Question

The normal melting point of substance A is 320C. Melting heat is a function of temperature. Determine the melting temperature of substance A at 10 bar.

Data :

Density of liquid A = 10200 kg/m³   

Density of Solid A = 12000 kg/m³

For liquid A C_p^L  = 30,0-3,1 x 10^-3T (J/mol.K)  

For solid A C_p^S = 20,0 + 9x10^-3T (J/mol.K)

T_ref = 320^oC     

ΔH_ref^erime = 4900 J/mol     

M_A = 250 kg/kmol

                  

                          

                                 

Answer 1

Solution:

Given Data :

at normal condition

• P1 = 1 atm = 1.01325 bar
• T1 = 320 0cel = 593 K

at High pressure

• P2 = 10 bar
• T2 = ?
• at Tref (593K) = Heat of melting = 4900 J/mol
• solid = 12000 kg/m3
• liquid = 10200 kg/m3
• Molecular Weight (MW) = 250 kg/kmol = 250 gm/mol

Now, we write Clausius - Claperyron Equation for Solid-Liquid equlibrium.

Where, = Vliq - Vsolid

   

.

= 3.6765*10-3 m3/kmol

= 3.6765*10-6 m3/mol

Put all values in Clausius - Claperyron Equation,

Integrating above equation with Boundary condition, At P = 1.01325 bar , T = 593 K & At P = 10 bar T = T2

[ 10- 1.01325] = 13327.89 * [ T2 - 593 ]

8.98675 = 13327.89*[ T2 - 6.3851 ]

T2 = 593.39 K = 320.4 0cel

Answer : Melting point at 10 bar = 320.4 0cel