Question

In: Chemistry

Consider the compression at 15.6 C of 0.35 mol H2O(l) from 0.10 atm to 1.00 atm....

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.

Solutions

Expert Solution

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.

queen_honey_blossom answered 2 months ago
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