In: Chemistry
Consider the compression at 15.6 C of 0.35 mol H2O(l) from 0.10
atm to 1.00 atm. Calculate the values of ΔG when the process is
carried out:
(a) isothermally and reversibly and (b) isothermally and
irreversibly under a constant external pressure of 1.00 atm. Take
the density of water at this temperature to be equal to 999.1
kg/m3.
0.35 mol x 18.015 g/mol / 999.1 g/L = 0.006311 L = 6.311x10-6 m3
1 atm = 101325 kg/(m.s2)
dG = -SdT + Vdp
dT = 0 (isothermal process) and V = constant and dG = dW (work)
by integration:
deltaG = V · delta p
= 6.311x10-6 m3 x 0.9 x 101325 kg/(m.s2)
= 5.755x10-1 J
For b)
Delta G = p delta V , assuming a small compressibility of liquid water.
But I don’t have further data to solve it.