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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.

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Examples of Chapter 4

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  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

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