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

The enthalpy of vaporization of water at 100oC is 40.63 kJ mol1. It's entropy change for the vaporization would be :

A
406.3 J K1 mol1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
108.9 J K1 mol1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
108.9 kJ K1 mol1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
4063 kJ K1 mol1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 108.9 J K1 mol1
We know that entropy change in phase transformation is given as:
Svap=HvapT
According to given data
Svap=40.63×103 J mol1373 K
so,
Svap=108.9 J K1 mol1

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon