8.314472K

310mol

101.4194p

TR

n

mol108.5161n

310mol

108.5161p

TR

0.0361L

0.2166V

V

P1.23) A gas sample is known to be a mixture of ethane and butane A bulb having a 200.0-cm3capacity is filled with the gas to a pressure of 100.0 × 103 Pa at 20.0°C If the weight of the gas

in the bulb is 0.3846 g, what is the mole percent of butane in the mixture?

With nRTpV= the total number of moles of moles of the mixture is:

m 0.0002 Pa

10 100 T

R

V p

1 1

3 -3

ethane butane

ethane

m M

m n

ethane butane butane

butane tot

tot

M M

M m M

m m n

g 30.08 mol

10 8.2055 M

M

m M

n

1 - 1

1 3

butane ethane

tot ethane tot

P1.24) A glass bulb of volume 0.136 L contains 0.7031 g of gas at 759.0 Torr and 99.5°C What

is the molar mass of the gas?

P1.25) The total pressure of a mixture of oxygen and hydrogen is 1.00 atm The mixture is

ig-nited and the water is removed The remaining gas is pure hydrogen and exerts a pressure of

0.400 atm when measured at the same values of T and V as the original mixture What was the

composition of the original mixture in mole percent?

We start by constructing the following table:

2

H 1



 +

n 2 n p

2

H 2

2

2

O O

O O

H O H

O O

H

H 1

n n

n 2 n n

n p

− +

( )

atm 0.400 1

3

1 p

p 1 3

1 x

P1.26) The photosynthetic formation of glucose in spinach leaves via the Calvin cycle involves

the fixation of carbon dioxide with ribulose 1-5 diphosphate C5H8P2O114−(aq) to form

3-phosphoglycerate C3H4PO73−(aq):

C5H8P2O114−(aq)+ H2O(l)+ CO2(g) →2C3H4PO73−(aq)+ 2H+(aq)

If 1.00 L of carbon dioxide at T = 273 K and P = 1.00 atm is fixed by this reaction, what mass of

P1.27) Calculate the pressure exerted by Ar for a molar volume of 1.42 L mol–1 at 300 K using

the van der Waals equation of state The van der Waals parameters a and b for Ar are 1.355 bar

dm6 mol–2 and 0.0320 dm3 mol–1, respectively Is the attractive or repulsive portion of the tial dominant under these conditions?

poten-To determine what portion of the potential is dominant we need to compare the pressure dicted by the van der Waals equation of state with that predicted by the ideal gas law The van der Waals equation of state yields:

mol dm 1.42

mol dm bar 1.355 mol

dm 0.0321 mol

dm 1.42

K 300 mol

K dm bar 10 8.314472 V

a b V

T R

1 3

2 6 1

3 1

3

1 1 3 -2

2 m m

K 300 mol

K dm bar 10 8.314472 V

T R

1 1 3 -2

m

Because pvdW < pideal, the attractive part of the potential dominates

P1.28) Calculate the pressure exerted by benzene for a molar volume of 1.42 L at 790 K using the Redlich–Kwong equation of state:

V m − b−

a T

1

V (V + nb)

The Redlich–Kwong parameters a and b for benzene are 452.0 bar dm6 mol–2 K1/2 and 0.08271

dm3 mol–1, respectively Is the attractive or repulsive portion of the potential dominant under these conditions?

The exerted benzene pressure is calculated using

( ) V (V b)

1 T

a b V

T R p

m m

mol K bar L 452.0

mol L 0.08271 mol

L 1.42

K 790 mol

K bar L 10 8.314 p

1 1

1

2 1/2 - 2

1 1

1 -1 -2