Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Chemistry
If for a given substance, melting point is TB and freezing point is TA then correct variation of entropy by graph between entropy change and temperature is
Q. If for a given substance, melting point is
T
B
and freezing point is
T
A
then correct variation of entropy by graph between entropy change and temperature is
3190
222
Thermodynamics
Report Error
A
19%
B
59%
C
14%
D
8%
Solution:
For a pure substance
T
A
and
T
B
are same. Above this temperature entropy will increase and below this temperature entropy remains constant.