Determine the entropy of vaporization of steam at 120 oC from
the Clapeyon equation.
Hints:
1. For saturated water at 120 oC, volume of gas phase = 0.89133 m3/kg, volume of liquid phase = 0.001060 m3/kg;
2. Pressure of water in gas phase = 232.23 kPa at 125 oC
3. Pressure of water in gas phase = 169.18 kPa at 115 oC
A. 1103.4 kJ/kg K
B. 11.22 kJ/kg K
C. 5.61 kJ/kg K
D. 2.81 kJ/kg K
E. Cannot be determined
Clausius Cleypron Equation is given as:
Given,
Pressure at 115 degree C or 388 K = 232.23 K
Pressure at 125 degree C or 398 K = 169.18 K,
considering the linear variation of pressure with temperature, pressure at 120 degree or 393 K will be
P2 = 169.18 + (232.23 - 169.18)/10 x 5 = 200.705 kPa
hence, taking T2 = 120 degree or 393 K
T1 = 115 C = 388 K
P1 = 169.18 kPa
we get:
hence
dHvap = -0.17087 x 8.314 / -0.00003279
dHvap = 43.324 KJ / kg
None of the given option comes close, only 'b' can be correct.
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