Introduction to the Thermodynamics of Materials

Transcription

1 Solutions Manual to accompany Introduction to the hermodynamics of Materials Sixth Edition David R. Gaskell School of Materials Engineering Purdue University West Lafayette, IN David E. Laughlin ALCOA Professor of Physical Metallurgy Carnegie Mellon University Pittsburgh, PA

2 INRODUCION his solutions manual provides worked-out answers to all problems appearing in Introduction to the hermodynamics of Materials, 6th Edition, with the exception of some of the problems in Chapter 5 and Problem 9.7), which are included in the answer section in the back of the book. Complete solutions to all the new problems to the 6 th edition are included and denoted by *. All solutions arc comprehensive, making this supplement a useful instructional tool for professors and students.

3 Problem 1.1* he plot of = (P, ) for a gas is shown in Fig Determine. the expressions of the two second derivatives of the volume of this plot. (note: the principle curvatures of the surface are proportional to these second derivatives). What are the signs of the curvatures? Explain. Solution: start with the defining equations of and. assuming is constant > 0 assuming is constant 0 P P P P Since all terms in the expressions are positive (, and ), both principle curvatures are positive. he surface is convex.

4 Problem 1.* he expression for the total derivative of with respect to the dependent variables P and is : d dp d P P Substitute the values of and obtained Qualitative Problem into this equation and obtain the equation of state for an ideal gas. Solution d dp d 1 d dp d P d dp d P ln c ln P c ln c 1 3 P (constant) he constant is nr for n moles of the ideal gas.

5 Problem 1.3* he pressure temperature phase diagram (Fig. 1.4) has no two phase areas (only two phase curves), but the temperature composition diagram of Fig. 1.5 does have two phase areas. Explain. Solution his must be due to the number of components in each system: he system displayed in Fig. 1.4 is unary and that in Fig. 1.5 is a binary. We will see more on this later in the text.

6 Problem 1.4* Calculate the value of the ratio volume. for an ideal gas in terms of its Solution 1 P R 1 P

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15 Problem.8* One mole of a monatomic ideal gas at standard temperature and pressure (SP) undergoes the following three processes: 1. at constant Pressure the emperature is doubled.. at constant emperature the Pressure is doubled 3. the gas is returned to SP via constant volume process Calculate U, H, q and w for each of the steps. Solution First calculate the values of, and P at the states A, B and C. Step 1 P A P A A A B P A A A C P A A A Step 1 A B U C C ( ) C A A A w P P ( ) P R A A A A A A A q U w C R ( C R) A A A Step BC

16 U C C ( ) 0 A A C R C B d C w d RA RA ln RA ln() B B q w R ln() A B Step 3 CA U C C ( ) C 3 A R w3 d 0 since C C q U C 3 3 A A A A A hus we see that for the entire process: U C 0 C 0 as it should for a state function A A w R R ln() 0 i A A q ( C R) R ln() C R R ln() i A A A A A

17 Problem.9* Paramagnetic salts often obey the Curie relation: M constant H C Obtain an expression for the work needed to change the magnetization from M 0 to M Mof such a material. Assume that the field and the magnetization are parallel. Solution M f M H M w dm dm 0 0 C C M f M f f f 0 H H f M From the equality w 0 f we see that to get to the same M at higher temperature, more work is needed. emperature works against the ordering of the moments. H f M From the equality w 0 f we also see that the work needed is the area under the linear H vs. M plot.

18 Problem.10 * One mole of a monatomic ideal gas is taken on the path A B C D A as shown in Figure.7. (1) A B is a reversible isothermal expansion of the gas; () B C is a reversible adiabatic expansion of the gas; (3) C D is a reversible isothermal compression of the gas (4) D A is a reversible adiabatic compression of the gas. a. Derive expressions for U, q and w during each step in terms of a, b, c, d, t1, t c and R. Determine the sign of each. b. Determine the values of wi), qi) and Ui) in terms of a, b, c, d, t1, t and R. Determine the sign of each. Figure.7

19 Solution b (1) A B U1 0; w1 q1 Rt ln( ) > 0 () B C q 0; U c (t -t ) 0; w = U so w > 0 1 d (3)C D U3 0; w3 q3 Rt1 ln( ) < 0 (4) D A q 0 ; U c (t -t ) > 0 - w U so w < a c d SUM wi Rt ln( ) Rt ln( ) 0 b 1 a d qi Rt ln( ) Rt ln( ) 0 U 0 b 1 a c c

20 Problem.11 * he change in enthalpy when one mole of solid water (ice) is melted at 73 K is 6008 J. a. Calculate the change in enthalpy when ice is melted at 98 K. Is this process possible at 1 atm? b. Calculate the change in enthalpy when supercooled water solidifies at 60 K. c. Sketch the H vs. plot for both solid and liquid water. For this problem take the heat capacity of liquid water to be J/k and that of solid water to be 38 J/K over the range in temperatures of the problem. he enthalpy of liquid water at 98 K may be set equal to zero. Solution: hus a. 73 S L S H 0( ) P m P H c d H c d 75.44(73 98) ( 73) J ( 73) J L H 0 98 L H ( ) c d 75.44( 98) J L S H H (98) H (98) M P 0 ( 7894) 38 (98 73) 6944 J. Yes, ice can be superheated b. S L H H (60) H (60) H H L S fusion (60) 75.44(60 98) (60) ((60 73) 8388 H ( 8388) ( 866.7) J fusion