Question

If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourisation of 1 mol of water at 1bar and 100°C is 41kJ molñ1. Calculate the internal energy change, when (i) 1 mol of water is vaporised at 1 bar pressure and 100°C. (ii) 1 mol of water is converted into ice. ( no spam )​

Answer 1

Answer:

(i) H2O(l) → H2O(g)

Δn(g) = 1 − 0 = 1

therefore, ΔH = ΔU + ΔngRT

or ∆U = ∆H − ∆ngRT

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.

∴ p∆V = ∆ngRT = 0

∴ ∆H = ∆U

Or ∆U = 41 kJ mol−¹

Explanation:

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