Suppose that 2.5 mmol n2 (g) occupies 42 cm3 at 300 k and expands isothermally to 600 cm3. calculate î´gm in kj/mol for the process. hint: boyle's law p1v1=p2v2

Given:

MilliMoles of N2 = 2.5 mmol = 0.0025 moles

Initial volume V1 = 42 cm3

Final volume V2 = 600 cm3

Temperature T = 300 K

To determine:

The change in Gibbs free energy, ΔG

Explanation:

The change in Gibbs free energy is related to the ration of the volumes:

ΔG = -nRTln(V2)/(V1)

      = -0.0025*8.314*300*ln(600/42) = -16.582 J/mol

Ans: The Gibbs free energy for the process is 0.0166 kJ/mol

answer; 4 electrons;