4.28 - An Introduction to Thermal Physics

Question

Imagine that your dog has eaten the portion of Table 4.1 that gives entropy data; only the enthalpy data remains. Explain how you could reconstruct the missing portion of the table. Use your method to explicitly check a few of the entries for consistency. How much of Table 4.2 could you reconstruct if it were missing? Explain.

Step-by-Step Solution

Answer

We can say that the missing portion of table 4.1 is reconstructed by applying the second law of thermodynamics and the table 4.2.

Step1 : Given information

Given table 4.1 

And Table 4.2

Step2: Explanation

First express the change in entropy for steam at zero temperature
$\Delta S_1=\frac{Q}{T}......\left(1\right)$
Where, Q is the heat change (absorbed or lost) and T is the temperature.

We know that the enthalpy is described as the energy that is required to produce a substance at fixed pressure.

Now write an expression of the change in entropy for water at given  temperature
$\Delta S_2=\frac{\Delta H}{T}......\left(2\right)$
Where, $\Delta H$ - change in enthalpy

Substitute Q = 2501 kJ/kg and T =273 K in equation (1), we get
$$\begin{gathered} \Delta S_1=\frac{\left(2501 \mathrm{kJ}·{\mathrm{kg}}^{-1}\right)}{(273 \mathrm{K})} \\ =9.156 \mathrm{kJ}·{\mathrm{kg}}^{-1}·{\mathrm{K}}^{-1} \end{gathered}$$
This is the same value as in Table 

Now, substitute $\Delta H=42 \mathrm{kJ}·{\mathrm{kg}}^{-1}$ and T=278 K in equation (2), we get

$$\begin{gathered} \Delta S_2=\frac{\left(42 \mathrm{kJ}.{\mathrm{kg}}^{-1}\right)}{(278 \mathrm{K})} \\ =0.151 \mathrm{kJ}·{\mathrm{kg}}^{-1}.{\mathrm{K}}^{-1} \end{gathered}$$

This value is the same as in given table 

Now write expression for the change in entropy for steam at given temperature
$\Delta S=\frac{\Delta H}{T}-nR\frac{\Delta P}{P}......\left(3\right)$
Where, n is number of moles, R is gas constant, P is average pressure and $\Delta P$ is change in pressure.

Substitute $\Delta H=19 \mathrm{kJ}·{\mathrm{kg}}^{-1}$ , T=278K, n= 55.55, R= 8.314 J/mole K,$\Delta P$ = 0.006 and P=0.009 bar in expression (3), we get

$$\begin{gathered} \Delta S=\frac{\left(19 \mathrm{kJ}·{\mathrm{kg}}^{-1}\right)}{(278 \mathrm{K})}-(55.55)(8.314 \mathrm{J}/\mathrm{mole}·\mathrm{K})\frac{(0.006\mathrm{bar})}{(0.009\mathrm{bar})} \\ =-0.239 \mathrm{kJ}·{\mathrm{kg}}^{-1}·{\mathrm{K}}^{-1} \end{gathered}$$

The  entropy for steam at finite temperature is:

$$\begin{gathered} \Delta S_1+\Delta S=\left(9.156 \mathrm{kJ}·{\mathrm{kg}}^{-1}·{\mathrm{K}}^{-1}\right)+\left(-0.239 \mathrm{kJ}·{\mathrm{kg}}^{-1}·{\mathrm{K}}^{-1}\right) \\ =8.917 \mathrm{kJ}·{\mathrm{kg}}^{-1}·{\mathrm{K}}^{-1} \end{gathered}$$

This value is very close to the value in the table. 

So we can say that the missing portion of table 4.1 is reconstructed by applying the second law of thermodynamics and the table 4.2.