Homework 3 Answer
(Due 02/04/2011 Friday)
P2.3, P2.7, P2.8, P2.11, P2.16, P2.19, P2.20, P2.21, P2.23, P2.25, P2.26
P2.3) 3.00 moles of an ideal gas are compressed isothermally from 60.0 to 20.0 L using a constant
external pressure of 5.00 atm. Calculate q, w, U, and H.
The work against a constant pressure:
w p external V 5 101325 Pa 20 10 3 m 3 60 10 3 m 3 2.03 10 4 J
U 0 and H 0 since T 0 and H U PV
q w 2.03 10 4 J
P2.7) For 1.00 mol of an ideal gas, Pexternal = P = 200.0 103 Pa. The temperature is changed from
100.0°C to 25.0°C, and CV,m = 3/2R. Calculate q, w, U, and H.
U n C V,m T
3
2
1.0 mol 8.314472 J K -1 mol -1 298 K - 373 K 935 J
H n C p,m T n C V,m R T
5
2
1.0 mol 8.314472 J K -1 mol-1 298 K - 373 K 1559 J
q P H 1559 J
w U - q P (935 J ) (1559 J ) 624 J
P2.11) Calculate H and U for the transformation of 1.00 mol of an ideal gas from 27.0°C and 1.00
atm to 327°C and 17.0 atm if
T 1 1
C P,m 20.9 0.042 in units of J K mol
K
For an ideal gas, H is given by:
1
Tf 600 K
T
ΔH n C p,m dT n 20.9 0.042 dT
Ti 300 K
K
ΔH 20.9 600 K 300K J 0.021 T 2 600 K
300 K J
6.27 10 J 5.67 10 J 11.94 10 J
3 3 3
ΔU ΔH ΔpV ΔH n R ΔT
11.94 10 3 J 1 mol 8.314472 J mol-1 K -1 300 K 9.45 103 J
P2.12) Calculate w for the adiabatic expansion of 1.00 mol of an ideal gas at an initial pressure of 2.00
bar from an initial temperature of 450. K to a final temperature of 300. K. Write an expression for the
work done in the isothermal reversible expansion of the gas at 300. K from an initial pressure of 2.00
bar. What value of the final pressure would give the same value of w as the first part of this problem?
Assume that CP,m = 5/2R.
3
2
w ad U n C p,m R T mol 8.314472 J K -1 mol-1 150 K 1.87 103 J
p p w reversible
w reversible n R T ln i and ln i
pf pf nRT
p
ln i
w reversible
1.87 103 J 0.7497
pf nRT
1 mol 8.314472 J K -1 mol-1 300 K
pi
2.12
pf
pi
pf 0.944bar
2.12
P2.15) An ideal gas undergoes an expansion from the initial state described by Pi, Vi, T to a final state
described by Pf, Vf, T in (a) a process at the constant external pressure Pf and (b) in a reversible process.
Derive expressions for the largest mass that can be lifted through a height h in the surroundings in these
processes.
2
p f V f Vi
w mgh p f V f Vi m
gh
Vf nRT V f
w mgh nRT ln m ln
Vi gh Vi
Alternate answer:
mg
pext
A
pext V f Vi
mg
Ah mgh
A
pext V f Vi
m
gh
If w is the average weight to be lifted
m g Vf
w Ah m gh nRT ln
A Vi
nRT V f
m ln
gh Vi
If m is the final weight
mg
pf Ah V V f Vi
A
A pf p f V f Vi
m
g gh
P2.19) 3.50 moles of an ideal gas are expanded from 450. K and an initial pressure of 5.00 bar to a
final pressure of 1.00 bar, and CP,m = 5/2R. Calculate w for the following two cases:
a. The expansion is isothermal and reversible.
b. The expansion is adiabatic and reversible.
3
Without resorting to equations, explain why the result for part (b) is greater than or less than the result
for part (a).
a) Calculating the initial and final volumina:
Vi
n R T 3.50 mol 8.314472 J K 1 mol 1 450 K
0.0262 m 3
pi 5.00 10 Pa
5
Vf
n R T 3.50 mol 8.314472 J K 1 mol 1 450 K
0.1310 m 3
pf 1.00 10 Pa
5
w for an isothermal, reversible process is then given by:
Vfinal
0.1310 m 3
3.50 mol 8.314472 J K mol 450 K ln 0.0262 m 3
1 1
w n R T ln
Vinitial
w -21076 J -21.1 kJ
b) For an adiabatic, reversible process:
T V
ln final - 1 ln final , where γ C P, m /C V,m equation 2.43
Tinitial Vinitial
T T p T p
ln final - 1 ln final initial - 1ln final ln initial
Tinitial Tinitial p final Tinitial p final
T
ln final
- 1 ln p initial
p
Tinitial final
Therefore:
γ - 1 p initial
Tfinal Tinitial Exp ln
γ p final
With C P,m 52 R, and C V,m 32 R , the final temperature is:
5bar
Tfinal 450 K Exp 0.4 ln 236 K
1bar
And finally w for an adiabatic process and for 3.5 moles of gas:
3
w qV nCV T (3.5mol ) (8.314472 JK 1mol 1 ) (236 K 450 K ) 9341.31J 9.34kJ
2
Less work is done on the surroundings in part b) because in the adiabatic expansion, the temperature
falls and therefore the final volume is less that that in part a).
4
P2.20) An ideal gas described by Ti = 300. K, Pi = 1.00 bar, and Vi = 10.0 L is heated at constant
volume until P = 10.0 bar. It then undergoes a reversible isothermal expansion until P = 1.00 bar. It is
then restored to its original state by the extraction of heat at constant pressure. Depict this closed-cycle
process in a P–V diagram. Calculate w for each step and for the total process. What values for w would
you calculate if the cycle were traversed in the opposite direction?
p 1 Vi
10
p [bar]
6
2
pi Vi p i V2
0
10 20 30 40 50 60 70 80 90 100 110 120 130
V [L]
First we calculate the number of moles:
n
p i Vi
1.00 bar 10.0 L 0.401 mol
R Ti
0.0831451 L bar K 1 mol 1 300 K
The process in the diagram above is described by the steps:
step 1: pi, Vi, Ti → p1,Vi, T1
step 2: p1, Vi, T1 → pi,V2, T1
step 3: pi, V2, T1 → pi,Vi, Ti
In step 1 (pi, Vi, Ti → p1,Vi, T1) w1 = 0 since V stays constant
In step 2 (p1, Vi, T1 → pi,V2, T1) we first calculate T1:
5
T1 Ti
p1
300 K
10.0 bar 3000 K
pi 1.00 bar
Then the work is:
V p
w 2 n R T ln f n R T ln f
Vi pi
10.0 bar
0.401 mol 8.314472 J K -1 mol -1 3000 K ln
1.00 bar
23.0 10 J
3
In step 3 (pi, V2, T1 → p1,Vi, Ti) we first calculate V2:
p 1 Vi
p 1 Vi p i V2 and V2 10Vi 100L
pi
And the work:
w 3 p externmal V 1.00 bar
10 5
Pa
10 L 100 L
10 3 m 3 9.00 10 3 J
1 bar 1 L
And for the entire circle:
w cycle w 1 w 2 w 3 0 J 23.0 10 3 J 9.00 10 3 J 14.0 10 3 J
If the cycle were traversed in the opposite direction, work of each step has the same value with
opposite sign.
P2.23) A pellet of Zn of mass 10.0 g is dropped into a flask containing dilute H2SO4 at a pressure of P
= 1.00 bar and temperature of T = 298 K. What is the reaction that occurs? Calculate w for the process.
The chemical equation for the process is:
2-
Zn 2 (aq) SO 4 (aq) H 2 (g)
Zn (s) H 2 SO 4 (aq)
First we calculate the volume of H2 that is produced:
6
10g
n H 2 n Zn 0.1529mol
65.39gmol -1
nH RT 0.1529mol 8.314472 JK 1mol 1 298 K
VH 2 2 3.79 10 3 m 3
P 1 10 Pa
5
Assuming that Vf Vi VH 2 , the work is:
w p ext Vf Vi 1 105 Pa 3.79 10 -3 m 3 - 379 J
P2.25) One mole of an ideal gas, for which CV,m = 3/2R, initially at 20.0°C and 1.00 106 Pa
undergoes a two-stage transformation. For each of the stages described in the following list, calculate
the final pressure, as well as q, w, U, and H. Also calculate q, w, U, and H for the complete
process.
a. The gas is expanded isothermally and reversibly until the volume doubles.
b. Beginning at the end of the first stage, the temperature is raised to 80.0°C at constant volume.
p1 V1 p1
a) p 2 0.500 10 6 Pa
V2 2
V
w n R T ln f
1.0 mol 8.314472 J K -1 mol -1 293.15 K ln 2 1.69 10 3 J
Vi
U H 0 because T 0
q w 1.69 10 3 J
b) constant volume, then
T1 T2
T p
, therefore p 2 2 1
0.500 10 6 Pa 353 K 6.02 10 5 Pa
p1 p 2 T1 293 K
3
U n C V,m T 1.0 mol 8.314472 J K -1 mol -1 353 K - 293 K 748 J
2
7
w = 0 because V = 0
q = U = 748 J
H n C p, m T n C V, m R T 1.0 mol
5
2
8.314472 J K -1 mol -1 353 K - 293 K 1.25 10 3 J
For the overall process:
q 1.69 10 3 J 748 J 2.44 10 3 J
w 1.69 10 3 J 0 J 1.69 10 3 J
U 748 J 0 J 748 J
H 1.25 10 3 J 0 J 1.25 10 3 J
P2.26) One mole of an ideal gas, for which CV,m = 3/2R, initially at 298 K and 1.00 105 Pa undergoes
a reversible adiabatic compression. At the end of the process, the pressure is 1.00 106 Pa. Calculate
the final temperature of the gas. Calculate q, w, U, and H for this process.
1 1 1
Tf Vf T pi
f
Ti Vi Ti pf
1
Tf p
i
Ti pf
1
Tf pi
Ti pf
5
1
3
Tf 1.00 10 5 Pa 5
0.100
0.4
2.51
3
Ti 1.00 10 6
Pa
8
Tf 2.51 298 K 749 K
q = 0 for an adiabatic process.
3
w U n C V,m T 1.0 mol 8.314472 J K -1 mol -1 749 K - 298 K 5.62 10 3 J
2
H U p V U R T 5.62 103 J 8.314472 J K -1 mol-1 749 K-298 K 9.37 103 J
P2.26) One mole of an ideal gas, for which CV,m = 3/2R, initially at 298 K and 1.00 105 Pa undergoes
a reversible adiabatic compression. At the end of the process, the pressure is 1.00 106 Pa. Calculate
the final temperature of the gas. Calculate q, w, U, and H for this process.
1 1 1
Tf Vf T pi
f
Ti Vi Ti pf
1
Tf p
i
Ti pf
1
Tf pi
Ti pf
5
1
3
Tf 1.00 10 5 Pa 5
0.100
0.4
2.51
3
Ti 1.00 10 6
Pa
Tf 2.51 298 K 749 K
q = 0 for an adiabatic process.
3
w U n C p,m T 1.0 mol 8.314472 J K -1 mol -1 749 K - 298 K 5.62 10 3 J
2
H U p V U R T 5.62 103 J 8.314472 J K -1 mol-1 749 K-298 K 9.37 103 J
P2.29) A cylindrical vessel with rigid adiabatic walls is separated into two parts by a frictionless
9
adiabatic piston. Each part contains 50.0 L of an ideal monatomic gas with CV,m = 3/2R. Initially, Ti =
298 K and Pi = 1.00 bar in each part. Heat is slowly introduced into the left part using an electrical
heater until the piston has moved sufficiently to the right to result in a final pressure Pf = 7.50 bar in the
right part. Consider the compression of the gas in the right part to be a reversible process.
a. Calculate the work done on the right part in this process and the final temperature in the right
part.
b. Calculate the final temperature in the left part and the amount of heat that flowed into this part.
The number of moles in each part is given by:
n
p i Vi
1.00 bar 50.0 L 2.02 mol
R Ti
8.314472 10 -2 L bar K 1 mol 1 298 K
a) We first calculate the final temperature in the right side:
1 1 1
Tf Vf T pi
f
Ti Vi Ti pf
1
Tf p
i
Ti pf
1
Tf pi
Ti pf
5
1
3
Tf 1.00 bar 5
3 2.24
Ti 7.50 bar
Tf 2.24 298 K 667 K
10
3
w ΔU n C V ΔT 2.02 mol 8.314472 J K 1 mol 1 667 K - 298 K 9.30 10 3 J
2
b) First we calculate the volume of the right part:
n R Trf 2.02 mol 8.314472 10 bar L K mol 667 K
-2 1 1
Vrf 14.9 L
p rf 7.50 bar
Therefore Vlf 1000 L 14.9 L 85.1 L , and
Tlf
p lf Vlf
7.50 bar 85.1 L 3800 K
nR 2.02 mol 8.314472 10-2 bar L K 1 mol 1
3
ΔU n C V ΔT 2.02 mol 8.314472 J K 1 mol 1 3800 K - 298 K 88.2 10 3 J
2
From part a) w = 9.30 10 3 J
q = U – w = 88.2 10 3 J + 9.30 10 3 J 97.5 10 3 J
P2.43) One mole of N2 in a state defined by Ti = 300. K and Vi = 2.50 L undergoes an isothermal
reversible expansion until Vf = 23.0 L. Calculate w assuming (a) that the gas is described by the ideal
gas law and (b) that the gas is described by the van der Waals equation of state. What is the percent
error in using the ideal gas law instead of the van der Waals equation? The van der Waals parameters
for N2 are listed in Table 1.3.
a) For an ideal gas:
V
w reversible n R T ln f 23.0 L
1.0 mol 8.314472 J K -1 mol -1 300 K ln 5.54 10 J
3
Vi 2.50 L
b) For a van der Waals gas:
11
Vf
RT
Vf
a Vf
Vf
w p external dV dV R T dV a dV
V b V 2
Vi
V Vm b V V 2
Vi m m i i m
The first integral can be solved by substituting y = Vm - b:
Vf y
RT f
RT
dV dy R TlnVf b lnVi b
V b
Vi m yi
y
Therefore, the work is given by:
V b 1 1
w reversible n R T ln f a
Vi b Vi Vf
23.0 L 0.0380 L
1.0 mol 8.314472 J K -1 mol-1 300 K ln
2.50 L 0.0380 L
10 Pa
10
5
6 1 1
1.366 L2 bar m 6 L2 5.52 103 J
1 bar 2.50 10 m
-3 3
23.0 10 -3 m 3
And the percent error is:
percent error 100
5.52 10 J 5.54 10 J
3 3
0.4%
5.52 10 J
3
12
1. Calculate dy/dx for the following y: (a) y =ax+b; (b) y =2x2; (c) y = 3x5 + 2(x+a)-3 (d) y = ln(x/3);
(e) y = C exp(-ax); (f) Acos(x) (g) Asin(bx) +Ccos(x/b) (h) y = x2exp(-ax2)
a)
dy d(ax+b)
a
dx dx
b)
dy d(2x 2 )
2 2 x1 4 x
dx dx
c)
dy d (3 x 2 x a ) d (3 x 5 ) d (2 x a )
5 -3 -3
3 5 x 4 2 (3) x a 15 x 4 6 x a
4 4
dx dx dx dx
d)
x x
d ln( ) d ln( )
dy a a 1a 1
dx dx x a x x
ad
a
e)
dy d [C exp(-ax)] Cd [exp(-ax)]
aC exp(ax)
dx dx 1
d (-ax)
a
f)
dy d [ A cos( x)] Ad [cos( x)]
A sin( x)
dx dx d ( x)
g)
dy d [ A sin(bx) C cos( x / b)] d [ A sin(bx)] d [C cos( x / b)]
Ab cos(bx) C / b sin( x / b)
dx dx dx dx
h)
dy d [ x 2 exp(-ax 2 )] d ( x2 ) d [exp(-ax 2 )] d [exp(ax 2 )]
exp(ax 2 ) x2 exp(ax 2 ) 2 x x 2
1
dx dx dx dx d (ax 2 )
a 2 x
2 3
2 x exp(ax ) 2ax exp(ax ) 2
2 x exp(ax 2 )(1 ax 2 )
{ d (uv) udv vdu }
2. Calculate dy/dx from the following equations
(a) ln(x) + y = xy (b) 1 = cos(x)y
13
a)
ln( x)
ln( x) y xy y
x 1
ln(x) d[ln(x)] d(x - 1) 1
d (x - 1) - ln(x) ( x 1) ln( x)
dy x 1 x ln( x)
x - 1 dx dx x
dx dx (x - 1) 2
( x 1) 2
x( x 1) 2
{ d ( u ) vdu 2 udv }
v v
b)
1
1 cos( x) y y
cos( x)
1
d
dy cos(x) cos(x) 0 - 1 (-sinx) sin( x)
2
dx dx (cosx) (cos x) 2
m C2 3
3. P (C ) 4C 2 exp
/ Z . (a) Calculate dP(C)/dC. (b) Show there are two solutions of
2kT
dP(C)/dC = 0 (C ≥ 0) and one solution gives the most probable speed, C =(2RT/M)1/2, where m is
the weight of a gas molecule, M is its molar mass, and k is the Boltzmann constant.
a)
4 2 mC 2
P (C ) 3
C exp
Z 2kT
mC 2 mC 2
d exp( ) 2 d exp( ) 2
dP(C ) 4 2 2kT mC d (C ) 4 2
2
2kT m2C mC d (C )
2
3 C exp 3 C exp
dC Z dC 2kT dC Z mC
2
2kT 2kT dC
d
2kT
4 mC 3
mC 2
mC 2
3 exp 2C exp
Z kT 2 kT 2kT
4 mC 2 mC 2
C exp
2kT 2
Z3 kT
b)
dP(C ) 4 mC 2 mC 2
0 3 C exp 2 0
dC Z 2kT kT
14
1/2
mC 2 2kT 2 N A kT 2 RT 2 RT
2 0C
2
C
C 0 or kT m N A m M M
2 RT
C C 0
M
4. Calculate ∂f/∂x and ∂f/∂y for the following f(x, y, z): (a) f = xy-2 (b) f = sin(3xy)
(c) f = ln(x2y) (d) f = yx4 + exp(-xy2)
a)
f x y 2y
2 3 3
x x x 3
f x y x y ) 3x
2 3 2 3
2
y2
y y y
b)
f sin 3 xy sin 3 xy
3 y cos(3 xy )
x x 1
3 xy
3y
f sin 3 xy sin 3 xy
3 x cos(3 xy )
y y 1
3 xy
3x
c)
f ln xy ln xy
3 3
1 1
y3 3
x x xy
3
xy x
y3
f ln xy ln xy
3 3
1 3
3 xy 2 3
y y xy 3 xy y
3 xy 2
d)
f y x y exp( xy ) y x y exp( xy )
2 4 2 2 2 4 2 2
4 y 2 x3 y 4 exp xy 2
x x x x
f y x y exp( xy ) y x y exp( xy )
2 4 2 2 2 4 2 2
2 x 4 y 2 y exp xy 2 2 xy 3 exp xy 2
y y y y
5. Calculate Cv = (∂U/∂T)V and Cp = {∂(U + PV)/∂T}P for n mole of ideal gas assuming that U =
5nRT/2 and n is a constant. What kind of molecular properties do they depend on?
5
U nRT
2
15
5
nRT
U 2 5 nR
CV
T V T 2
V
5 7
nRT nRT nRT
U PV 2 2 7 nR
CP
T P T T 2
P P
CV and CP depend on moles n.
16
