Question

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

Answer #1

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

P_{2} = 169.18 + (232.23 -
169.18)/10 x 5 = 200.705 kPa

hence, taking T_{2} = 120
degree or 393 K

T_{1} = 115 C = 388 K

P_{1} = 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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