Question

18 gram of liquid water vapour is the 81 bore and 373 kelvin

Answer 1

Answer:

Explanation:

(i) The change : H2O (l) → H2O(g)

ΔH = ΔU + ΔngRT

or ΔU = ΔH - ΔngRT

Substituting the values, we get ΔU = 41.00 kJ/mol - 8.3 J/mol/K X 373 K/1000

= 41.00 kJ mol−1 − 3.096 kJ mol−1

= 37.904 kJ/mol

(ii) The change : H2O (l) → H2O(s)

There is negligible change in volume, so, we can put pΔV = ΔngRT ≈ 0.

So, ΔH ≅ ΔU

or ΔU = 41.00kJ mol−1

Answer 2

Answer:

heya..

83983 - If water vapour is assumed to be a perfect ... (i ) The change : H2O (l) → H2O(g) ... we get ΔU = 41.00 kJ/mol - 8.3 J/mol/K X 373 K ...