1. Examples of Chapter 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
= J kilomole-1K-1
cv (50K) = 19.4 x 105 J kilomole-1K-1 x (50K/320K)3
= 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