Question

Melting and boiling points of a substance are 150 K and 250 K respectively. Cp values for solid-phase is 1.5 J/mol-K at 15 K and 20 J/mol-K for greater than 15 K. For liquid phase Cp=0.2T 5 J/mol-K. For gas-phase Cp== 50+ 5* 10-3T J/mol-K. Enthalpies of freezing and condensation are 7.5 kJ/mol and 25 kJ/mol respectively. Determine specific entropies of the substance in J/mol-K at 200 K and 400 K. The temperature T in Cp equations is in K.

Answer 1

30

Explanation:

T(m) = 150K

T(b) = 250K

Entropy at 0K is zero.

ds = da/T

For T < 15K, C = 1.5J/molK

dq = CdT

ds [range 0 to s1] = CdT/T [range 0 to 15]

For 15 < T < 150

ds [range s1 to s2] =  CdT/T [range 15 to 150]

s2 - s1 = 20 * en(T2/T1)

   = 10 en(150/15)

For T = 150K, s = a/T = 7.5 * 10^3 / 150 = 30