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
0 votes
69 views
in Chemistry by (82.0k points)
closed by
Enthalpy is equal to
A. `T^(@)[(del(G//T))/(delT)]_(P)`
B. `-T^(2)[(del(G//T))/(delT)]_(P)`
C. `T^(2)[(del(G//T))/(delT)]_(V)`
D. `-T^(2)[(del(G//T))/(delT)]_(V)`

1 Answer

0 votes
by (84.4k points)
selected by
 
Best answer
Correct Answer - B
`G=H-TS` . . . (i)
`G=U+PV-TS` ltbr `DeltaG=DeltaU+PDeltaV+VDeltaP-TDeltaS-SDeltaT`
From the first and second laws,
`T DeltaS=DeltaU+PDeltaV`
`thereforeDeltaG=VDeltaP-SDeltaT`
At constant `DeltaP=0`
`(DeltaG)/(DeltaT)=S` . . .(ii)
From eq.s (i) and (ii)
`G=H+T(DeltaT)/(DeltaT)`
or `G=H+T((delG)/(delT))_(P)`
`-(H)/(T^(2))=-(G)/(T^(2))+(1)/(T)((delG)/(delT))_(P)`
`=[(del(G//T))/(delT)]_(P)`
`H=-T^(2)[(del(G//T))/(delT)]_(P)`.

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.

Categories

...