wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Calculate the entropy change accompanying the following change of state.

H2O(s,10C,1atm) H2O(l,10C,1atm)
CP for ice = 9 cal deg1mol1

CP for H2O = 18 cal deg1 mol1

Latent heat of fusion of ice = 1440 cal mol1 at 0C


A

8.88 cal deg1 mol1

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

7.77 cal deg1 mol1

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

6.3 cal deg1 mol1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C

6.3 cal deg1 mol1


The total process involves three stages and entropy change can be calculated for each stage as follows:

[ΔS1For changing 1 mole of ice from -10C,1 atm to 0C1 atm]

ΔS1=010n.CpTdT

= n × CP × 2.303 × log 273263

= 1 × 9 × 2.303 × 0.0162

= 0.336 cal deg1mol1

ΔS2 for melting 1 mol of ice at 0C will be as

ΔS2 = qrevT = 1440273 = 5.25 cal deg1mol1

ΔS3 for heating 1 mol of H2O from 0C to 10C at 1 atm

ΔS3=100nCpTdT

=1 × 18 × 2.303 log283273

= 0.647 cal deg1 mol1

ΔS = 0.336 + 5.276 + 0.647

=6.258 cal deg1 mol1


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Third Law of Thermodynamics
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon