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

If â–³Hvaporisation of substance X(l) (molar mass : 30 g/mol) is 300 J/g at it's boiling point 300 K, then molar entropy change for reversible condensation process is

A
30J/molK1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
60 J/molK1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
30 J/molK1
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 30 J/molK1
Vapurisation and condensation are reversible process,
Enthalpy of vapourisation = 300 J/g
Enthalp of condensation = 300 J/g
Molar enthalpy of condensation = 300×30 g/mol
Molar entropy, S=HcondensationT
S=300×30300
=30 J/mol/K

flag
Suggest Corrections
thumbs-up
21
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Second Law of Thermodynamics
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon