Question

If water vapour is assumed to be a perfect gas, molar enthalpy change for vaporization of 1 mol of water at 1 bar and 100°C is 41 kJ mol^-1 Calculate the internal energy change, when: 

(i) 1 mol of water is vaporized at 1 bar pressure and 100°C. 

(ii) 1 mol of water is converted into ice.

Answer 1

(i) H₂O(l) → H₂O(g)

Δn_(g) = 1 − 0 = 1

\(\because\) ΔH = ΔU + Δn_gRT

or ∆U = ∆H − ∆n_gRT

Given, ∆H = 41 kJ mol⁻¹,

T = 100 + 273 = 373 K

\(\therefore\) ∆U = 41 kJ mol⁻¹ − 1 × 8.3 J mol⁻¹K⁻¹ × 373 K

= 37.904 kJ mol⁻¹

(ii) H₂O(l) → H₂O(s)

There is negligible change in volume.

\(\therefore\) p∆V = ∆n_gRT = 0

\(\therefore\) ∆H = ∆U

Internal energy ∆U = enthalpy of fussion of ice.