Molar Entropy Of Water

Strings (SiPjAjk) = S7P5A51     Base Sequence = 12735     String Sequence = 12735 - 5 - 51

Expressions Of Pj Problems
Molar Entropy Of Water
Math The molar entropy of ice at 0oC is given as 51.84 J deg-1 mole-1.

(a) What is the molar entropy of water at 0oC?
(b) What is the molar entropy of water at 25oC ?

The strings:

S7P5A51 (Change - Physical Change)

The math:
Pj Problem of Interest is of type change (physical change). Entropy is a state variable and the problem of interest is a change in entropy (ΔS). Phase changes are physical changes hence the change is physical change. (a) Heat of fusion during the melting of one mole of ice = 6010 J deg-1
So, system absorbs 6010 J deg-1 without change in temperature.
So, entropy increases by 6010/273.15 = 22 J deg-1 mole-1
where 273.15 is from the Kelvin temperature scale.
So, molar entropy of water at 0oC = 51.84 + 22 = 73.84 J deg-1 mole-1

(b) To obtain the molar entropy of water at 250, determine increase in entropy due to increase in temperature from 0oC to 25oC (273oK to 298oK)
So increase in entropy = ΔS = ∫273298 Cp(dT/T).
where T is temperature and Cp is heat capacity at constant pressure.
So, ΔS = Cpln(298/273) = 0.0880Cp
Cp = 75.3 J deg-1 mole-1 (from commonly available heat capacity /specific heat table).
So, ΔS = 0.0880 x 75.3 = 6.63.
So, molar entropy of water at 25oC = 73.84 + 6.63 = 80.47 J deg-1 mole-1 The point . is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.
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