Solution Manual For Introduction to the Thermodynamics of Materials, 7E David R. Gaskell David E. L

Solution Manual Introduction to the Thermodynamics of Materials, 7E David R. Gaskell David E. Laughlin Chapters 1-15 Chapter 1

Problem 1.1 The plot of V = V (P, T) for a gas is shown in Fig. 1.1. Determine. the expressions of the two second derivatives of the volume of this plot. (note: the principal 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 T and P.

 V     V  T  P T

assuming  T is constant

  2V  2  2   TV > 0  P T  V     VP  T  P

assuming  is constant

  2V  2  2    PV  0  T  P Since all terms in the expressions are positive (V, P2 and 2), both principal curvatures are positive. The surface is convex.

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Problem 1.2 The expression for the total derivative of V with respect to the dependent variables P and T is :

 V   V  dV    dP    dT  P T  T  P Substitute the values of P and  obtained in Qualitative Problem 2 into this equation and obtain the equation of state for an ideal gas.

Solution

dV    TVdP   PV dT V 1 dP  V dT P T dV dP dT   V P T

dV  

ln V  c1   ln P  c2  ln T  c3 PV  (constant)  T

The constant is nR for n moles of the ideal gas.

Problem 1.3 The 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:

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This is due to the number of components in each system. The 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.

Problem 1.4* Calculate the value of the ratio

for an ideal gas in terms of its volume.

Solution: 1

P T P R    T 1 T V P

Problem 1.5* Obtain an expression for the isobaric thermal expansion of a van der Waals gas.

Solution: We will start with the equation:

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 P   P  dP    dT    dV  T V  V T  P    T V  V   Now: we set dP  0 :     P    T  P    V T P

RT a  2 for a van der Waals gas (V  b) V

R  P  for a van der Waals' gas     T V V  b RT 2a  P   3 for a van der Waals gas    2 (V  b) V  V T  P  R R   T V  V   V b V b       P  RT 3 2 2 a RTV  2 a ( V  b )  T    P  3 2   (V  b) 2V 3  V T (V  b) V R (V  b)V 3 RTV 3  2a (V  b) 2

P 

1  V  R (V  b)V 2    V  T  P RTV 3  2a (V  b) 2

Problem 1.6* Obtain an expression for the isothermal compressibility of a van der Waals gas.

Solution:

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P

RT a  2 (V  b) V

We need to find:  T   P

1  V    V  P T

RT a  2 (V  b) V

RT 2a 2a (V  b) 2  RTV 3  P        (V  b) 2 V 3 (V  b) 2V 3  V T

T  

1  V  (V  b) 2V 2     V  P T 2a (V  b) 2  RTV 3

Note: for an ideal gas, a = b = 0: Thus: (V  b) 2V 2 V 1 T     2 3 2a (V  b)  RTV RT P

Problem 1.7* Show that the expression

P  P is valid for an ideal gas. T  V

Solution  V   V   P         1  T  P  P T  T V V  P (   TV )1( P V )1  1 1  V  R    P V  T  P P   P 1  V  1  P  T  V   RT   1 R     2      V  P T P  T V  P  P V 

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