Solutions for Physics 1201 Course Review (Problems 18 through 24) 18) In order to raise the balloon, the buoyancy force of the helium, B , must exceed the weight, Mg , of the equipment, balloon body, and the helium itself. For lifting a balloon through the air, the buoyancy force will equal the weight of a volume of air equal to the volume of the balloon, V ; hence, B = !

air gV , where the volume of this balloon can be

treated as

!

V =MHe"He

. If we write the total mass as M = MP + M

He , where

MP = 450 kg. is the mass of the payload plus balloon body, then the force equation is

!

B " Mg = Ma # $air % g %MHe$He

&

' (

)

* + " (MP + MHe ) % g = (MP + MHe ) % a

!

" MHe#air#He

$

% &

'

( ) g * g * a

+

, -

.

/ 0 = MP (g + a ) .!

!So the relationship between the desired lift-off acceleration and the required mass of helium is

!

MHe = MP(g + a )

"air"He

# $ %

& ' ( ) 1

* + ,

- . / 0 g ) a

1

2

3 3 3 3

4

5

6 6 6 6

;

for a desired lift-off acceleration of a = 0.05 m./sec.2 , we will need

!

MHe = (450 kg.)(9.81+ 0.05 m.

sec.2)

1.3 kg.m.3

0.18 kg.m.3

"

#

$ $ $

%

&

' ' ' ( 1

)

* +

, +

-

. +

/ + 0 9.81 m.

sec.2( 0.05 m.

sec.2

1

2

3 3 3 3 3 3 3

4

5

6 6 6 6 6 6 6

7 72.8 kg.!

The number of moles of helium required is

!

72.8 kg. " 1000 gm.1 kg. "1 mole4 gm.

!

" 18,200 moles . At standard temperature and pressure (STP), one mole of ideal gas has a volume of 22.4 liters, so the volume of helium under these conditions is

!

VHe = 18,200 moles " 22.4 l.mole "

1 m.31000 l. # 408 m.3

At the destination altitude, the ambient temperature is "12º C. ! 261 K and the ambient pressure is 0.29 kPa , as compared with 0º C. ! 273 K and 100 kPa at STP* . From the “ideal gas law”, PV = nRT , we can obtain the volume comparison ratio

!

V 'V =

nRT ' P'nRT P = T '

T" # $

% & ' PP'"

# $

%

& ' = 261 K

273 K"

# $

%

& ' 100 kPa0.29 kPa"

# $

%

& ' ( 330 .

* current IUPAC definition (other definitions for STP exist, so we would need to adjust our solution here accordingly)

The helium expands by a factor of about 330 during the balloon ascent, causing the balloon to attain a volume of V ‘ ! 330 V ! 330 · 408 m.3 ! 134,000 m.3 . This corresponds to the volume of a sphere over 63 meters across (!) . When it is at 40 km. altitude, the balloon will subtend an angle of about 0.09º , nearly a fifth the diameter of the full moon (these exploratory balloons have at times been the cause of UFO reports). 19)

a) Within the container of water, we can apply Bernoulli’s Equation, p + " !v 2 + !gh = p‘ + " !v’ 2 + !gh’ . The atmospheric pressure is essentially the same at the top and the bottom of the container, so p = p’ . The speed of the water flow at the top is very nearly zero, so we will call v ! 0 and v’ = v

0 . We will set h = H at the

top of the container, making the bottom h’ = 0 (so we have made “upward” positively implicitly for the container). We have now reduced the Equation simply to !gH = " !v

0 2

, which tells us that the water leaves the bottom of the container at a speed

!

v0 = 2gH

(the same as an object in “free-fall”). We can also use Bernoulli’s Equation to describe the flow in the falling stream of water. Again, the atmospheric pressure acting on the stream is almost exactly the same everywhere, so we may cancel the pressure terms. We will label the distance below the bottom of the container y , setting y = 0 at the bottom (so we have y increasing “downward” ). Finally, we will use v = v

0 at the bottom of the container and will call the

speed of the water further downward v’ = v ( y ) . Since gravitational potential energy decreases downward, we have !gh = 0 for the term at the bottom of the container and !gh’ = "!gy for this term downstream. This leads us to write

" !v0 2 = " ! [ v ( y ) ] 2 " !gy ; combining this with our earlier result for v

0 then gives

us

!

" # 2gH = 12 " [ v( y ) ]

2 $ "g y % v( y ) = 2g (H + y ) (still basically a “free-fall”

result). Since water is an “incompressible” fluid, we can apply the equation of continuity, which relates the cross-sectional area of a stream to its velocity. (The density of an incompressible fluid remains constant, so since mass is conserved in a flow, volume must be conserved as well. So for a constant flow rate, the mass and volume passing a fixed point in a given time interval is constant.). Consequently, Av = A’ v’ , where A is the

cross-sectional area of the stream. The hole through which the water leaves the container is circular with a radius of R , so A = #R2 ; we are interested in the cross-section of the

stream below the container, A’ = A ( y ) = # [ r ( y ) ]2 . We can now incorporate our results for the stream velocity into the equation of continuity to write

!

Av = A' v' " # R2 $ v0 = # [ r( y ) ]2 $ v( y )

!

" R2 # (2gH )1/ 2 = [ r( y ) ]2 # [ 2g (H + y ) ]1/ 2 !!

! ! !

!

" [ r( y ) ]2 = R2 # (2gH )1/ 2

[ 2g (H + y ) ]1/ 2= R2 H

H + y$

% &

'

( ) 1/ 2

!

!

! ! !

!

" r( y ) = R HH + y#

$ %

&

' ( 1/ 4

.

Let us call Y the distance below the bottom of the container at which r ( Y ) = 0.75 R . We find that

!

r(Y ) = R HH + Y"

# $

%

& ' 1/ 4

= 0.75R ( HH + Y = 3

4" # $ % & ' 4

= 81256 !

!! !

!

" Y = 17581 H # 2.16 H .

b) As with the container in part (a) , atmospheric pressure is essentially equal at the top of the drum and at the puncture, so p = p’ . The speed at which the water moves at the top of the drum is close to zero, while the water escapes through the hole in its side at speed V . With the puncture at a height Y above the floor, Bernoulli’s Equation gives us p + " ! · 0 2 + !gH = p + " !V 2 + !gY $ V 2 = 2g ( H – Y ) .

(continued)

The water leaves the drum horizontally, so its initial vertical speed is zero. Thus, the time T required for the water to fall to the floor is given from the kinematic equation

!

y f = y0 + vy 0T + 12 ay T

2 " 0 = Y + 0 # T + 12 ( $ g )T

2 % T = 2Yg

& ' (

) * + 1/ 2

.

This tells us that the distance from the foot of the drum at which the water lands is

!

x = V " T = [ 2g (H # Y ) ]1/ 2 " 2Yg

$ % &

' ( ) 1/ 2

= 2 [ Y (H # Y ) ]1/ 2 .

We see from this that the predicted horizontal distance that the escaping water can travel is x = 0 for both Y = 0 and Y = H ; the maximum range occurs for Y = " H .

For convenience in differentiating, let us square both sides of this equation. If we then differentiate implicitly with respect to time, we obtain

!

x 2 = 4 [ Y (H " Y ) ] = 4Y H " 4Y 2 # ddt ( x

2 ) = ddt (4Y H " 4Y 2 )

!

" 2x dxdt = 4Y dHdt # 0 " dx

dt =2Yx $ dHdt .!

! ! ! ! ! !!!Y is constant

The question at this point is: what is

!

dHdt !? Water obeys the equation of

continuity, since it maintains a constant density (as discussed in the solution to part (a) ), so the relation between the rate at which the water level in the drum falls and the speed at which the water flows through the puncture is given by

!

A " dHdt = a "V # dH

dt = aA$ % & ' ( ) "V .

Since the water level is falling,

!

dHdt < 0!!, so we will use

!

dHdt = " a

A# $ % & ' ( )V . If we now

combine all of our results, we find that the speed with which the point where the water lands moves along the floor is

!

dxdt =

2Yx " dHdt =

2YV T " # a

A$ % & ' ( ) "V*

+ , - . /

= #2YT " a

A$ % & ' ( ) = #

2Y2Yg

$

%

& &

'

(

) )

" aA$ % & ' ( ) = # a

A$ % & ' ( ) " 2gY ,

which turns out to be a constant! So the rate at which the place where the water hits the floor moves is independent of how far above the puncture the water level is, even though the speed at which the water escapes does depend on this difference.

20)

a) At the top of the cup, atmospheric pressure p

0 is applied to the soda. By the

equation for hydrostatic pressure, the soda at the bottom, which we will call depth D , is p = p

0 + !gD . The pressure being produced at the upper end of the straw is

p‘ = 0.97 p0 . The speed of the liquid entering the straw is very nearly zero; we shall call

the speed at which it leaves the straw V . If we declare the height at the bottom of the straw to be zero, then the height at the upper end is close to D . From Bernoulli’s Equation, we then have ( p

0 + !gD ) + " ! · 0 2 + !g · 0 ! 0.97p

0 + " !V 2 + !gD

$ " !V 2 ! p

0 " 0.97p

0 ! 0.03p

0 .

So the soda enters the drinker’s mouth at a speed given by

!

V 2 "2 # 0.03 p0

$ "2 # 0.03 # 101,300 Pa

1000 kg.m.3

" 6.08 m.2

sec.2% V " 2.5 m.sec. .

With a straw having a cross-sectional area of 0.25 cm.2 , the student can draw out soda

from the cup at a rate of

!

Acs "dVdt # 0.25 cm.2 " 2.5 m.sec. "

100 cm.1 m. # 63 cm.

3

sec. . At this rate, the student could drink a liter of soda in about 16 seconds. However, one cannot drink through a straw quite this fast, since 0.97 atmosphere is the lowest pressure a typical person can produce with their lungs and this is not sustainable for very long.

b) We will once again employ Bernoulli’s Equation. Since, in this situation, the tube is horizontal, we have !gh = !gh’ , that is, there is no change in the gravitational potential energy of the liquid. From the information in the problem, we find p + " !V 2 + !gh = p‘ + " !v 2 + !gh’ $ p + " !V 2 = 0 + " !v 2 . The water in this tube is incompressible and so it obeys the equation of continuity,

AV = av $

!

A "V = ( 19 A ) " v # v = 9V . If we now substitute this into our result

from Bernoulli’s Equation, we have p + " !V 2 = 0 + " ! · ( 9V )2 $ p = " ! · ( 81 – 1 ) · V 2

!

" V 2 =p40# =

2 $101,300 Pa40 $ 1000 kg.

m.3

% 5.07 m.2

sec.2

!

" V # 2.25 m.sec. " v = 9V # 20.3 m.sec. So the water leaves the constriction as a high-speed jet. The rate at which a volume of water flows through this tube is

!

Q = AV = a v . Since the diameter of the wide end is 0.30 cm. , the area of that cross-section is

A = # ( " · 0.003 m. )2 ! 7.07 · 10"6 m.2 . Hence,

!

Q " 7.07 #10$6 m.2 # 2.25 m.sec. " 1.59 #10$5 m.3sec. #

100 cm.1 m.

%

& '

(

) * 3

" 15.9 cm.3

sec. .

21) a) We start from the ideal gas law, PV = nRT ; in view of the available information in the Problem, it will be convenient to re-cast this as a comparison ratio:

!

P'V 'PV =

n'RT 'n RT =

n'T 'nT . It is reasonable to assume that the mass of methane, and

thus the number of moles, in the bubble does not change as it rises, so n = n’ and,

consequently,

!

V 'V = P

P' "T 'T .

Now we will need expressions to describe the ways in which the temperature and pressure of the methane change with depth. We assume in this that the bubble is small enough that it can very rapidly come into thermal equilibrium with the water surrounding it, so we will rely upon the description of the lake water given in the Problem statement. At y = 22 meters depth, the water temperature is 6º C. = 279.2 K , while the surface ( y = 0 ) temperature is 32º C. = 305.2 K . If we assume that the water (and thus the bubble) temperature changes linearly with depth, then the temperature falls at the

constant rate of 26 K in 22 meters, or

!

" 2622Km. . We can thus write

!

T ' = 305.2 " 1311 y K .

The pressure at the surface of the lake is given as 102.7 kPa . Within the body of the lake, the pressure will follow the law for hydrostatic equilibrium:

!

P' = P ( y = 0) + " g y = 102,700 Pa + (1000 kg.m.3

) (9.81 m.sec.2

) ( y m.)

!

= 102.7 + 9.81 y kPa We can use this formula immediately to find the hydrostatic pressure at the bottom of the lake, P = 102.7 + 9.81 · 22 kPa = 318.5 kPa . For a bubble with a volume of 0.6 cm.3 at depth y = 22 m. , its volume as a function of depth in the lake is then

!

V ' = V " PP' "

T 'T = 0.6 cm.3 318.5 kPa

102.7 + 9.81y kPa# $ %

& ' ( 305.2 ) 13

11y K

279.2 K

#

$

% %

&

'

( ( .

Halfway up to the surface, at y = 11 m. , the volume of the bubble would be

!

V ' = 0.6 cm.3 318.5 kPa102.7 + 9.81 "11 kPa# $ %

& ' ( 305.2 ) 13

11"11 K

279.2 K

#

$

% %

&

'

( (

!

" 0.6 cm.3 318.5 kPa210.6 kPa# $ %

& ' ( 292.2 K279.2 K#

$ %

&

' ( " 0.950 cm.3 ,

while, at the surface ( y = 0 ) , the volume becomes

!

V ' " 0.6 cm.3 318.5 kPa102.7 kPa# $ %

& ' ( 305.2 K279.2 K#

$ %

&

' ( " 2.03 cm.3 .

b) The ideal gas law will once again be useful. Subject only to atmospheric pressure, the chamber containing the gas has a cross-section of A = 140 cm.2 and a height of h = 18.0 cm., giving it a volume of

!

V = Ah " 2520 cm.3 # 1 m.100 cm.$

% &

'

( ) 3

" 0.00252 m.3

The temperature of the gas is T = 15º C. = 288.2 K . From the ideal gas law, we have

!

PV = nRT " n =PVRT #

(101,300 Pa) (0.00252 m.3 )(8.314 J

mole $ K ) (288.2 K )# 0.107 mole .

Since we assume the chamber containing this gas does not leak, n does not change in any of the situations described in the remainder of this Problem. As a result, we’ll find that we don’t actually make much use of this value.

(continued)

Let us now look at the situation where the gas temperature is raised to 35º C. ! 308.2 K ; we want to arrange to keep the volume unchanged. If we re-write the ideal

gas law as a comparison ratio, as we did for part (a), we have

!

P'V 'PV =

n' T 'nT . With

the volume of the chamber and the number of moles of gas fixed ( V’ = V , n’ = n ), this

reduces to

!

P'P =

T 'T . The pressure in the chamber under these circumstances

becomes

!

P' = P T 'T

" # $

% & ' ( 101,300 Pa ) 308.2 K

288.2 K"

# $

%

& ' ( 108,300 Pa . To provide this

pressure on the gas without allowing the volume to change, we place a weight on top of the piston closing the chamber, in order to provide the necessary pressure difference %P = P’ " P ! 108,300 " 101,300 ! 7000 Pa . Because pressure is the force delivered per unit area, the additional mass that must be added to the piston is given by

%F = %P · A = Mg

!

" M =#P $ Ag %

7000 Nm.2

$ 140 cm.2 $ 1 m.100 cm.& ' (

) * + 2

9.81 m.sec.2

% 10.0 kg. ,

as described in the Problem statement. Now, with the 10 kg. weight remaining on the piston, we will raise the temperature of the gas to 45º C. = 318.2 K . This will keep the pressure fixed at P’ = 108,300 Pa , so

we will need a new comparison ratio,

!

P''V ''P'V =

n'' T ''nT ' "

V ''V =

T ''T ' . The volume of the

gas at 45º C. will be

!

V ' ' = V T ''T '

"

# $

%

& ' ( 2520 cm.3 ) 318.2 K

308.2 K"

# $

%

& '

!

" 2600 cm.3 This volume is

V’’ = A · h’’ , so the height of the piston increases to

!

h' ' =V ''A "

2600 cm.3140 cm.2

" 18.6 cm.

A general expression for the height of the chamber as a function of temperature is given by

!

V ''V =

A h''A h =

T ''T ' " h' ' = 18.0 T ''

308.2# $ %

& ' ( cm.

Finally, we return to the previous arrangement at 35º C., with V = 2520 cm.3 and P’ = 108,300 Pa. We now wish to keep this volume constant by increasing the pressure on the gas in the chamber through the use of additional weight placed on the piston. The comparison ratio here will be similar to the one used in part (a),

!

P'''V '''P'V =

n''' T '''nT ' "

P'''P' =

T '''T ' . We needed M = 10.0 kg. to hold the volume

constant at 35º C. , for which %F = ( P’ – P ) · A = Mg , so we will require a larger total mass M’’’ to maintain this volume at higher temperature, given by

( %F )’’’ = ( P’’’ – P’ ) · A = M’’’g " Mg

!

" M ' ' ' = M +(P''' # P' ) $ A

g . the additional weight required to keep the chamber volume constant

We have already determined that

!

P' ' ' = P' T '''T '" # $

% & ' !, so we can write

!

M ' ' ' = M +[ P' T '''

T '"

# $ $

%

& ' ' ( P' ] ) A

g = M +P' ) Ag

" # $

% & ' [ T '''

T '" # $

% & ' ( 1] .

Upon inserting all of the known quantities into this expression, we obtain

!

M ' ' ' = M + 108,300 Pa " 0.014 m.2

9.81 m.sec.2

#

$

% % %

&

'

( ( ( [ T '''308.2 K# $ %

& ' ( ) 1] kg.

!

" M + 154.6 [ T '''308.2 K# $ %

& ' ( ) 1] kg.

At 45º C., the mass that should be placed atop the piston is

!

M ' ' ' " 10.0 + 154.6 [ 318.2 K308.2 K# $ %

& ' ( ) 1] kg. " 15.0 kg.

As a check, we find that the gas pressure in the chamber at 45º C. is

!

P' " 101,300 Pa # 318.2 K288.2 K$

% &

'

( ) " 111,800 Pa .

The excess in the chamber above atmosphere pressure is 111,800 – 101,300 Pa ! 10,500 Pa , which must be countered by downward pressure on the piston supplied by additional mass placed atop the piston. The earlier 10.0 kg. sufficed to provide a needed 7000 Pa , so the total mass now required to be added to the piston is given by

!

M '''10.0 kg. =

10,500 Pa7000 Pa " M ' ' ' # 15.0 kg.

22)

a) To assess the rate at which the blade rotor of the blender transfers mechanical work to the surrounding water, we will need to calculate the moment of inertia of the blades about the rotation axis (we assume that the core of the rotor contributes very little rotational inertia to the total). We will treat each blade as a rod being rotated about one end (the blades are neither rectangular nor flat, but we will keep matters simple here). As there are three blades, the total moment of inertia of the

rotor will be

!

I " 3 # 13m l2 = m l 2 " (0.003 kg.) (0.02 m.)2 = 1.2 #10$6 kg.- m2 .

The angular speed of the rotor is given by

!

" = 2# f = 2# rad.rev. $ 800rev.min. $

1 min.60 sec. % 83.8

rad.sec.

We estimate the rate at which the mechanical work of the blades goes into heating the

water as

!

P ~ 100 I " 3 # 100 (1.2 $10%6 kg.- m2 ) (83.8 rad.sec. )3 # 71W . This

estimates, in turn, the total heat input to the water in the blender to be

!

Q = P " # t

!

~ 71W " 20 seconds # 1400 J "1 cal.4.184 J # 330 calories !!. This quantity of heat will

cause a temperature change in the water given by

!

Q = mcW " #T $ #T =Q

mcW~ 330 cal.(710 gm.) (1 cal.

gm.%ºC.)& 0.5 ºC. ,

using the facts that 1 ml. of water has a mass of one gram and that 1 K = 1 ºC. absolute. Thus, the final temperature of the water in the blender will be about 12.5 ºC. (this answer is approximate to the degree that our estimate of the amount of heating is approximate). b) For an object made of a combination of materials with specific heat capacities c

1

and c2 , the overall specific heat for that object is c = f

1c

1 + f

2c

2 = f c

1 + ( 1 – f

) c

2 ,

where f1 = f is the fraction of the total mass made up of material 1 and f

2 = 1 – f is the

fraction made up of material 2 . As regards our object, 64% is copper by mass ( c

1 = 0.385 J/gm.-K ) and 36% by mass is an unknown substance; thus, we may write

c = 0.64 · 0.385 + 0.36 c2 ! 0.246 + 0.36 c

2 J/gm.-K .

The calorimetry equation for heat transfer between the object and the water in the calorimeter is

!

Q = mW cW (T f " TiW ) = "Mc (T f " Ti ) . The object has a mass

heat passing into water heat leaving object

M = 96 gm. and is placed into the calorimeter at a temperature of T

i = 72.0º C. The

calorimeter contains mw = 240 gm. of water initially at T

iw = 18.0º C. The final equilibrium

temperature of the object and the water in the calorimeter is Tf = 20.0º C. Inserting this

information into the calorimetry equation gives us

!

(240 gm.) (4.184 Jgm."ºC. ) (20.0º " 18.0ºC.) # " (96 gm.) (0.246 + 0.36c2

Jgm."ºC. ) (20.0º " 72.0ºC.)

!

" 2010 J # 1230 + 1800c2 J " c2 #2010 $ 1230 J1800 gm.-ºC. # 0.433 J

gm.$ºC. .

The material chosen in devising this Problem is in fact nickel, which has c ! 0.44 J/gm.-K .

c) The mass of this sheet of ice in this temperature range is given by M = !

ice V = !

ice · A · & = ( 0.917 gm./cm.3 ) ( A cm.2 ) ( 1.0 cm. ) ! 0.92 A gm. ,

where A is the (unspecified) area of the ice-sheet and & is its thickness. The total amount of heat required to warm the ice from "5º C. to 0º C. and then to melt the 0º C. ice is Q = M · c

ice · ( T

f – T

i ) + M · L

f

! ( 0.92 A gm. ) [ ( 0.49 cal./gm.-ºC. ) ( 0º " [ "5º ] C. ) + 80 cal./gm.-ºC. ! 75.9 A calories . At the indicated time of year, sunlight on a clear day in Minneapolis is supplying energy to the ground at the aveage rate of 300 W/m.2 around the middle of the daylight hours. We will take it that 60% of this energy actually goes into heating the ice-sheet: a large fraction simply reflects off the surface without effect and a small amount passes through the ice to heat the pavement beneath it. So we estimate that the ice is being heated at the rate

!

Parea ~ 0.6 " 300 J sec.

m.2"1 cal.4.184 J "

1 m.100 cm.# $ %

& ' ( 2) 0.006 cal sec.

cm.2.

This winter sunlight is thus heating the ice-sheet with a power of

!

P " 0.006 cal sec.cm.2

#

$ %

&

' ( (A cm.2 ) !!. So the time interval in which the sunlight alone would

melt this ice is given by

!

Q = P " # t $ # t ~ 75.9 A calories0.006 A cal.

sec.

% 13,000 sec. (3.6 hours ) .

Since the warming ice will in fact lose some of this heat to the now cooler surroundings, it will in fact take longer than this amount of time for all of the ice to melt. This is why winter ice layers can be so persistent. 23) a) To find the entropy change in this ice-and-water calorimetry system, we need to know both the amount of heat transferred between the two components and the resulting temperature changes in those components. We’ll first look at the warming of the ice from "8º C. to 0º C. by using the calorimetry equation:

!

Q1 = mice cice (T f ice " Ti ice ) = "mwater cwater (T f water " Tiwater )

!

" (115 gm.) (0.49 cal.gm#ºC. ) ( 0º # [ # 8º ] C.) = # (685 gm.) (1 cal.

gm#ºC. ) (T f water # 19º C.)

!

" Q1 # 451 cal. = ($ 685T f water + 13,015) cal. " T f water # 18.34º C.

Next, we will follow the melting of the ice at 0º C. :

!

Q 2 = mice Lf = "mwater cwater (T ' f water " T ' iwater )

!

" (115 gm.) (80 cal.gm. ) = # (685 gm.) (1 cal.gm#ºC. ) (T ' f water # 18.34º C.)

!

" Q2 # 9200 cal. = ($ 685T f water + 12,563) cal. " T ' f water # 4.91º C. Finally, we consider the melt-water from the ice coming into thermal equilibrium with the rest of the water in the calorimeter:

!

Q 3 = mice cwater (T f " T ' i ice ) = "mwater cwater (T f " T ' ' iwater )

!

" (115 gm.) (1 cal.gm#ºC. ) (T f # 0º C.) = # (685 gm.) (1 cal.

gm#ºC. ) (T f # 4.91º C.)

!

" Q3 # 115T f cal. = ($ 685T f + 3363) cal.

!

" T f #3363 cal.

115+ 685 cal.ºC.

# 4.20º C. " Q3 # 115 $ 4.20 cal. # 483 cal.

We can now evaluate the entropy changes in the ice and the water. For a temperature change in a substance with constant specific heat capacity, we have

Q = mc %T $ dQ = mc dT

!

" dS =dQT

= mc # dTT

. The entropy change in this

material while passing from an initial temperature Ti to a final temperature T

f is then

!

"S =dQTTi

T f# = mc dTTTi

T f# = mc ( lnT ) TiT f = mc ( lnT f $ lnTi ) = mc % ln

T fTi

&

' (

)

* + .

Let us assemble a table of the relevant physical quantities for the process described in this part of the Problem. for the ice:

ice warms from "8º to 0º C.

ice melts at 0º C. melt-water warms from 0º to 4.20º C.

Q 451 cal. 9200 cal. 483 cal.

Ti "8º C. = 265.16 K 0º C. = 273.16 K 0º C. = 273.16 K

Tf 0º C. = 273.16 K 0º C. = 273.16 K 4.20º C. = 277.36 K

%S

!

(115 gm.) (0.49 cal.gm"K ) # ln

273.16 K265.16 K$

% &

'

( )

! +1.675 cal./K!

!

Q 2T ice

"9200 cal.273.16 K

!

" + 33.680 cal. K

!

(115 gm.) (1 cal.gm"K ) # ln

277.36 K273.16 K$

% &

'

( )

! +1.755 cal./K

for the calorimeter water:

ice warms from "8º to 0º C.

ice melts at 0º C. melt-water warms from 0º to 4.20º C.

Q "451 cal. "9200 cal. "483 cal.

Ti 19º C. = 292.16 K 18.34º C. = 291.50 K 4.91º C. = 278.07 K

Tf 18.34º C. = 291.50 K 4.91º C. = 278.07 K 4.20º C. = 277.36 K

%S

!

(685 gm.) (1 cal.gm"K ) # ln

291.50 K292.16 K$

% &

'

( )

! "1.550 cal./K!!

(685 gm.) (1 cal.gm"K ) # ln

278.07 K291.50 K$

% &

'

( )

!

" # 32.309 cal. K

!

(685 gm.) (1 cal.gm"K ) # ln

277.36 K278.07 K$

% &

'

( )

! "1.751 cal./K The calculations need to be made with a slight excess of decimal places and holding off “rounding” until the very end, in order to be able to see that the net changes in entropy actually are positive, particularly in the third stage of this process. The total entropy change for the ice is

!

"Sice # 1.68 + 33.68 + 1.76 cal.K # + 37.12 cal.K $ 4.184 J

cal. # +155.3 JK , while the change for the water in the calorimeter is

!

"Swater # ($1.55) + ($ 32.31) + ($1.75) cal.K

!

" # 35.61 cal.K $ 4.184 Jcal. " #149.0 JK .!

!The net change in entropy within the calorimeter due to this process is thus %S

tot

= %Sice

+ %Swater

! +6.3 J/K . If the ice is instead put into a huge body of water, the behavior of the ice (and the change in its entropy) will be the same as it was in the calorimeter. However, the mass of Lake Superior is so vast by comparison that it can lose the heat to the ice with

only an immeasureable drop in temperature (on the order of 10"15 ºC. ) So the total

heat flow from the Lake is QSup

! ( "451 ) + ( "9200 ) + ( "483 ) ! "10,130 cal. , while its temperature remains at T = 13º C. = 286.2 K . This makes its entropy change

!

"SSup =QSupT #

$10,130 cal.286.2 K % 4.184 J

cal. # $148.1 JK , and the net entropy change of the ice plus the Lake is then %S

tot = %S

ice + %S

Sup !

+7.2 J/K .

b) For this Problem, we need to turn to the defining equation for entropy, S = k ln ' , where ' is the number of possible “configurations” of the particles in a specific “state” of the gas. It is discussed in the course (and so will not be proven here) that for N particles in a volume V , the entropy is given by S = k ln ( V N ) = kN ln V . In each chamber of volume V

0 , there are n moles of a particular gas (oxygen

or nitrogen); the number of molecules in each chamber is then N = n · NA , where N

A

is Avagadro’s Number. Thus, each gas has an initial entropy of S0 = k · nN

A · ln V

0 ;

the total entropy while the two gases are in their separate chambers is just twice this amount. Using the relationship between physical constants, kN

A = R , we can write

the initial total entropy as 2S0 = 2nR · ln V

0 .

When the partition between the chambers is abruptly removed, each gas is now free to expand into the full available volume of the box, which is 2V

0 . From our

expression for entropy of a gas, we find the final entropy of each gas to be S’ = k · nN

A · ln ( 2V

0 ) = nR · ln ( 2V

0 ) , making the final total entropy of the gases in

the box twice this much. The entropy change in the container, once equilibrium is re-established, is then

!

"S = 2S' # 2S0 = 2 nR ln (2V0 ) # 2nR ln V0 = 2nR $ [ ln (2V0 ) # ln V0 ]

!

= 2nR " ln2V0V0

#

$ %

&

' ( = 2 nR " ln 2 .

This determines what is referred to as the “entropy of mixing” in the container. Now suppose both chambers contained n moles of nitrogen each. On one hand, it seems like the answer for the entropy change caused by removing the partition should be the same as the result we just found. On the other hand, how is the situation for the gas molecules physically different when there are 2n moles of nitrogen in the volume 2V

0 of the container both before and after the partition is

taken away? This problem with the definition of entropy was pointed out in 1875 and is known as Gibbs’ Paradox. The resolution comes from the difference in the mechanics of the gas molecules in the two situations. With two distinct gases in the chambers, the collisions among the molecules of each gas are different from the collisions between the molecules of the gases when they mix. In the box containing only nitrogen, there is essentially no difference in how the molecules behave in collisions within each chamber or while mixing, except in the immediate vicinity of the partition. So if all of the

molecules in the box are identical (physicists use the term “indistinguishable”), the “entropy of mixing” is taken to be zero.

c) First, we’ll consider the possible orientations of a single die. Any of the six faces can be on top. Once one face has been selected, for instance the “6”, there are four ways the die can be turned so that a particular face, say the “4”, can be to the left. So altogether, there are 6 · 4 = 24 possible orientations for this die. For the first “state” of the twenty dice in a row, we want all of them to have the same orientation. We have seen that there are 24 possible choice for one die and, once that choice is made, all of the rest of the dice are to be turned the same way. So, for this state, the number of possible “configurations” of the dice is '

1 = 24 , making

the entropy of this state S1 = k ln 24 .

As for the second “state” of the twenty dice, we want arrangements where each die has a different orientation from any of the others. We can tote these up in the following way. This first die in the row may be given any of the 24 possible orientations. Once that is selected, however, there are now only 23 available choices for the second die; one that is taken, there are only 22 choices left for the third die; and so on. The number of “configurations” for the dice in the row is thus

!

"2 = 24 # 23 # 22 # K # 520 dice

1 2 4 4 4 3 4 4 4 or 24 # 23 # 22 # K #1

4 # 3 # 2 #1 =24!4! .

So the entropy of this second state is S2 = k ln

!

24!4!

" # $

% & ' . The entropy change between

the first and second states of the row of dice is then

!

"S = S 2 # S1 = k ln 24!4!

$ % &

' ( ) # k ln 24 = k ln

24 * 23 * 22 * K * 14 * 3 * 2 * 1

$ % & '

( )

24

+

,

- - -

.

/

0 0 0

!

= k ln 23 " 22 " K "14 " 3 " 2 "1

#

$ %

&

' ( = k ln 23!

4!#

$ %

&

' ( = k " [ ln (23!) ) ln (4!) ] !

!! ! !

!

" k # [ ln (2.59 #1022 ) $ ln 24 ] " k # (51.61$ 3.178) " 48.43 k . It is worth mentioning that the second state, with its requirement that each die of the twenty have a unique orientation, is still a long way from “complete disorder”. If we describe a third state simply by permitting each die to have any of its 24 possible orientations, the

number of possible “configurations” is

!

"3 = 24 # 24 # 24 # K # 2420 dice

1 2 4 4 4 3 4 4 4 = 2420 !, making the

entropy of this state S3 = k ln ( 2420 ) = k · 20 ln 24 ! 63.56 k . This is the state with the

largest possible entropy for the row of dice.

24) Before we can analyze the processes in the thermodynamic cycle described in this Problem, we need to establish where each endpoint lies on the p – V diagram. Points A and D lie on an isotherm, as do B and C . From the ideal gas law, pV = nRT , we see that if we have a fixed amount of gas ( n = 1 mole here ), all the points on an isotherm ( T is constant ) will have the same value for the product pV . For point A , p

A V

A = ( 16 atm. ) ( 1.5 L. ) = 24 L.-atm. , so we also have p

D V

D = p

D · ( 12 L. )

= 24 L.-atm. $ pD = 2 atm. By the same token, we have p

B V

B = ( 16 atm. ) ( 3.5 L. )

= 56 L.-atm. , so pC V

C = p

C · ( 12 L. ) = 56 L.-atm. $ p

C = 4.67 atm.

We can also determine the absolute temperature on each isotherm, once we have put pV into SI units:

!

pAVA = pDVD = 24 L." atm. # 1 m.31000 L. #

101,300 N m.21 atm. $ 2431 J

!

" TA = TD =2431 J

1 mole # 8.314 Jmole$K

% 292.4 K ;

!

pBVB = pCVC = 56 L." atm. # 5673 J $ TB = TC # 682.3 K . This completes the information for the first table in the Problem:

point p ( atm. ) V ( L. ) T ( K ) A 16 1.5 292.4 B 16 3.5 682.3 C 4.67 12 682.3 D 2.00 12 292.4

We will now want to use the First Law of Thermodynamics, %U = Q " W , where %U is the change in the internal energy of the system, Q is the heat flow into the system, and W is the work done by the system. (Note: chemists often use W to

denote the work done on the system and write %U = Q + W as a result.) We will start with the internal energy changes due to each process. Processes B ! C and D ! A are isothermal, so for these %U = 0 . For processes A ! B and C ! D ,

which cause a change in the temperature of the system, %U = CV %T . We have a

monatomic gas in this cycle, for which

!

CV = 32 nR

JK , giving us

!

"UA#B = 32 nR (TB $ TA) %

32 (1 mole) (8.314

Jmole$K ) ( 682.3$ 292.4 K) % 4862 J

and

!

"UC#D = 32 nR (TD $ TC) = 3

2 nR (TA $ TB) = $ "UA#B % $ 4862 J . The next quantity that can be calculated directly is the work done by the system,

!

W = p dVinitial

final

" . The easiest of the four processes to evaluate this for is C ! D :

since it is a constant-volume (isochoric) process, dV = 0 , and thus

!

WC"D = 0 !!. The next simplest process for which to calculate work is A ! B : because it occurs at constant pressure (an isobaric process), the work integral becomes

!

WA"B = p dVA

B

# = pA dVVA

VB

# = pA $ V VA

VB = pA (VB % VA )

!

= (16 atm.) (3.5" 1.5 L.) = 32 L." atm. # 3242 J . Lastly, we’ll need to find the work done by an isothermal process. From the ideal gas

law, we have

!

p =nRTV , making the work integral

!

W =nRTV

" # $

% & ' dV

initial

final

( ; with the

temperature being constant, this yields

!

W = nRT dVV

Vi

Vf

" = nRT ( ln V ) ViVf = nRT ( ln Vf # ln Vi ) = nRT ln Vf

Vi

$

% &

'

( ) .

For our isothermal processes, we obtain

!

WB"C = nRTB lnVCVB

#

$ %

&

' ( = (1 mole) (8.314 J

mole)K ) ( 682.3 K) ln12 L.3.5 L.#

$ %

&

' ( * 6990 J

and

!

WD"A = nRTD lnVAVD

#

$ %

&

' ( = (1 mole) (8.314 J

mole)K ) ( 292.4 K) ln1.5 L.12 L.#

$ %

&

' ( * ) 5055 J .

The foregoing now puts us in a position to readily evaluate the heat input to the system by applying the First Law:

!

QA"B = #UA"B + WA"B = 4862 + 3242 J = 8104 J ;

!

QB"C = #UB"C + WB"C = 0 + 6990 J = 6990 J ;

!

QC"D = #UC"D + WC"D = $ 4862 + 0 J = $ 4862 J ;

!

QD"A = #UD"A + WD"A = 0 + ($ 5055) J = $ 5055 J . If we add up each of these thermodynamic quantities for all the processes around the cycle, we see that %U

net = 4862 + 0 + ( "4862 ) + 0 = 0 J , as we would expect for the state

variable U ; W

net = 3242 + 6990 + 0 + ( "5055 ) = +5177 J , positive work done

by the system; and Q

net = 8104 + 6990 + ( "4862 ) + ( "5055 ) = +5177 J , a net heat input

into the system . Some of these results will now be of use to us as we compute the thermodynamic efficiency for this cycle, ( . This efficiency is defined as the ratio of the net work done by the system to the amount of heat put into the system. The sum of the process works we have computed is the total work done by the system, W

net .

The amount of heat put into the system is the sum of only the positive Q terms among the processes (the negative terms, which represent heat output from the cycle, are not counted because they cannot be turned back into fuel or returned to the cycle in a useful way). Therefore, we write the thermodynamic efficiency of our cycle as

!

" =WnetQinput

=5177 J

(8104 + 6990 J ) # 0.343 . By contract, the ideal or Carnot efficiency is

the largest possible efficiency a thermodynamic cycle could have operating in this temperature range. That value is given by

!

"C = 1 #TcoldThot

= 1 #TATB

$ 1 #292.4 K682.3 K $ 0.571 .

Finally, we will calculate the entropy changes caused by each process by

applying

!

"S =d QT

initial

final

# !!; the little cross-stroke through the “d ” serves as a reminder

that the result of this integral using the variable Q depends on the process that takes the system from the initial to the final state. Here, the easiest processes to make this calculation for are the isothermal processes, since T is constant. Thus, we find that

!

"S = 1T d Qinitial

final

# =Qi$ fT , and so

!

"SB#C =QB# CT =

+6990 J682.3 K = +10.245 JK and

!

"SD#A =QD# AT =

$5055 J292.4 K = $17.288 JK .

For the other two processes, we will proceed from the First Law equation, Q = %U + W

$

!

d Q = dU + d W . Since %U = CV %T , we can write

!

dU = 32 nR dT for a

monatomic gas. The differential for heat flow for the constant-pressure process A ! B

is thus

!

d Q = 32 nR dT + p dV ; by also applying the ideal gas law, we obtain

!

dS =d QT

= 1T

32 nR dT + nRT

V" # $

% & ' dV

(

) * +

, - = 3

2 nR dTT + nR dV

V .

The entropy change for the process A ! B is thus

!

"SA#B = dSA

B

$ = 32 nR

dTT

TA

TB

$ + nR dVV

VA

VB

$

!

= nR 32 " ( lnT ) TA

TB + ( ln V ) VAVB[ ] = nR 3

2 " lnTBTA

#

$ %

&

' ( + ln VB

VA

#

$ %

&

' (

)

* +

,

- . !

!

! !

!

= (1 mole) (8.314 Jmole"K )

32 ln

682.3 K292.4 K#

$ %

&

' ( + ln 3.5 L.

1.5 L.#

$ %

&

' (

)

* +

,

- .

!

" 8.314 JK # 32 # 0.8473 + 0.8473[ ] " +17.611 JK .

For the constant-volume process C ! D ,

!

d W = 0 , so

!

d Q = 32 nR dT + 0 . The

entropy change for this process is then just

!

"SC#D = dSC

D

$ = 32 nR

dTT

TC

TD

$ = 32 nR ln

TDTC

%

& '

(

) *

!

= 32 (1 mole) (8.314

Jmole"K ) ln

292.4 K682.3 K#

$ %

&

' (

!

" 32 # 8.314

JK # ($ 0.8473) " $10.567 JK .

We note here that %Snet

= 17.611 + 10.245 + ( "10.567 ) + ( "17.288 ) J/K = 0 for the cycle (allowing for round-off error), again as we would expect, since S is a state variable. (A careful examination of the abstract thermodynamic terms for %S shows that they all cancel out exactly for the cycle.) This completes the information compiled in the second table for this Problem:

Process %U ( J ) W ( J ) Q ( J ) %S ( J/K ) A ! B +4862 +3242 +8104 +17.611 B ! C 0 +6990 +6990 +10.245 C ! D "4862 0 "4862 -10.567 D ! A 0 "5055 "5055 -17.288

totals for

cycle

0

+5177

+5177

0

-- G.Ruffa

original notes developed during 2007-09

revised: September 2010, May 2011