HW6_key - HW6_Chapter 21 21-2 Q The molar heat capacity of...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: HW6_Chapter 21 21-2: Q. The molar heat capacity of H2O(l) has an approximately constant value of from 0 to 100 . Calculate if two moles of H2O(l) are heated from 10 constant pressure. to 90 at A. W e use Equation 21.9 to write 21-7: Q. The results of Problems 21-4 and 21-5 must be connected in the following way. Show that the two processes can be represented by the diagram where paths A and B represent the processes in Problem 21-5 and 21-4, respectively. Now, path A is equivalent to the sum of paths B and C. Show that is given by and that the result given in Problem 21-6 follows. A. In Problem 21-4, ethane is heated at constant volume, so (assuming ideal behavior) Likewise, in Problem 21-5, the system is kept at constant pressure and so These values correspond to those shown in the diagram. Now, path C represents an isothermal process. Because we are assuming ideal behavior, , which means that . Then we can write as (Equation 20.22) Note that path A is equivalent to the sums of paths B and C, so and , we can write . Because and the result given in Problem 21-6 follows. 21-12: Q. W hy is ? A. because gases are essentially completely unordered; the molecules of a gas travel more or less randomly within the gas’s container. Liquids, however, are much more cohesive and structured, and solids are very structured. The difference between the entropy of a liquid and that of a solid is less than the difference between the entropy of a liquid and that of a gas. 21-27: Q. Show for an ideal gas that A. W e express the partition function of an ideal gas as (Equation 17.38) and so We substitute into Equation 21.19 to write We use Stirling’s approximation and divide both sides of the equation by n to find 21-40: Q. In each case below, predict which molecule of the pair has the greater molar entropy under the same conditions (assume gaseous species). a. CO CO2 b. CH3CH2CH3 c. CH3CH2CH2CH2CH3 A. a. CO2 (more atoms) b. CH3CH2CH3 (more flexibility) c. CH3CH2CH2CH2CH3 (more flexibility) 21-42: Q. Arrange the following reactions according to increasing values of references). a. S(s) + O2(g) SO2(g) (do not consult any b. H2(g) + O2(g) H2O2(l) c. CO(g) + 3 H2(g) CH4(g) + H2O(l) d. C(s) + H2O(g) CO(g) + H2(g) A. Recall that molar entropies of solids and liquids are much smaller than those of gases, so we can ignore the contributions of the solids and liquids to when we order these reactions. Considering only the gaseous products and reactants, we can find for each reaction to be a. b. c. The correct ordering of the reactions is therefore d. . ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online