Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
+1 vote
2.4k views
in Chemical Thermodynamics by (81.4k points)
recategorized by

In an isobaric process, when temperature changes from T1 to T2ΔS is equal to

(a) 2.303 Cp log (T2toT1)

(b) 2.303 Cp ln (T2/T1)

(c) Cp ln (T1/T2)

(d) Cv In (T2/T1)

1 Answer

+1 vote
by (92.4k points)
selected by
 
Best answer

Correct option   (a) 2.303 Cp log (T2toT1)

Explanation:

The entropy change for a process, when T and P are the variables is given by

For an isobaric process P1= P2 Hence the above equation reduces to

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...