Question

What will be the change in entropy if 400gm of water is evaporated at a temperature of 100°C? The latent heat of water evaporation is 2260000J/kg. ​

Answer 1

Answer:

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)

Answer 2

Answer:

26 cal/(mol.K)

Explanation:

Solution:- (A) 26 cal/mol−K

As we know that,

ΔS=TΔH

ΔH=540 cal/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=TΔH

Given T=100℃=(273+100)=373K

∴ΔS=3739720

=26 cal/(mol.K)