Answer:
The tabulated values at 120 C of <u><em>hfg = 2202.1</em></u><u> kJ/Kg </u>and <u><em>sfg </em></u><u>= 5.6 </u>
<u>kJ/Kg.K</u>
Explanation:
According to Clapeyron Equation :
![h_{fg}=TV_{fg}\left (\frac{\partial P}{\partial T} \right )_{sat}](https://tex.z-dn.net/?f=h_%7Bfg%7D%3DTV_%7Bfg%7D%5Cleft%20%28%5Cfrac%7B%5Cpartial%20P%7D%7B%5Cpartial%20T%7D%20%20%5Cright%20%29_%7Bsat%7D)
![=T(V_{g}-V_{f})_{120^{0}}\left ( \frac{\Delta P}{\125^{0}-115^{0}} \right )](https://tex.z-dn.net/?f=%3DT%28V_%7Bg%7D-V_%7Bf%7D%29_%7B120%5E%7B0%7D%7D%5Cleft%20%28%20%5Cfrac%7B%5CDelta%20P%7D%7B%5C125%5E%7B0%7D-115%5E%7B0%7D%7D%20%5Cright%20%29)
Taking P2 at temperature 125 and P1 at 115 degree Celsius .
T = 120 + 273.15 = 393.15 K
P2 = 232.23 kPa and P1 = 169.18kPa
(these are experimental values given in the Data)
![= 393.15(0.88133 -0.001060\ m^{3}/Kg)\frac{232.23-169.18 kPa}{10}](https://tex.z-dn.net/?f=%20%3D%20393.15%280.88133%20-0.001060%5C%20m%5E%7B3%7D%2FKg%29%5Cfrac%7B232.23-169.18%20kPa%7D%7B10%7D)
![h_{fg} = 2206.8 kJ/Kg](https://tex.z-dn.net/?f=h_%7Bfg%7D%20%3D%202206.8%20kJ%2FKg)
![S_{fg}=\frac{h_{fg}}{T}](https://tex.z-dn.net/?f=S_%7Bfg%7D%3D%5Cfrac%7Bh_%7Bfg%7D%7D%7BT%7D)
![\frac{2206.8}{393.15}](https://tex.z-dn.net/?f=%5Cfrac%7B2206.8%7D%7B393.15%7D)
= 5.6131 kJ/Kg.K
The tabulated values at 120 C of <em>hfg = 2202.1</em> kJ/Kg and <em>sfg </em>= 5.6131
kJ/Kg.K