Problem Set 1
Solutions
3.20 MIT
Professor Gerbrand Ceder
Fall
2001
Fall 2002
LEVEL 1 PROBLEMS
Problem 1.1
Gas is heating in a rigid container from To = 40C to T1 = 315 C
U = U (T1 ) U (To ) = W + Q
(First Law)
(a) W = 0 Since only PV work is possible & V = 0 since the container is rigid.
Q = U = (0.17T1 + C ) (0.17To + C )
Q = U = 0.17(T1 To ) = 46.75 kJ
kg
(b) Entropy change:
dS =
since ∂W = 0
Q
T
Q = dU = 0.17dT
0.17dT
T
0.17dT
T1
kJ
= 0.17 ln
= 0.107 kgK
To
T
dS =
or,
T1
S =
To
1
(for reversible heating)
Problem 1.2
Calculate entropy change by going from 1 to 2 according to:
(1 → 1 ) A reversible isothermal compression
(1 → 2) Followed by an isobaric heating
Entropy change 1 → 1
U = 0
S 1→1
(U = U (T ))
nRT
Q = W or Q = pdV =
dV
V
nR
Q
dS =
dV
=
T
V
V1
P1 J
= nR ln
= nR ln
= 36.53
V1
P1
mole K
Entropy change 1 → 2
Q = cp dT
7
Cp = n R
2
Q
cp dT
dT
7
dS =
=
= nR
T
T
T
2
T
J
7
2
S 1 →2 = nR ln
= 15.59
T1
mole K
2
Now for the total process
S tot = S 1→1 + S 1 →2 = 20.94
2
J
mole K
(isobaric heating)
Problem 1.3
(a)
H F e =
H F e = 34
912 C
1100 C
cp, dT + H → +
0 C
912 C
cp, dT
J
J
J
(912C 1100 C ) 900
+ 38
(0 C 912C )
mole K
mole
mle K
HF e =-41.95kJ
(b) System Fe+ice water is adiabatic → H = 0
HF e = Hicewater
HF e = (mice H melting ) , mice = mass of ice transformed
mice = mol
41950 moleJK 100g 156
HF e
g
=
j
1mol
H melting
6048 moleK 18g
mice =223g
Problem 1.4
Diatomic ideal gas =⇒ cp = 72 R and cv = 52 R
3
Assume the process is reversible.
1 → 1
There are two methods one can use to find the work done along this path. For both approaches it will be useful to
find p, V, T for the starting and ending points. We are given the starting pressure, volume and temperature and the
final pressure. From this we can calculate the final volume and temperature.
p1 V1 = p1 V1
(with =
cp
cv
= 75 )
(with c =
cv
R
= 52 )
V1 =
p1 V1
p1 V1 = 6.919m3
The final temperature can be found with a similar relation
V1 T1c = V1 T1c
V1 T1c
T1c =
V1
T1 = 137K
It will also be useful to calculate the number of moles from pV = nRT giving n = 615 moles. In summary the
conditions at each point are
p (P a)
V (m3 )
T (K )
1
101300
1
298
Method I
Start with
pV
with =
cp
cv
=
7
5
= p1 V1
p=
So,
1
1519500
6.919
137
p1 V1
V
pdV = p1 V1
w=
w=
P1 V1
( 1)
1
1
V ( 1)
=
1
P1 V1
1
( 1)
1
dV
V
1
( 1)
V1
1
( 1)
V1
Inserting in the values for P, V, we get
w = 2.03M J
Method II
Start with the known fact the the internal energy of an ideal gas is only a function of temperature.
U = Q + W = ncv dT
However, since this is an adiabatic path, Q = 0 and the work done is simple ncv dT
w = ncv (T1 T1 )
w = 2.03M J
4
1 → 2
This process is at constant pressure, so the work is given by
w = pdV = p2
2
dV
1
w = p2 (V2 V1 )
We just have to fine V2 . That can be done easily by using the equation of state of an ideal gas along an isotherm
P2 V2 = P1 V1
V2 = 15m3
So,
w = 101300 P a 15 6.919m3
w = 819kJ
The total work done is then
wtot = 2030 + 819 = -2849kJ
Problem 1.5
Function 1: df1 = y(3x2 + y2 )dx + x(x2 + 2y2 )dy
(1) Integrate along y = x (dy = dx)
1
x(3x2 + x2 )dx + x(x2 + 2x2 )dx
0
1
1
7x3 dx =
(4x3 + 3x3 )dx =
0
0
7x4
4
1
=
0
(2) Integrate along y = x2 (dy = 2xdx)
1
x2 (3x2 + x4 )dx + x(x2 + 2x4 )2xdx
0
1
5x7
(5x + 5x )dx = x +
7
6
0
4
1
5
=
12
7
0
Function 2: df2 = y(3x2 + y)dx + x(x2 + 2y)dy
(1) Integrate along y = x (dy = dx)
1
x(3x2 + y)dx + x(x2 + 2x)dx
0
1
(4x3 + 3x2)dx =
0
4x4
3x3
+
4
3
1
= 2
0
(2) Integrate along y = x2 (dy = 2xdx)
1
x2 (3x2 + x2 )dx + x(x2 + 2x2 )2xdx
0
1
10x4 dx =
0
10x5
5
1
Since function 2 is path independent it is an exact differential.
5
= 2
0
7
4
Problem 1.6
dU = w + Q
This is an ideal gas so dU = 0 along an isothermal path.
(a)
RT
w = pdV , p =
V
1
RT
V2
p2
w=
dV = RT ln
= RT ln
V
V
p1
1
2
w = RT ln(15) = 6754J
(b)
dU = 0 =⇒ Q = w
Q = 6754J
(c)
H1 = U1 + P1 V1
H2 = U2 + P2 V2
For an ideal gas at constant temperature (T1 = T2 ) =⇒ U1 = U2
Also, when T1 = T2 =⇒ P1 V1 = P2 V2
Therefore
H1 = H2
(d)
Q
pdV
nR
dV
=
=
T T
V V2
P2
S = R ln
= R ln
V1
P1
dS =
J
S = 22.5 K
6
Problem 1.7
(a) Work done in the process is:
final
pdV
w=
initial
Define water as the system
Initial State = Liquid
Final State = Vapor
So we write
w=
Vvapor
pdV
Vliq
w = p
Vvapor
dV
(Since p is constant)
Vliq
w = p(V vap V liq )
Since the specific volume of a vapor is much larger than for a liquid
w ≈ pV vap = p
#vapor
This is about 7.5% of the heat of evaportation
=
105 P a
J
= 167504
= 167 Jg
kg
kg
0.592 m3
(b)
H = U + pV
H = U + (pV ) = U + pV
U = H p V
J
J
U = 2261 167 = 2094 Jg
g
g
Conclusion: The heat one has to transfer to water to evaporate it is partly used for increasing the internal energy of
J
water (2094 g,
breaking bonds) but aslo for the work required by the vapor expansion.
Problem 1.8
7
Given:
N
p1 = 1 atm = 101300 m
2
V1 = 1 liter = 0.001m3
V2 = 2 liters = 0.002m3
T1 = T2 = 373K
Need to know how many moles of gas:
n=
1 → 2 (isothermal)
w 1→2
p1 V1
= 0.033 moles
RT1
1
N
p1 V1 = p2 V2 =⇒ p2 = atm = 50650 2
2
m
V2
V2
nRT
V2
pdV =
dV = nRT ln
=
V
V1
V1
V1
w 1→2 = nRT ln 2 = 70.9J
2 → 3 (isobaric)
p3 = p2
with =
cp
cv
p3 V3 = p1 V1
=
5
3
for monoatomic ideal gas
1
p1 V1 V3 =
= 1.52liters = 0.00152m3
p3
N 2→3
2→3
w
= p2 V
= 50650 2
0.00152 0.002m3
m
w 2→3 = 24.3J
3 → 1 (reversible adiabatic)
w 3→1 =
V1
pdV
V3
p1 V1
V
V1
p1 V1
p1 V1
1
1
1
dV =
=
1 V1 1
V
V3
V3
pV = p1 V1 =⇒ p =
w 3→1
w 3→1 = 37.1J
So the total work done is
w total = w 1→2 + w 2→3 + w 3→1 = 70.9 + 24.3 + 37.1
w total = 9.5J
Problem 1.9
(i) The amount of work done
by the gas is zero, since the gas does no work on the surrouindings outside of the
chamber. The expression w = pdV cannot be applied since the process is not reversible.
(ii) The walls of the chamber are insulating, thus Q = 0 and from (i) w = 0. Thus U = Q + W = 0
(iii) For an ideal gas U is only a function of temperature. Since for a free expansion in an insulated chamber
U = 0 =⇒ Tinitital = Tfinal
(iv) For a non-ideal gas where U = f (T, V ) , T will change after a free expansion in an insulated chamber since
V changes and U = U (Tfinal , Vfinal ) U (Tinitial , Vinitital ) = 0
8
(Most gasses at low pressures can be well approximated as being ideal gasses)
(v) If now the walls of the chambers conduct heat, for an ideal gas U = U (T ) is still true. The initial and final
states are in equilibrium with the environment. Thus
Tinitial = Tfinal = Tenvironment
(vi) For non-ideal gases U = 0 since U = U (T, V ) and V changes.
LEVEL 2 PROBLEMS
Problem 2.1
(a) S = 0
(b) S > 0
(c) S > 0
Problem 2.2
Given information
Vi = 1m3
Pi = 15 atm
Ti = 298K
Vf =?
Pf = 2 atm
Tf = 298K
(a) Find the final volume after the expansion
Pi Vi = Pf Vf
15 1
Pi Vi
=
Vf =
2
Pf
Vf = 7.5m3
Finding the number of moles, n, will be useful for parts b & c
L atm
, 1m3 = 1000L
R = 0.08205
K mole
15 1000
PV
=n=
0.08205 298
RT
n = 613.5 moles
(b) Find the work done if the process is isothermal
w=
w = nRT ln
Vf
Vi
pdV
= 8.314 613.5 298 ln
w = 3006 kJ
(c) Find the work in the multi-step process
9
7. 5
1
Two parts:
(I) Adiabatic expansion from 1 to 2
(II) Heating at constant P from 2 to 2
(wI + wII ) + (QI + QII ) = Usys
Since state 1 and 2 are at the same temperature and U for an ideal gas is only a function of temperature, Usys = 0
and
(wI + wII ) = (QI + QII )
So we can calculate w ’s or Q’s. Since QI = 0 (adiabatic process) let’s calculat QII
Need to find T2
QII = ncp T = ncp (T2 T2 )
7
J
n = 613.5 moles, cp = R = 29
2
mole K
T2 = T1
T
2
= 298
2
15
P2 P1
27
cR
p
= 168K
Now for QII
QII = 613.3 moles 29
J
(298 168K )
mole K
QII = 2313kJ
So,
wtotal = (QI + QII ) = (0 + 2313kJ )
wtotal = 2313kJ
Problem 2.3
U = Ap2 V
dU = w + Q
Along a reverisble path, ∂W = pdV when only p-V work is possible.
Along a reversible adiabatic path, ∂Q = 0 and therefore
dU = w
10
(A = positive constant)
(First Law)
∂U
∂U
dp +
dV
∂p V
∂V p
∂U
= 2ApV
∂p V
∂U
= Ap2
∂V p
dU =
Going back to equation 1
(2ApV )dp + (Ap2 )dV = pdV
2AV dp = (Ap + 1)dV
dp
dV
=
(Ap + 1)
2AV
After intergration
ln(Ap + 1)
ln V
=
+ constant
2A
A
1
ln (Ap + 1) V 2 = new constant
1
(Ap + 1)V 2 = even newer constant
Problem 2.4
(a) Using the first law
U sys = Q(500C ) + Q(300C ) + W + Q(25 C ) = 0
1.5M W + 0.5M W 1M W + Q(25 C ) = 0
Q(25C ) = 1M W
(b) Chack to see if the second law is satisfied
dSmachine Q
T
But dSmachine has to be zero since we are operating at steady state
Q(500C ) Q(300C ) Q(25C )
+
+
0
773
573
298
1. 5M W
0. 5M W
1M W
J
= 542.6
+
573K
298K
773K
Ks
542.6 KJs 0
11
(1)
Problem 2.5
(a)
U = Q + w
Since the process is adiabatic, Q = 0
(10 moles)
U = ncv T = w
5
J
T = 80000J
8.314
2
mole K
T = 385K
Tf = 388K
(b)
S1→3 = S1→2 + S2→3
S1→2 = 0 since this process is reversible and adiabatic
∂S S2→3 is at constant pressure, temperatre varies→ ∂T
p
What is T2 though? The temperatre that would be reached if expansion was reversible.
cR
2
2 7
p2 p
= 773
= 253K
T2 = T1
100
p1
388
dT
388
7
S2→3 =
= 8.314 ln
cp
2
253
T
253
J
S2→3 = 12.4
mole K
Problem 2.6
In case (b) extra work is done on the gas so that its internal energy will remain higher than in (a)
Tb > Ta
Problem 2.7
(a) Total amount of energy required
Necessary heat input is the total enthalpy change of the material (since p is constat)
Determined by the final and initial state
12
Enthalpy change is only due to the 10 L (the 50 L remains in exactly the same state)
dH = ncp dT
1000g 1 cal
H = 10L 75K
L
gK
H = 750 kcal = 3140kJ
H = 0.87kW h
(b) Entropy change
Again,entropy change is soley determined by the final and initial state
dS = ncp
dT
T
1000g 1 cal
S = 10L ln
L
gK
353
278
= 2.38
kcal
K
S = 9993 molJK
Problem 2.8
w1 = dV
pdV = RT1
= RT1 ln
V
V4
w3 = RT ln
V3
The second step where Q2 = 0 gives dU = dw
dV
BdT = pdV = RT
V
T3
V3
B ln
= R ln
T1
V2
and the fourth step
B ln
T1
T3
= R ln
V1
V4
Thus
V3
V4
V1
V4
or
=
=
V2
V1
V2
V3
and
13
V2
V1
w3
w1
V1
V4
= R ln
= R ln
=
T1
V2
V3
T3
T3
w1 (T3 T1 )
w3 (T3 T1 )
Q(T3 T1 )
Q3 (T3 T1 )
w = w1 w3 = w1 + w1
=
=
=
=
T1
T1
T3
T1
T3
By drawing the heat Q3 from a heat souce at T3 one may ths produce mechanical work in the amount of
if the rest of the energy, QT33T1 can be disposed of as heat to a heat sink at T1
Q3 (T3T1 )
T3
Problem 2.9
Suniverse = Ssys + Senvironment
There is no heat flow into the environment, so
Senvironment = 0
But for the system (the gas) →
The gas goes from state 1 at Pi , Vi , Ti to a state 2 at Po , Vo , To (= Ti ) with Po < Pi
To compute the Sgas we need to how the entropy changes with pressure (or volume) at constant T
and for an ideal gas
∂S
∂P
=
T
∂V
∂T
(Maxwell Relation)
P
Sgas
R
∂V
=
∂T P
P
Po
Po
= R
= R ln
Pi
Pi
since Pi > Po
R ln
Pi
Po
>0
Since Suniverse > 0 the process is irreversible
Problem 2.10
To calculate U (P, V ) for any (P, V ) given a value U (Po , Vo ) at some (Po , Vo ) it is sitable for this system to follow
the path sketched below.
14
First go from Vo , Po → Vo , P (i.e. constant volume)
A constant colume, if only P-V work is possible
dU = Q
U = U (P , Vo ) U (Po , Vo ) = Q = A(P Po )
P is unknown, but we do know because of the path we chose from (Vo , Po) → (V, P ) that (Vo , P ) and (V, P ) are
connected by a reversible adiabat. Therefore,
V
P Vo = P V or P = P
Vo
So,
U (P , Vo ) U (Po , Vo ) = A(P r Po )
(with r =
V
)
Vo
Second go from Vo , P → V, P
Along the reversible adiabat Q = 0 and therefore
V
U = U (P, V ) U (P , Vo ) = P dV Vo
with P and V representing values along the adiabat going through (V, P ). To calaculate this integral use
P V = P V PV P = V
So,
V
U = U =
PV
( 1)
V
1
PV Vo
=
dV V PV
( 1)
1
1
( 1)
V ( 1)
Vo
V
1
U = U (P, V ) U (P , Vo ) = (P 1) 1 r
V ( 1)
Vo
(with r =
V
)
Vo
(with r =
V
)
Vo
Combining the first and second step
Utot = U (P, V ) U (Po , Vo ) = A(P r Po ) +
PV ( 1)
1 r 1
LEVEL 3 PROBLEMS
Problem 3.1
The reaction at 1200K is
2N H3 → N2 + 3H2
P =
nRT
V
, T and V are constant but n increases by a factor of 2 in the reaction. Therefore
Pf = 2Pi → Pf = 20atm
(b) Heat flow
The system is underconstant volume (dV = 0) so W = 0 and
U = Q
We can compute U from H because
U = H PV
15
U = H (P V ) = H (nRT ) = H RT n
kJ
H = 87
and n = 2 moles
mole
J
J
U = 87000
(1200K )
(2) 8.314
mole
mole K
J
U = Q = 67046 mole
(c) System is adiabtic and under constant volume=⇒ U = 0
So,
Ta
Ureaction +
(cv ,N2 +3cv ,H2 )dT = 0
1200
Ta = 1200 use cp cv = R
Ta = 1200 Ureaction
cv ,N2 +3cv ,H2
J
67046 mole
J
4 24.686 mole
K
Ta = 521K
The system cools down which is expected given that the reaction is endothermic.
Problem 3.2
We can define the system as the gas in the tank (no moles) and the gas that will be squeezed into the tank (nc
moles). It will helpful to define some variables
Po
Pf
To
Tf
V
Vf
= The initial pressure in the tank
= The external pressure pushing gas in
= The initial temperature of the gas
= The final temperature of the gas (what we want to solve for)
= The volume of the tank
= The volume of the gas pushed in
We can write the following relationships
Po V
RTo
Pf V
no + nc =
RTf
nc RTo
Vc =
Pf
no =
Since this is an ideal gas we know that the internal energy change is only a function of temperature, given by
U = (no + nc )cv (Tf To )
Givem that the process is adiabatic (isolated),
U = Q + W = W
But for this case we know that
W = Pf Vc = nc RTo
16
(Q = 0 )
or the external pressure (which is constant) times the volume of gas pushed in. Now we can just work through
eliminating variables to find the final temperature.
nc RTo = (no + nc )cv (Tf To )
Pf V
nc RTo =
cv (Tf To )
RTf
Pf V
Po V
Pf V
RTo
cv (Tf To )
=
RTf
RTo
RTf
Pf RTo Po RTf = Pf cv (Tf To )
Pf To (R + cv ) = Tf (Po R + Pf cv )
Tf =
Pf To (R + cv )
P o R + P f cv
Treating air as a diatomic ideal gas, we can use cv = 52 R and solve for Tf
298 1 (R + 72 R)
7
Tf =
=
298
(0.5R + 1 cv )
6
Tf = 347.6K
Problem 3.3
Assume that the volume change resulting from the reaction A + B → AB is V and the heat of reaction is Q.
During the reaction, the system will perform work against the environment which has a constant pressure Po . Since
the pressure of the environment Po is constant, the external work done by the system is
w = Po V
Using the first law
U = Q Po V
Q = U + Po V
In the initial and final states of the reaction (both equilibrium states), the pressure of the gas has to equal Po
(otherwise they wouldn’t be equilibrium states). Therefore if P is the pressure of the gas
P = Po
or
Q = U + (P V )
Q = H
where H is the difference of enthalpies in the initial and final states.
Problem 3.4
Closed system solution
System is gas flowing into the tank
U2 U1 = Q + w = w
U2 U1 = P1 V1
For an ideal gas,
U2 U1 = ncv (T2 T1 )
P1 V1 = nRT1
17
(Q = 0 (adiabatic))
So,
ncv (T2 T1 ) = nRT1
cv T2 = (R + cv )T1 = cp T1
cp
T2 = T1
cv
Open system solution
System is the tank
H i mi = Uf Ui = U f mf
Uinitial = 0 because minitial = 0
min = mf
H i = U f = U in + Pin Vin
U f U i = Pin Vin
cv (Tf Ti ) = RTi
cp
Tf = Ti
cv
Problem 3.5
(a) Gas in tank I is undergoing an adiabatic reversible expansion
cR
T2
P2 p
=
T1
P1
7R
5 2R
T2 = 300
10
T2 = 246K
(b) Try to get TII from the fact that
Utot = UI + UII = 0
because the total system is isothermal. UI is the internal energy change of gas that remains in I and UII is the
internal energy change of gas that ends up in tank II.
UII = nII cv (TII T1 )
UI = nI cv (TI T1 )
need to find nI
PV
RT
nI
PI T1
5 atm 300K
=
=
= 0.61
10 atm 246K
n1
TI P1
n=
nII = 1 nI = n1 (1 0.61) = 0.39n1
now from UI + UII = 0
UII = UI
0.39n1cv (TII 300) = 0.61n1cv (246 300)
TII = 384K
(c) Entropy change → S varies with T and P
18
∂S
dS =
dT +
dP
∂P T
P
∂S
cp
∂S
∂V
R
and
=
=
=
∂T P
T
∂P T
∂T P
P
∂S
∂T
cp
R
dT dP
T
P
For the gas remaining in tank I dSI = 0 as it is undergoing an adibatic reversible expansion.
For the gas ending up in tank II
TII
PII
R ln
S = nII cp ln
To
Po
dS =
Po = 10 atm, To = 300K
nII = 0.39 (per mole of gas initial in tank I)
TII = 384K
nII RTII
nII TII
PII =
=
= 5 atm
V
n1 To
7
384
5
R ln
S = 0.39n1 R ln
2
300
10
J
S = 5.05n1 K mole
LEVEL 4 PROBLEMS
Problem 4.1
This is a very tricky question which is why it is alone in Level 4.
First let’s determine the equilibrium conditions. Assume that the system has some how reached equilibirum with
final values T1f , P1f , V1f for compartment 1 and T2f , P2f , V2f for compartment 2. We can ssk what criteria must there
variable satisfy for the system to be in equilibrium.
Well we know that for any reversible change around equilibrium
dU = dU1 + dU2 = 0
i.e. U (V, S ) must be minimal at equilibrium, and
dS = dS1 + dS2 = 0
i.e. S (U, S ) must be maximal at equilibrium.
Since there is no heat flow between compartments and between the chamber and the environment (since all walls
are adiabatic)
dS1
=
0
dS2
=
0
So at equilibrium, the only reversible changes we can make are in the volume of the compartments, i.e.
dV1 = dV2
Then we can write (for a reversible change
dU1 = P1f dV1 = +P1f dV2
dU2 = P2f dV2
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dU = dU 1 + dU 2 = (P1f P2f )dV2 = 0
since this should hold for arbitrary dV2 ,
P1f = P2f
i.e. equilibrium is characterized by mechanical equilibrium.
NOTE: Since dS1 = 0 and dS2 = 0, the equilibrium condition that dU = 0 does not yield any information about
the final temperatures T1f and T2f . This means that the final temperatures of this system depend on the path followed
by the system to attain equilibrium.
The question remains: can we obtain equilibrium from the initial state T1 = To , T2 = To , P1 =
P2 , and n1 =
n2
by following a reverible path?
The answer is: probably not.
One candidate reversible path is to allow the piston to very slow move until the equilibrium pressures are attained
(i.e. P1f = P2f ). This path, however, cannot proceed reversible for the following reason:
Assume that is can proceed reversibly. Then an infinitesimal change in the volume of compartment 1 by dV1 will
result in an increase of the internal energy of compartment one by
dU1 = P1 dV1
The change in internal energy of compartment 2 is
dU2 = P2 dV2 = P2 dV1
But since both compartments are adiabatically sealed from the environment,
dU = 0 =⇒ dU1 = dU2
or
P1 dV1 = P2 dV1
or
P1 = P2
But this is not true except in the final equilibrium state. In the initial state P1 =
P2 by assumption.
=⇒Therefore the suggested path cannot proceed reversibly. The suggest path must be accompanied by dissipation
(through friction between the piston and the walls for example). This dissipation will be accompanied by a production
of heat. We cannot predict a-priori how much of that heat evolves to compartment 1 or compartment 2 though.
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