NEET Class neet Answered
the enthalpy change for the transition of liquid water to steam is 40KJ at 27degree centegrade; the change for the process would be
Asked by mistymouni2507 | 30 Oct, 2023, 10:27: AM
Expert Answer
Dear Student,
The transition under consideration is:
H2O (l) → H2O (g)
We know that, ΔSvapour = ΔHvapour / T
Given, ΔHvapour = 40 kJ mol−1
= 40 ⨯ 1000 J mol
T = 27 + 273 = 300 K
Thus, ΔSvapour = (40 ⨯ 1000) / 300 = 133.33 J K−1 mol−1
Answered by | 31 Oct, 2023, 02:33: PM
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