Question 11.11: A new substance has the following equations of state corresp...

For the saturation tables, let A = A_1 – A_2/T_\text{sat} and B = B_1 – B_2/T/, then p_\text{sat}= \text{exp}[A] and

p = \frac{RT}{v} – (\frac{T}{v^2})\text{exp }[B]

then

(\frac{dp}{dT})_\text{sat} = (\frac{A_2}{T^2_\text{sat}} )\text{exp }[A]

so that

T_\text{sat}(\frac{dp}{dT})_\text{sat}  –  p_\text{sat} = T_\text{sat} (\frac{A_2}{T^2_\text{sat}} )\text{exp }[A] – \text{exp }[A] = (\frac{A_2}{T_\text{sat}} -1) \text{exp }[A]

and

\int_{v_0}^{v_g} [T_\text{sat} (\frac{dp}{dt})_\text{sat})  –  p_\text{sat}] dv =\left\{(\frac{A_2}{T_\text{sat}} -1) \text{exp }[A]\right\} (v_g − v_0)

Since Eq. (b) must be valid to the saturated vapor line, it can be solved to find v_g. For this problem, Eq. (b) is a quadratic equation in v_g with the following solution:

v_g = \frac{R T_\text{sat}}{2p_\text{sat}} + \sqrt{(\frac{R T_\text{sat}}{2p_\text{sat}})^2 – \frac{T_\text{sat}}{p_\text{sat}}\text{exp }[B] }

From Eq. (c) and the preceding result for v_g, we can now calculate v_{fg} = v_g – v_f.

Then, with c^0_v equal to a constant (D_1), we get

u_g − u_0 + \int_{T_0}^{T_\text{sat}} [T (\frac{dp}{dT})_\text{sat})  –  p] dv = u_0 +c^0_v (T_\text{sat} – T_0) +\left\{(\frac{A_2}{T_\text{sat}} -1) \text{exp }[A]\right\} (v_g − v_0)

then, with vg from the preceding equation, we can easily find h_g = u_g + p_\text{sat} v_g.

To find the saturated vapor entropy, we need to determine

\int_{v_0}^{v_g}(\frac{dp}{dt})_\text{sat} dv = \int_{v_0}^{v_g}(\frac{A_2}{T^2_\text{sat}}) \text{exp }[A] dv = \left\{(\frac{A_2}{T^2_\text{sat}}) \text{exp }[A] \right\} (v_g − v_0)

Then, with c^0_v equal to a constant, we get

s_g = s_0 + \int_{T_0}^{T_\text{sat}} \frac{c^0_v}{T} dT + \int_{v_0}^{v} (\frac{dp}{dt})_\text{sat} dv

= s_0 + c^0_v  \text{ln} (\frac{T_\text{sat}}{T_0} ) + \left\{(\frac{A_2}{T^2_\text{sat}} )\text{exp }[A]\right\} (v_g − v_0)

Now, we can determine the remaining saturated liquid properties as

h_f = h_g −h_{fg} = h_g −T_\text{sat} v_{fg} (\frac{dp}{dt})_\text{sat} = h_g −v_{fg} (\frac{A_2}{T_\text{sat}})\text{exp }[A]

u_f = u_g − u_{fg} = u_g − (h_{fg} − p_\text{sat} v_{fg}) = u_g \left\{v_{fg} (\frac{A_2}{T_\text{sat}})  \text{exp }[A] –  p_\text{sat} v_{fg}\right\}

and

s_f = s_g − s_{fg} = s_g − \frac{h_{fg}}{T_\text{sat}} = s_g − v_{fg}  \text{exp }[A] (\frac{A_2}{T^2_\text{sat}})

Inserting the empirical values for the constants A_1 through D_1 into these equations provides the desired set of property relations for this material in the saturated region.