Themodynamics 2Calculate the total entropy change in 100 g o

Themodynamics

2.Calculate the total entropy change in 10.0 g of water as it is converted from the liquid phase at 25 ^oC to the gas phase at 220 ^oC. For the calculation, use this data :

C_P(H₂O_(l)) = 75.29 J×K^-1×mol^-1 and DH_vap (H₂O) = +44.01 kJ×mol^-1 Treat the vapor phase as ideal, and use: C_V(H₂O_(g)) = 25.27 J×K^-1×mol^-1.

Solution

Mass of water m = 10 g

                         = 10 x10 ^-3 kg

                         = 10 ^-2 kg

                         = 0.01 kg

Initial temprature of water T = 25 o C

                                        = 25 + 273 = 298 K

Vaporisation temprature of the water T \' = 100 ^o C = 100 + 273 = 373 K

Final temprature of the water T \" = 220 ^o C = 220 + 273 = 493 K

Specific heat water C = 4186 J / kg K

Specific heat of steam C \' = 1996 J / kg K

Latent heat of vaporisation of water L = 2260x10 ³ J / kg

Total change in entropy of water dS = [mC ln(T \' / T) ]+[mL/T \'] +[mC \' ln ( T \" / T \')]

                   dS = [0.01x4186xln(373/298)] +[0.01x2260x10 ³ /373 ] +[0.01x1996xln(493/373)]

                        = 9.396 +60.58 + 5.567

                        = 75.54 J / K