Question

8) Find the molar entropy change when H20 (1) at 1.00 atm and 300 K is changed to H2O (g) at 1.00 atm and 500 K. The enthalpy of vaporization of H2O is 40.66 kJ mol"! Cp [H2O(1)]=[12.16-1.943 x 10-2T +3.042 x 10-ST?]R Cpm[H,0(8)]=[3.633–1.195 x 10-T +1.34 x 10-?T?]R

Answer 1

We have conversion H0 (0) H0 (9) Book, lath 5ook, latm we have to find out molar entropy change, (As) for this conversion at constant pressure Given, - enthalpy of Vapourication. Allwer 40.66 kJ/moe Cp, on [the, 8] = [12.16 - 1.94341029 + 3,042x10-5 92] R J/mol cpim (1,0,9] - [3.633 -- 1.195 X 70-34 + 1.34 X10-? ] R 9/mol Now, we know that liquid to gas transformation occur at 373 16 for Reg molar entropy change, AS= Com [Hope]. dT + Alluap + pm [H, O, g} .dT A 3138 5ook 300K According to Novo the data provided, we get As. - 1.943x10 3 ,042-X10 J/molt & 40660 /mol 343K Stephen curry 3.633 - 1.1958 10 37 1 1348 107 4. R. dT I mol-k ET 300 500 plan e-formeninga - fisve"mate malen 15 - 16* In 20** (1 hoje * 2.02 (po - 10m* 27. + tseklo"e (! = 12.1648013) XR lauk lot x (312-00) XR + 3,042 *lo*, (373"- 300*) *R 313 + 109.088 1/moe-k + 3.633 In 500 XR - 1.195X10 [ 50 3150X R + 1.34419 [$€-578]e

- 2.648 R – 1.418R & 0.7472 & 109.088 g/molik at 1.064R – 0.152 R of 0.007 R 1,977 R + 109,088 g/molok + 0,99 R 1984 [2.896 R +109.088) J/mol k IR2 834-5/mol-k] > (21896 X 8.314 +10.9,088) olmalk [24.077 + 109.088 ) Ilmolok 133.185 g/mol.k Required men tropy change je 133.165 g/ melek Molar

Here, molar entropy change is calculated in terms of J/mol. K and thus calculations are done accordingly.