Chemistry, asked by yaswanthchowdary2003, 6 months ago

- The entropy change associated with the
conversion of 1 kg of ice at 273 K to water
vapours at 383 K is:
(Specific heat of water liquid and water
vapour are 4.2 kJ K-1 kg and 2.0 kJ K-1 kg 1;
heat of liquid fusion and vapourisation of
water are 344 kJ kg-' and 2491 kJ kg-1,
respectively).
(log 273=2.436, log 373=2.572, log 383=
2.583)
1) 7.90 kJ kg-1 K-1 2) 2.64 kJ kg- K-
3) 8.49 kJ kg- K-1 4) 9.26 kJ kg-' K-

Answers

Answered by mrityaunjaykumar5164
0

Answer: 2

Explanation:

ANSWER

ΔS=  

T

ΔH  

Transition

​  

 

​  

 

ΔS=nC ln  

T  

i

​  

 

T  

f

​  

 

​  

=ms ln  

T  

i

​  

 

T  

f

​  

 

​  

 

n= no. of moles, m= mass, C=molar heat capacity, s= specific heat capacity

Phase change path

H  

2

​  

O(s)  

ΔS  

1

​  

 

​  

H  

2

​  

O(l)  

ΔS  

2

​  

 

​  

H  

2

​  

O(l)  

ΔS  

3

​  

 

​  

H  

2

​  

O(g)  

ΔS  

4

​  

 

​  

H  

2

​  

O(g)

Δs  

1

​  

=  

273

ΔH  

fusion

​  

 

​  

=  

273

334

​  

=1.22

Δs  

2

​  

=4.2 ln  

273

363

​  

=1.31

Δs  

3

​  

=  

2491

ΔH  

vap

​  

 

​  

373=6.67

Δs  

4

​  

=2.0 ln  

373

383

​  

=0.05

Δs  

total

​  

=Δs  

1

​  

+Δs  

2

​  

+Δs  

3

​  

+Δs  

4

​  

=9.25kJkg  

−1

K  

−1

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