Question

If the value of latent heat of vaporization, lvap/T for common substances is about 83.74 J/g mol K

(Trouton 's rule) and heat capacity (CP) is 41.87 J/g mol K, prove the specific heat at constant

saturation, Csat

’’’ is negative.​

Answer 1

Explanation:

Trouton's rule states that the entropy of vaporization is almost the same value, about 85–88 J/(K·mol), for various kinds of liquids at their boiling points.[1] The entropy of vaporization is defined as the ratio between the enthalpy of vaporization and the boiling temperature. It is named after Frederick Thomas Trouton.

Answer 2

Explanation:

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Class 11

>>Chemistry

>>Thermodynamics

>>Enthalpy Change of a Reaction

>>The latent heat of vapourization of wate

Question

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The latent heat of vapourization of water at 100

o

C is 540 cal.g

−1

. Calculate the entropy increase when one mole of water at 100

o

C is evaporated.

Hard

Solution

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Correct option is A)

Solution:- (A) 26cal/mol−K

As we know that,

ΔS=

T

ΔH

ΔH=540cal/g

Mol. wt. of water =18g

ΔH

(per mole)

=ΔH

(per gm)

×Mol. wt.

ΔH

(per mole)

=540×18=9720cal/mol

Now, as we know that,

ΔS=

T

ΔH

Given T=100℃=(273+100)=373K

∴ΔS=

373

9720

=26cal/(mol.K)

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