1 / 8

Examples of Chapter 4

Examples of Chapter 4. Example (Problem 4.1) The specific heat capacity c v of solids at low temperature is given by the Debye T 3 law: The quantity A is a constant equal to 19.4 x 10 5 J Kilomole -1 K -1 and θ is the Debye temperature, equal to 320K for NaCl.

mei
Télécharger la présentation

Examples of Chapter 4

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Examples ofChapter 4

  2. Example (Problem 4.1) The specific heat capacity cv of solids at low temperature is given by the Debye T3 law: The quantity A is a constant equal to 19.4 x 105 J Kilomole-1K-1 and θ is the Debye temperature, equal to 320K for NaCl. • What is the molar specific heat capacity at constant volume of NaCl at 10K and at 50K? • How much heat is required to raise the temperature of 2 kilomoles of NaCl from 10K to 50K at constant volume? • What is the mean specific heat capacity at constant volume over this temperature range?

  3. Solution • The calculation is straight forward, inserting the temperature to the given equation cv (10K) = 19.4 x 105 J kilomole-1K-1 x (10K/320K)3 = 59.204 J kilomole-1K-1 cv (50K) = 19.4 x 105 J kilomole-1K-1 x (50K/320K)3 = 7400.5 J kilomole-1K-1 • Using the definition of The unit of heat is J kilomole-1K-1 (c) The mean specific heat capacity can be calculated by dividing the sum of (cv (10K)+ cv (50K) ) by 2

  4. Example 2 (problem 4.2) The equation of state of a certain gas is (P + b)v = RT and its specific internal energy is given by u = aT + bv + u0 • Find cv, • Show that cp – cv = R Solution: (a) cv can be found via its definition:

  5. (b)

  6. Example 3: (Problem 4.10): The equation of state for radiant energy in equilibrium with temperature of the walls of a cavity of volume V is P = aT4/3, where a is a constant. The energy equation is U = aT4V. (a) Show that the heat supplied in an isothermal doubling of the volume of the cavity is (4/3)aT4V. (b) Show that in an adiabatic process, VT3 is constant.

  7. Solution (a)

  8. (b) for adiabatic process dq = 0, thus du + PdV = 0

More Related