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
280 views
in Chemistry by (93.2k points)
closed by
The work done during the expansion of a gas from a volume of `4dm^(3)` to `6dm^(3)` against a constant external pressure of 3 atm is (1L atm = 101.32 J)
A. `+ 304 J`
B. `- 304 J`
C. `- 6 J`
D. `-608 J`

1 Answer

0 votes
by (91.8k points)
selected by
 
Best answer
Correct Answer - D
`W=-pDeltaV, W=-3xx(6-4)`
`W=-6xx101.32(therefore "1 L atm=101.32 J")`
W=-608 J

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

...