Question

Estimatethe total entropy change when 100.g of liquid water, initially at 298 K and 0.50 atm,is first heated to boiling at 373 K and 1.00atm, and the steamcollected at 1.00 atmsubsequently heated and compressed to 500Kand20.atm. UseDHvap(at 100 ̊C, 1 atm)= 40.6 kJ/mol and the following values for the heat capacity:CP,m(H2O,l) = 75.2 J/(K mol), CP,m(H2O,g) = 35.0J/(K mol)

Answer 1

100 g of H₂O = 100/18 = 5.56 mole

We can divide this into four steps. Keeping in mind phase changes occur at constant temperature, the four steps are:

(i) due to warming liquid from 298 K or 25 ^oC (T₁) to 373 K or 100 ^oC (T₂).

(ii) due to vaporizing liquid at 100 ^oC (T₂) or 373 K

(iii) due to warming vapour from 100 ^oC or 373 K(T₂) to 500 K or 227 ^oC (T₃)

(iv) due to pressure change from 1 atm to 20 atm

Therefore, the entropy change for the first step is

The liquid is converted to steam at 373 K, another isothermal process. This time, the latent heat of vaporization or Heat of Vaporization at 100 ̊C, 1 atm = 40.6 kJ/mol = 40600 J/mol. The entropy change for this step is:

For the last two process, we will consider the Maxwell relation, dQ = TdS + VdP

According to ideal gas equation, PV = nRT, Hence V/T = nR/P. Therefore, you can write the above equation as

Gas constant R = 8.314 J.mol^-1.K^-1. Therefore, the change of entropy will be

Therefore, total entropy change will be