Question 11.8: In Chapter 3, an equation of state developed in 1903 by Pier...

The Berthelot equation is given in Eq. (3.46) as

p(v −b) = RT − a(v −b)/(Tv^2)

where a and b are constants. Solving this equation for p gives

p = RT/ (v −b) −a/ (Tv^2)

a. The change in specific internal energy is given by Eq. (11.20), for which we need

(\frac{∂p}{∂T} )_v = R/ (v −b) −a/ (T^2v^2)

Then, for an isothermal process (T_1 = T_2), Eq. (11.20) gives

(u_2 − u_1)_T = \int_{v_1}^{v_2} [RT/ (v −b) −a/ (Tv^2) – RT/ (v −b) −a/ (Tv^2)]dv

= – (2a/T)(1/v_2 −1/v_1) = 2a (v_2 − v_1)/(Tv_1v_2)

b. To find the change in specific enthalpy, we could use Eq. (11.23). However, to evaluate this equation, we need to be able to determine the relation (∂v/∂T)_p. Since the Berthelot equation is not readily solvable for v = v (T, p), we choose instead to use the simpler approach, utilizing only the definition of specific enthalpy, h = u + pv. Then,

(h_2 − h_1)_T = (u_2 − u_1)_T + p_2v_2 −p_1v_1 = \frac{2a(v_2 − v_1)}{Tv_1 v_2} +p_2v_2 −p_1v_1

= \frac{3a(v_2 − v_1)}{Tv_1 v_2} +RT (\frac{v_2}{v_2 − b} – \frac{v_1}{v_1 − b} )

c. Finally, since we already evaluated the relation (∂p/∂T)_v. , we choose to use Eq. (11.25) for the isothermal specific entropy relation:

(s_2 − s_1)_T = \int_{v_1}^{v_2} (\frac{∂p}{∂T})_v dv

= \int_{v_1}^{v_2} [R/(v −b) +a/ (T^2v^2)]dv

=R  \text{ln} [(v_2 −b)/ (v_1 −b) + a (v_2 − v_1)/(T/^2v_1v_2)]