Unformatted text preview: 50 Section 2 Solutions and Comments on Problems
Similarly, integrating both sides of Eq. 8 yields The righthand sides of Eqs. 9 and 10 must be equal, since both are equal to z(x, y). This can be hue only if we have x F(y) = C a nd G(x) =  + 3x 2 + C, where C is a constant. The required function is, therefore, Homework 2 Spring 2006 2.2 Consider the differential dZ = [ 2xy + s in(y)] dx + [ x2 + x cos(y)] dy. (A) The state of the system is changed from (x = 0, y = 0) to (x = 1, y = 1 ) along the path x = y. Calculate the total change in Z along the path. (B) The change in (A) is now made along the path x = y 2.Compute the total change in Z along this path. (C) Show that d Z is an exact differential. Solution
(A) Along the path, x becomes
= y, we have x = y a nd dx = dy, so the integral of dZ = [ x3  cos x + cos x + x sin x]; = [x3 + x sin x]; In executing the integrals in Eq. 1, we have used the fact that Jx cos(x)dx = cos x + x sin x, which can be obtained by integrating by parts one time. (6) Along the path x = y2,we have x may be written as = y 2and dx = 2y dy, so the integral of dZ [5y4 + 2y sin y + y2 cos y] dy
2y cos y + 2 sin y = [ Y5 y 2sin y]: +
= [y'  2y cos y
=1 + (y2  2)sin y];
(2) + sin(1) = 1.84147... , which is the result obtained in Part (A).The integrals in Eq. 2 can be done by parts. The results are J2y sin y dy = 2 sin y  2y cos y and J y 2 cos y dy =2y cosy + (y2  2) sin y. These results can also be found in most integral tables. (C) Applying the Euler criterion, we obtain Chapter 2
We further have P,,,
11he t The First Law of Thermodynamics 53 = 0 ; therefore,
W, = process  IV, V2
P,JV
= 0, because P,,, = 0 at all points during the expansion. Since A U , = 0 , we must have q , =  w2 Therefore, q 2 = 0 . The total results are I AU
I AH = = AU, AH, + A U, = o ( ; + A H , = 01; (8)
( 9) 2.6 Two moles of an ideal gas at 300 K and 10 atm pressure are expanded isother isothermal P zero and the = 1 atm and iI
r mally against a constant external pressure of 5 atm until the internal pressure reaches a value of 7 atm. At this point, the external pressure is reduced to zero and the gas is further expanded into a vacuum until a n internal pressure of 1 atm is reached with T = 300 K . The gas is then compressed reversibly and isothermally back to its initial state of T = 300 K and P = 10 atm. Compute, if possible, w, q , A U, and A H for the total process. After the entire process is finished, has heat been gained or lost by the system? You may ignore acceleration effects. (See explanatory note in Example 2.3.) Solution
Since we have dT
= 0 for the process in the first step and the gas is ideal,
(1) dH = C,dT
= = 0. 0 . The work is given by (2) Therefore, in this step, we have A U , AH, = = nRTP,,,[P;'  P ;'], (3) = 10.55 L atm
= = 1,069 J .
= = (4)
 w,. Thus, Since A U, 0 , we must have q , q, 1,069 joules.
= (5) In the second step, we also have dT AH, We further have P,,,
= = 0 ; thus,
= A U2 0. ( 6) 0; therefore, Section 2 Solutions and Comments on Problems
because P,,, = 0 at all points during the expansion. Since A U2 = 0, we must have q , =  w2. herefore, q , = 0. The third step T is an isothermal, reversible compression. The work for this process is given by Eq. 2.33 in the text. We have since the gas is ideal. Thus, The total work done is zq,,,, = W, + T U, + W , = 1,069 + 1.149 x 10'J = 1.042 X l o4J. (10) Since there is no change in T throughout the entire process, we must have and The first law then gives AUtotai
so that = qtotal + 7()tota1 = 0, qtotal=  wtotal  1 .042 X 10' J. =
The system has, therefore, lost heat to the surroundings. 2.7 Two moles of an ideal gas at 300 K a nd 10 atm are expanded isothermally to a final pressure of 2 atm with T = 300 K. (A) What is the minimum magnitude of the work for this expansion? (B) What is the maximum magnitude of the work for the expansion? Solution
(A) The minimum magnitude for the work is zero. Since we have  if P,,, is zero, the work will be zero.
(€3) The maximum magnitude is obtained for a reversible process with P,,, This gives  P. or ?urnax  (2 rnoi)(8.314J mol' K')(300 K)In[;] =
Therefore, the maximum magnitude of the work is = 8,029 J (3) 58 Section 2 Solutions and Comments on Problems
From the first law, we have AU
Solving for w yields
ul = q + so.
J. (8) = A~I = 14,858 T  20,794 J = 5,936 ( 9) (B) If the heating is done at constant volume, the values for AH and AU will not change, since the initial and final temperatures are still 300 and 500 K, respectively, a nd d U and d H depend only upon T for an ideal gas. Therefore, we have AU
and =  14,858 J  (10) AH = 20,794 J . (11) However, the path is now different, so q and w change. For a constantvolume path, we have q = q, = AU = 14,858J. (12) The volume is constant, so the amount of work done must be zero. This is also clear from the first law: AU
But AU = a: s ow
= = q + w. (13) 0. the N,gas to 500 K in Part A , we must urovide enough enerev not onlv to heal reason C p is larger than C , is the work done in the constantpressure process that is not present in the constantvolume heating. I
I 2.12 A gas obeys the equation o state PV = RT + aP, where a is a function o T only. f f (A) Show that the reversible work done whenever the gas is heated at constant pressure from T I to T 2 is given by w = R(T,  T ,) + ( a ,  a ,) P , where a , and a , are the values of a at temperatures T , and T,, respectively ( 6 ) Show that if the gas is expanded isothermally from V , to V 2, he reversible t work will be
u = RT h [ p ] . i Solution
(A) The reversible work at d P
= 0 is given by Substituting the PV product from the equation of state gives Chapter 2
(B) The isothermal, reversible work is The First Law of Thermodynamics 59 (3) Using the equation of state, we find that the pressure is ~d U will not A p = RT va' (4) SubstitutingEq. 4 into Eq. 3 produces which may be written in the form nstantvolume $0 This is also Equation 6 is the desired expression. 2.13 The oxygen in a highpressure cylinder is at 300 K a nd 100 atm pressure. The valve of the cylinder is opened, and the gas is allowed to expand to an atmospheric pressure of 1 atm. Assuming that the expansion is adiabatic and conducted under conditions of constant enthalpy, estimate the final temperature of O 2after expansion. Do not assume O 2to be an ideal gas. Under the stated conditions, we have shows that the ressure process
(1) aT = pap. (2) Integrating both sides between corresponding limits gives I we may regard the JouleThomson coefficient as being nearly independent f of pressure, we may estimate T 2 = TI +p ll
"2 dP = T I + p (P2  P I ) . (4) Using the value of p for 0 ,given in Table 2.3, we obtain T 2 = 300 K + 0.31 K atm' ( 1  100) atm = 269 K. (5) As expected, the expansion of a gas with a positive JouleThomson coefficient results in a cooling of the gas. 2 4e ." 1 mole of an ideal gas at a temperature of 500 K a nd a pressure of 6 atm is subjected to the following changes: I
I I I 1 68 Section 2 Solutions and Comments on Problems
Substituting of values into Eq. 5 gives the solution: (10 atm)(8.206 L )  (500  a tm)  14.19 moles of gas. L n2 = (0.08206 L atm molI K')(500 K )
(7' I I 1 2.20 In Chapter 1, it was shown that the second virial coefficient for a gas described by a van der Waals equation of state is C,(T) = 6  a/(RT), where a and b are the van der Waals parameters. (A) Obtain an expression giving the work done in an isothermal, reversible expansion from volume V , to volume V , for a gas whose equation of state is a virial expansion. (B) If the virial expansion 1s truncated after the second virial term, compute the work done for the isothermal expansion of' 1mole of CO, at 300 K from 10 llters to 50 liters. The van der Waals parameters for C O, are given in Table 1.1 (C) Using the van der Waals equation of state, determine the work done for the process in (B). What is the percent difference between the answers in (B) and (C)? Solution
(A) The virial equation of state is P = nRT[V ' + ~ c , ( T ) v+ ? z'C,(T)V~ ...I. ~ + (1) The isothermal, reversible work for this gas is given b y nRT n'a Chapter 2 The First Law of Thermodynamics 69 gas. (7) I Substituting this expression for P into Eq. 2 gives as described ? a and b are
 I, reversible I
I o state is a f compute the rom 10 liters l 1.1. e 'done for the I (B) and (C)? I I z =  [ nRT I/'  nb  n2a ]dV v2 = V nb nRT ln  nb] ;[  'a : l[  i] (7) Substituting values yields the work: zu =  (1 mo1)(0.08206L atm molI K')(300 K) 50 L ln[ 10 L
 ( 1 mo1)(0.04286L m o l l ) ( 1 mo1)(0.04286L m ol5] (1mo12)(3.658 ' bar mo1') L L   Ll o = 39.706 L atrn 39.706 L atrn 1 atrn + 0.2926 L bar [1.01325 b ar] = + 0.2888 L atrn = 39.417 L atrn = 3,994 J. ( 8) (1) ...]dV The results of Parts (B) and (C) are the same to four significant digits. Obviously, truncating the virial expansion after the second virial coefficient gives good approximations for the temperature and volumes of this problem. It should be noted that it is very important to be careful with the units in evaluating Eq. 8. The second term has units of L bar. Therefore, we need to convert this to L atm, add the two terms, and then convert to joules. Of course, we could equally well convert both terms to L bar and then convert the sum to joules. Alternatively, both terms can be converted to joules and lin Table 1.1, 300K)
(3) 2.21 More about the thermodynamics of hell. (Note: This problem is dependent upon the analysis and assumptions contained in Problems 1.17 and 1.18). In Problems 1.17and 1.18, we found that if hell is expanding at a slower rate than the rate at which additional souls enter the place, the temperature of hell will decrease with time. With certain assumptions we made in those problems, it could be determined that the temperature of hell is given by T  3.326 X l o3" exp 5,000 in degrees K , where b is a constant that determines the rate of expansion of the volume of hell and f is the fraction of people whose souls will enter hell. In Chapter 1, we employed the value b = 0.0020723 years'. We also found that the number of moles of souls in hell is given by where A = 1.66 X lo'" mol and a = ln(lOLOf 1)/5,000 years1 + (6) (A) If the average molar heat capacity of hell at constant pressure is 3.5R, determine the total heat change q of hell since the Devil entered it 5,000 years ago (by assumption). (B) Given that hell is expanding at constant pressure, compute the total amount of pressurevolume work done by the expansion of hell since the Devil entered it. (C) Compute AH a nd A U for hell over the last 5,000 years. Chapter 2
(C) Since the heating is done at constant pressure, we have
atm, we have The First Law of Thermodynamics 71 q p = AH
AU may now be computed from A H = AU = 5.078 X 1012J. (12) (1) + A ( P V ) = AU + P A V ,
X (13) because P is constant. Therefore, we have AU
nce we know to time. From = = AH  P A V = AH
6.529
X + work = 5.078 1012J + (1.451 x 1012J )
(14) 10" J . 2.22 Two moles of an ideal gas at 500 K and 10 atm pressure are contained in an Insulation insulated pistoncylinder arrangement such as the one to the right. The gas is allowed to expand reversibly and adiabatically.During the expansion, gas at a temperature equal to the current instantaneous temperature of the gas in the cylinder is added through an inlet valve in sufficient quantity to maintain the internal pressure constant at 10 atm throughout the expansion. When the total work done equals 50 L atm, the expansion is halted. Calculate the final temperature of the gas inside the piston and determine the number of moles of gas present after the expansion is complete. You may assume that the molar heat capacity of the gas, C r , is constant at the value 1.5R. inlet Solution
The general equation for a reversible, adiabatic expansion is Eq. 2.81 in the text. Since we are dealing with an ideal gas under...
View
Full Document
 Spring '08
 PRESTONSNEE
 Physical chemistry, Thermodynamics, pH, First Law Of Thermodynamics

Click to edit the document details