Download Course Solution Set 18-24

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
M
V = " He . If we write the total mass as M = M + M , where
P
He
He
MP = 450 kg. is the mass of the payload plus balloon body, then the force equation is
&M )
B " Mg = Ma # $ air % g % ( He + " ( M P + M He ) % g = ( M P + M He ) % a
' $ He *
!
!
+$ # '
.
" M He -& air ) g * g * a0 = M P ( g + a ) . !
,% # He (
/
!
So the relationship between the desired lift-off acceleration and the required mass of
helium is
!
M He
1
4
3
6
(g + a)
6 ;
= MP 3
3 *# " air & 6
3 +%$ " (' ) 1. 0 g ) a 6
/
2 , He
5
for a desired lift-off acceleration of a = 0.05 m./sec.2 , we will need
M He
1
4
3
6
!
3
6
( 9.81 + 0.05 m.2 )
3
6
sec.
= ( 450 kg.) 3
6 7 72.8 kg.!
3 )+" 1.3 kg. % -+
6
m.3 ' ( 1. 0 9.81 m. ( 0.05 m. 6
3 *$
sec.2
sec.2 6
3 +$$ 0.18 kg. '' +
/
m.3 &
2 ,#
5
The number of moles of helium required is
72.8 kg. "
1000 gm. 1 mole
"
1 kg.
4 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
V He = 18,200 moles " 22.4
!
!
1 m.3
l.
"
# 408 m.3
mole 1000!l.
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
" 261 K % " 100 kPa %
V'
nRT ' P'
"T ' % " %
=
= $ '$ P ' = $
'$
' ( 330 .
V
nRT P
# T & # P' &
# 273 K & # 0.29 kPa &
* 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’ = v0 . 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 = " !v0 2
, which tells us that the water leaves the bottom of the container at a speed v 0 =
(the same as an object in “free-fall”).
2gH
We can also use Bernoulli’s Equation to describe the flow in the falling stream of
! exactly the same
water. Again, the atmospheric pressure acting on the stream is almost
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 = v0 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 v0 then gives
us " # 2gH =
result).
!
1
" [ v( y ) ]2 $ " g y % v( y ) =
2
2g ( H + y ) (still basically a “free-fall”
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' " # R 2 $ v 0 = # [ r( y ) ]2 $ v( y )
" R 2 # ( 2gH )1/ 2 = [ r( y ) ]2 # [ 2g ( H + y ) ]1/ 2 !
!
!
!
!
!
1/ 2
( 2gH )1/ 2
2 $ H '
" [ r( y ) ] = R #
= R &
) !
%H + y (
[ 2g ( H + y ) ]1/ 2
!
# H &1/ 4
" r( y ) = R %
.
(
$H + y'
!
!
!
!
!
2
2
Let us call Y the distance below the bottom of the container at which
r ( Y ) = 0.75 R . We find that
!
" H %1/ 4
r(Y ) = R $
' = 0.75R (
#H + Y &
" %4
H
81
= $ 3' =
#4 &
H +Y
256 !
!
!
!
" Y =
175
H # 2.16 H
81
.
!
!
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 0 T +
1
2
ay T 2
" 0 = Y + 0# T +
1
2
( $ g )T
2
& 2 Y )1/ 2
+
% T = (
' g *
.
This tells us that the distance from the foot of the drum at which the water lands is
$ 2Y '1/ 2
x = V " T = [ 2g ( H # Y ) ]1/ 2 " & ) = 2 [ Y ( H # Y ) ]1/ 2 .
% g (
!
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 #
!!
!
!
d
d
( x2 ) =
( 4Y H " 4Y 2 )
dt
dt
2Y dH
dx
dH
dx
" 2x
= 4Y
# 0 "
= x $
dt
dt
dt
dt
!
!
!
!!!Y is constant
The question at this point is: what is
dH
dt !?
.!
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 " dH = a " V
dt
Since the water level is falling,
#
dH = $& a ') " V
%A(
dt
.
# &
dH
dH
= " % a ( ) V . If we now
< 0 !!, so we will use
$A'
dt
dt
combine all of
! our results, we find that the speed with which the point where the water
lands moves along the floor is
! * $a'
2Y dH
2Y
2Y !
2Y
$ '
$ '
$ '
dx
= x "
=
" ,#& )"V / = #
" &a ) = # $
" & a ) = # & a ) " 2gY
' % A(
% A(
dt
dt
V T + % A( .
T %A(
2Y
&
)
&
%
,
g )(
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 p0 is applied to the soda. By the
equation for hydrostatic pressure, the soda at the bottom, which we will call depth D , is
p = p0 + !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
( p0 +
!gD ) + " ! · 0 2 + !g · 0 ! 0.97p0 + " !V 2 + !gD
$
" !V
2
! p0 " 0.97p0 ! 0.03p0
.
So the soda enters the drinker’s mouth at a speed given by
V2 "
2 # 0.03 p0
2 # 0.03 # 101,300 Pa
m.2
"
"
6.08
$
sec.2
1000 kg.3
m.
% V " 2.5 sec. .
m.
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 "
100 cm.
dV
m.
cm.3
# 0.25 cm.2 " 2.5 sec. "
# 63 sec.
dt
1 m.
.
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
2
= 0 + " !v
.
The water in this tube is incompressible and so it obeys the equation of continuity,
AV = av $
1
A " V = ( 9 A ) " v # v = 9 V . If we now substitute this into our result
from Bernoulli’s Equation, we have
!
p + " !V
2
= 0 + "
" V2 =
! · ( 9V )2 $ p = " ! · ( 81 – 1 ) · V 2
2 $101,300 Pa
p
m.2
=
% 5.07
kg.
40 #
sec.2
40 $ 1000 3
m.
m.
m.
" V # 2.25 sec. " v = 9V # 20.3 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 = 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
= av .
3
%100 cm. (3
m.
cm.3
$5 m.
!
m. # 2.25 sec. " 1.59 #10 sec. # '
* " 15.9 sec.
& 1 m. )
2
.
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 '
n'RT '
n'T '
=
=
PV
n RT
n T . 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'
P T'
=
"
V
P' 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 "
T ' = 305.2 "
13
y K.
11
!
!
26 K
. We can thus write
22 m.
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
13
#
&
& % 305.2 ) 11 y K (
318.5 kPa
P T'
3 #
V' = V "
"
= 0.6 cm. %
.
(
P'
T
$ 102.7 + 9.81y kPa ' %
$
279.2 K
(
'
Halfway up to the surface, at y = 11 m. , the volume of the bubble would be
13
#
&
#
& % 305.2 ) 11 "11 K (
318.5 kPa
V ' = 0.6 cm. %
(
(
279.2 K
$ 102.7 + 9.81 "11 kPa ' %
$
'
3
!
# 318.5 kPa & # 292.2 K &
3
" 0.6 cm.3 %
(%
( " 0.950 cm. ,
$ 210.6 kPa ' $ 279.2 K '
!
while, at the surface ( y = 0 ) , the volume becomes
!
!
# 318.5 kPa & # 305.2 K &
3
V ' " 0.6 cm.3 %
(%
( " 2.03 cm.
$ 102.7 kPa ' $ 279.2 K '
.
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
$ 1 m. '3
3
V = Ah " 2520 cm.3 # &
) " 0.00252 m.
%100 cm. (
The temperature of the gas is T = 15º C. = 288.2 K . From the ideal gas law, we have
!
PV = nRT
" n =
PV
(101,300 Pa) ( 0.00252 m.3 )
#
# 0.107 mole .
J
RT
(8.314 mole $ K ) ( 288.2 K )
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 '
n' T '
=
PV
nT . With
the volume of the chamber and the number of moles of gas fixed ( V’ = V , n’ = n ), this
P'
T'
=
P
T . The pressure in the chamber under these circumstances
" 308.2 K % !
" T '%
' ( 108,300 Pa . To provide this
becomes P' = P $ ' ( 101,300 Pa ) $
#T&
# 288.2 K &
reduces to
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 $ A
%
g
7000
as described in the Problem statement.
N
m.2
& 1 m. )2
$ 140 cm.2 $ (
+
'100 cm. *
% 10.0 kg. ,
9.81 m.2
sec.
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
P'' V ''
n'' T ''
V ''
T ''
=
"
=
P' V
nT '
V
T ' . The volume of the
" T '' %
" 318.2 K %
3
' ( 2520 cm. ) $
' " 2600 cm.3 This volume is
gas at 45º C. will be V ' ' = V $
# T'&
# 308.2 K &
V ''
2600 cm.3
!
"
" 18.6 cm.
V’’ = A · h’’ , so the height of the piston increases to h' ' =
A
140 cm.2
! a function of temperature is given
A general expression
for the height of the chamber as
!
we will need a new comparison ratio,
by
V ''
A h''
T ''
=
=
V
Ah
T'
# T '' &
!
( cm.
" h' ' = 18.0 %
$ 308.2 '
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 '''
n''' T '''
P'''
T '''
=
"
=
P'V
nT '
P'
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 +
the additional weight required to
keep the chamber volume constant
!
( P''' # P' ) $ A
g
.
" T ''' %
P' ' ' = P' $
' !, so we can write
#T' &
We have already determined that
" T''' %
''
# T' &
[ P' $$
M ' ' ' = M!+
( P' ] ) A
g
" P' ) A % " T''' %
= M + $ g ' [ $ ' ( 1] .
#
& # T' &
Upon inserting all of the known quantities into this expression, we obtain
!
#
&
%108,300 Pa " 0.014 m.2 ( # T''' &
M ''' = M + %
(( [ %$ 308.2 K (' ) 1] kg.
9.81 m.2
%
$
'
sec.
# T''' &
" M + 154.6 [ %
( ) 1] kg.
$ 308.2 K '
!
At 45º C., the mass that should be placed atop the piston is
! M ' ' ' " 10.0 + 154.6 [ #% 318.2 K &( ) 1] kg. " 15.0 kg.
$ 308.2 K '
As a check, we find that the gas pressure in the chamber at 45º C. is
$ 318.2 K '
P' " 101,300 Pa # &
) " 111,800 Pa .
% 288.2 K (
!
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 '''
10,500 Pa
=
" M ' ' ' # 15.0 kg.
10.0 kg.
7000 Pa
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
1
2
2
rotor will be I " 3 # 3 m l = m l " ( 0.003 kg.) ( 0.02 m.)
2
= 1.2 #10 $6 kg.- m2 .
The angular speed of the rotor is given by
1 min.
rad.
rev.
rad.
" = 2 # f = 2 # rev. $ 800
$
% 83.8 sec.
min. 60 sec.
!
We estimate the rate at which the mechanical work of the blades goes into heating the
rad.
P ~ 100 I " 3 # 100 (1.2 $ 10 %6 kg. - m2 ) (83.8 sec. ) 3 # 71 W . This
estimates, in turn, the total heat input to the water in the blender to be Q = P " #t
1 cal.
~ 71W " 20 seconds # 1400 J "
# 330 calories !!. This quantity of heat will
4.184 J
!
cause a temperature change in the water given by
!
water as
!
Q = m cW " #T
!
330 cal.
Q
$ #T = m c ~
W
( 710 gm.) ( 1
cal.
)
gm.%ºC.
& 0.5 º C. ,
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 c1
and c2 , the overall specific heat for that object is c = f1c1 + f2c2 = f c1 + ( 1 – f ) c2 ,
where f1 = f is the fraction of the total mass made up of material 1 and f2 = 1 – f is the
fraction made up of material 2 . As regards our object, 64% is copper by mass
( c1 = 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 c2 J/gm.-K .
The calorimetry equation for heat transfer between the object and the water in the
calorimeter is Q = mW cW (T f " T i W ) = " M c (T f " T i ) . 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 Ti = 72.0º C. The
!
calorimeter
contains mw = 240 gm. of water initially at Tiw = 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
J
J
( 240 gm.) ( 4.184 gm."ºC. ) ( 20.0º " 18.0º C.) # " ( 96 gm.) ( 0.246 + 0.36c 2 gm."ºC. ) ( 20.0º " 72.0º C.)
" 2010 J # 1230 + 1800c 2 J " c 2 #
2010 $ 1230 J
J
# 0.433
.
1800 gm.-ºC.
gm.$ºC.
The material chosen in devising this Problem is in fact nickel, which has c ! 0.44 J/gm.-K .
!
!
c) The mass of this sheet of ice in this temperature range is given by
M =
!ice V = !ice · A · & = ( 0.917 gm./cm.3 ) ( A cm.2 ) ( 1.0 cm. ) ! 0.92 A gm. ,
where A is the (unspecified) area of the ice-sheet and & is its thickness. The total
amount of heat required to warm the ice from "5º C. to 0º C. and then to melt the
0º C. ice is
Q = M · cice · ( Tf – Ti ) + M · Lf
! ( 0.92 A gm. ) [ ( 0.49 cal./gm.-ºC. ) ( 0º " [ "5º ] C. ) + 80 cal./gm.-ºC.
! 75.9 A calories
.
At the indicated time of year, sunlight on a clear day in Minneapolis is supplying
energy to the ground at the aveage rate of 300 W/m.2 around the middle of the
daylight hours. We will take it that 60% of this energy actually goes into heating the
ice-sheet: a large fraction simply reflects off the surface without effect and a small
amount passes through the ice to heat the pavement beneath it. So we estimate that the
ice is being heated at the rate
# 1 m. &2
J sec.
1 cal.
cal sec.
P
.
area ~ 0.6 " 300 m.2 " 4.184 J " %$ 100 cm. (' ) 0.006 cm.2
This winter sunlight is thus heating the ice-sheet with a power of
#
cal sec. &
2
! P " %$ 0.006 cm.2 (' ( A cm. ) !!. So the time interval in which the sunlight alone would
melt this ice is given by
Q = P " #t $ #t ~
!
!
75.9 A calories
cal.
0.006 A 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:
Q 1 = mice c ice ( T f ice " T i ice ) = " mwater c water ( T f water " T iwater )
" (115 gm.) ( 0.49
!
cal.
)
gm#ºC.
cal.
( 0º # [ # 8º ] C.) = # ( 685 gm.) (1 gm#ºC. ) ( T f water # 19º C.)
" Q1 # 451 cal. = ($ 685 T 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 c water ( T ' f water " T ' iwater )
" (115 gm.) (80
!
cal.
)
gm.
cal.
= # ( 685 gm.) (1 gm#ºC. ) ( T ' f water # 18.34º C.)
" Q 2 # 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 c water ( T f " T ' i ice ) = " m water c water ( T f " T ' ' iwater )
cal.
cal.
" (115 gm.) (1 gm#ºC. ) ( T f # 0º C.) = # ( 685 gm.) (1 gm#ºC. ) ( T f # 4.91º C.)
" Q 3 # 115T f cal. = ($ 685T f + 3363) cal.
!
!
" Tf #
3363 cal.
# 4.20º C. " Q 3 # 115 $ 4.20 cal. # 483 cal.
115 + 685 cal.
ºC.
!
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 =
dQ
dT
= mc #
. The entropy change in this
T
T
material while passing from an initial temperature Ti to a final temperature Tf is then
"S =
#
Tf
Ti
&T )
T f dT
dQ
T
= mc !
= mc ( ln T ) T f = mc ( ln T f $ ln T i ) = mc % ln ( f + .
i
Ti
T
T
' Ti *
#
Let us assemble a table of the relevant physical quantities for the process described in
this part of the Problem.
!
for the ice:
Q
Ti
Tf
%S
ice warms
from "8º to 0º C.
451 cal.
"8º C. = 265.16 K
0º C. = 273.16 K
(115 gm.) ( 0.49
cal.
gm"K
ice melts at 0º C.
$ 273.16 K '
) # ln &
)
% 265.16 K (
! +1.675 cal./K!
9200 cal.
0º C. = 273.16 K
0º C. = 273.16 K
Q2
9200 cal.
"
T ice
273.16 K
" + 33.680 cal. K
!
!
!
!
melt-water warms
from 0º to 4.20º C.
483 cal.
0º C. = 273.16 K
4.20º C. = 277.36 K
$ 277.36 K '
cal.
(115 gm.) (1 gm"K ) # ln &
)
% 273.16 K (
! +1.755 cal./K
for the calorimeter water:
Q
Ti
Tf
%S
ice warms
from "8º to 0º C.
"451 cal.
ice melts at 0º C.
"9200 cal.
melt-water warms
from 0º to 4.20º C.
"483 cal.
19º C. = 292.16 K
18.34º C. = 291.50 K
18.34º C. = 291.50 K
4.91º C. = 278.07 K
4.91º C. = 278.07 K
4.20º C. = 277.36 K
$ 291.50 K '
cal.
( 685 gm.) (1 gm"K ) # ln &
)
% 292.16 K (
$ 278.07 K '
cal.
( 685 gm.) (1 gm"K ) # ln &
)
% 291.50 K (
! "1.550 cal./K!
( 685 gm.) (1
cal.
gm"K
$ 277.36 K '
) # ln &
)
% 278.07 K (
! "1.751 cal./K
" # 32.309 cal. K
! with a slight excess of decimal
! calculations need to be made
The
places and holding off
!
“rounding” until the very end, in order to be able to see that the net changes in entropy
! in the third stage of this process.
actually are positive, particularly
The total entropy change for the ice is
"Sice # 1.68 + 33.68 + 1.76
cal.
cal.
J
J
# + 37.12
$ 4.184
# + 155.3
K
K
cal.
K
,
while the change for the water in the calorimeter is
!
"Swater # ($ 1.55) + ($ 32.31) + ($ 1.75)
" # 35.61
!!
cal.
K
cal.
J
J
$ 4.184
" #149.0
K
cal.
K
.!
The net change in entropy within the calorimeter due to this process is thus
=
%Sice + %Swater ! +6.3 J/K .
!
%Stot
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 =
QSup
$10,130 cal.
J
J
#
% 4.184
# $ 148.1
T
286.2 K
cal.
K
and the net entropy change of the ice plus the Lake is then
+7.2 J/K .
!
,
%Stot = %Sice + %SSup !
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 V0 , there are n moles of a particular gas (oxygen
or nitrogen); the number of molecules in each chamber is then N = n · NA , where NA
is Avagadro’s Number. Thus, each gas has an initial entropy of S0 = k · nNA · ln V0 ;
the total entropy while the two gases are in their separate chambers is just twice this
amount. Using the relationship between physical constants, kNA = R , we can write
the initial total entropy as 2S0 = 2nR · ln V0 .
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 2V0 . From our
expression for entropy of a gas, we find the final entropy of each gas to be
S’ = k · nNA · ln ( 2V0 ) = nR · ln ( 2V0 ) , 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 ) # 2 nR ln V0 = 2 nR $ [ ln ( 2V0 ) # ln V0 ]
!
# 2V &
= 2 nR " ln % 0 ( = 2 nR " ln 2 .
$ V0 '
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 2V0 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 # 5 or
144
42444
3
20 dice
24 # 23 # 22 # K #1
24!
=
.
4 # 3 # 2 #1
4!
" 24! %
' . The entropy change between
# 4! &
So the entropy of this second state is S2 = k ln $
! and second states of the row of dice is then
the first
"S = S 2
!
+ $ 24 * 23 * 22 * K
- &%
$ 24! ' !
4 * 3* 2 * 1
# S1 = k ln &
#
k
ln
24
=
k
ln
)
4!
%
(
24
-,
* 1 '.
)0
(
0
0/
# 23 " 22 " K "1 &
# 23! &
= k ln %
( = k ln % ( = k "[ ln ( 23!) ) ln ( 4!) ] !
$ 4 " 3 " 2 "1 '
$ 4! '
!
!
!
!
" k #[ ln ( 2.59 #10 22 ) $ 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
# 24 # 24 # K # 24 = 24
number of possible “configurations” is " 3 = 24
144424443
20
!, making the
20 dice
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 ,
pA VA = ( 16 atm. ) ( 1.5 L. ) = 24 L.-atm. , so we also have pD VD = pD · ( 12 L. )
= 24 L.-atm. $ pD = 2 atm. By the same token, we have pB VB = ( 16 atm. ) ( 3.5 L. )
= 56 L.-atm. , so pC VC = pC · ( 12 L. ) = 56 L.-atm. $ pC = 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. #
" TA = TD =
!
1 m.3
101,300 N m.2
#
$ 2431 J
1000 L.
1 atm.
2431 J
1 mole # 8.314
J
mole$K
% 292.4 K ;
pBVB = pCVC = 56 L. " atm. # 5673 J $ T B = T C # 682.3 K .
!
This completes
the information for the first table in the Problem:
!
point
A
B
C
D
p ( atm. )
16
16
4.67
2.00
V ( L. )
1.5
3.5
12
12
T(K)
292.4
682.3
682.3
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,
monatomic gas in this cycle, for which
CV
%U = CV %T . We have a
3
J
= nR
2
K , giving us
3
3
J
nR ( T B $ T A ) % (1 mole) (8.314 mole$K ) ( 682.3 $ 292.4 K ) % 4862 J
2
2
!
3
3
"U C# D = nR ( T D $ T C ) = nR ( T A $ T B ) = $ "U A# B % $ 4862 J .
2
2
"U A# B =
and
!
The next quantity that can be calculated directly is the work done by the
system,
!
final
W =
"
p dV . The easiest of the four processes to evaluate this for is C ! D :
initial
since it is a constant-volume (isochoric) process, dV = 0 , and thus W C" 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
!
VB
B
W A" B =
#
A
p dV = pA
#
!
V
dV = pA $ V VB = pA (VB % VA )
A
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
final
" nRT %
nRT
$
' dV ; with the
law, we have! p =
, making the work integral W =
V
# V &
(
temperature being constant, this yields
!
W = n RT
Vf
"
Vi
initial
$V '
dV
V
= n RT ( ln V ) Vf = n RT (!
ln Vf # ln Vi ) = n RT ln & f ) .
i
V
% Vi (
For our isothermal processes, we obtain
!
#V &
# 12 L. &
J
W B" C = nRT B ln % C ( = (1 mole) (8.314 mole)K ) ( 682.3 K ) ln %
( * 6990 J
$ 3.5 L. '
$ VB '
and
!
#V &
#1.5 L. &
J
W D" A = nRT D ln % A ( = (1 mole) (8.314 mole)K ) ( 292.4 K ) ln %
( * ) 5055 J .
$ 12 L. '
$ VD '
The foregoing now puts us in a position to readily evaluate the heat input to the
system by applying the First Law:
!
QA" B = #U A" B + W A" B = 4862 + 3242 J = 8104 J ;
QB" C = #U B" C + W B" C = 0 + 6990 J = 6990 J ;
!
QC" D = #U C" D + W C" D = $ 4862 + 0 J = $ 4862 J ;
!
QD" A = #U D" A + W D" 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
!
%Unet = 4862 + 0 + ( "4862 ) + 0 = 0 J , as we would expect for the state
variable U ;
Wnet = 3242 + 6990 + 0 + ( "5055 ) = +5177 J , positive work done
by the system;
and
Qnet = 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, Wnet .
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
" =
W net
5177 J
=
# 0.343 . By contract, the ideal or Carnot efficiency is
Qinput
(8104 + 6990 J )
the largest possible efficiency a thermodynamic cycle could have operating in this
temperature range. That value is given by
!
"C = 1 #
T cold
T
292.4 K
= 1 # A $ 1 #
$ 0.571 .
T hot
TB
682.3 K
Finally, we will calculate the entropy changes caused by each process by
final
!
applying
"S =
#
initial
dQ
!!; the little cross-stroke through the “d ” serves as a reminder
T
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
!
1
"S =
T
!
final
#
initial
dQ =
Q
i$f
T
, and so
"SB# C =
"SD# A =
Q
B# C
T
Q
D# A
T
!
=
+6990 J
J
= +10.245
682.3 K
K
=
$5055 J
J
= $ 17.288
292.4 K
K
and
.
For the other two processes, we will proceed from the First Law equation, Q =
$ dQ = dU + dW . Since
%U = CV %T , we can write dU =
!
3
2
%U + W
n R dT for a
monatomic gas. The differential for heat flow for the constant-pressure process A ! B
3
is thus dQ = 2 n R dT + p dV ; by also applying the ideal gas law, we obtain
!
!
dQ
dT
dV
" nRT % +
1 (3
3
' dV - = 2 nR
dS =
=
nR dT + $
+ nR
*
2
T )
T
V
# V & ,
T
!
.
The entropy change for the process A ! B is thus
TB
B
!
"SA# B =
$ dS
=
A
= nR
!
[
3
2
$
3
2
TA
dT
nR
+
T
T
VB
$ nR dV
V
VA
V
" ( ln T ) TB + ( ln V ) VB
A
A
)
] = nR +*
3
2
#T &
#V & ,
" ln % B ( + ln % B ( . !
$T A '
$ VA ' -
!
!
) 3 # 682.3 K &
# 3.5 L. & ,
J
= (1 mole) (8.314 mole"K ) + 2 ln %
( + ln %
(.
$ 292.4 K '
$ 1.5 L. ' *
!
!
" 8.314
J
#
K
[ 23 # 0.8473 + 0.8473 ] " +17.611 KJ
!
For the constant-volume process C ! D ,
3
dW = 0 , so dQ = 2 nR dT + 0 . The
entropy change for this process is then just
!
D
"SC# D =
$
C
TD
%T (
dT
3
3
dS = ! 2 nR
=! 2 nR ln ' D *
T
&T C )
$
TC
# 292.4 K &
3
J
= 2 (1 mole) (8.314 mole"K ) ln %
(
$ 682.3 K '
!
3
J
J
" 2 # 8.314 # ($ 0.8473) " $ 10.567
K
K
!
!
.
.
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
A!B
B!C
C!D
D!A
totals for
cycle
+4862
0
"4862
0
W(J)
+3242
+6990
0
"5055
Q(J)
+8104
+6990
"4862
"5055
0
+5177
+5177
%U ( J )
% S ( J/K )
+17.611
+10.245
-10.567
-17.288
0
-- G.Ruffa
original notes developed during 2007-09
revised: September 2010, May 2011