Question

4. The heat of vaporization (AH) of water at 100C is 40.6 k]/mol and when 1mol of water is vaporized at 100C and 1 atm pressure the volume increase is 30.19dm3. Calculate the work done by the system, the change in internal energy Δυ, the change in Gibbs energy ΔG and the entropy change AS.

Answer 1

Work done by the system (W)

Work done by the system (w) W = P Delta V = 101325 times 30.19 times 10^- 3 = 3.059 kJ The change in internal energy (Delta U) Delta U = Q - W where Q is the heat added here it is 40.6 KJ, W is work done Delta U = Q - W = 40.6 KJ - 3.059 kJ = 37.541 KJ Entropy change (Delta S) Delta S = Q/T = 40.6 kJ/373.15 K = 108.80 J k^- 1 Gibbs free energy change (Delta G) Delta G = Delta H - T Delta S when the same temperature is maintained then T Delta S = Delta H and Delta G = 0

The change in internal energy ( $\Delta U$ )

$\Delta U = Q-W$

where Q is the heat added here it is 40.6kJ, W is work done

$\Delta U = Q-W =40.6kJ - 3.059kJ = 37.541 kJ$

Entropy change ( $\Delta S$ )

Gibbs free energy change ( $\Delta G$ )

When the same temperature is maintained then

$T\Delta S = \Delta H$

and

$\Delta G = 0$