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
58 views
in Chemistry by (80.6k points)
closed by
There is `1 mol` liquid (molar volume `100 ml`) in an adiabatic container initial, pressure being `1` bar Now the pressure is steeply increased to `100` bar, and the volume decreased by `1 ml` under constant pressure of `100` bar. Calculate `Delta H` and `Delta E`. [Given `1 "bar"=10^(5)N//m^(2)`]

1 Answer

0 votes
by (80.9k points)
selected by
 
Best answer
Correct Answer - `Delta H=990, Delta E=19 J`
`Delta U= q+W`
for adiabatic process `q=0`, hence `Delta U=W` and `W= -p(Delta V)= -P(V_(2)-V_(1))`
Now `Delta H= Delta U+Delta (PV)`
Here `Delta U` already calculated abobe and
`Delta PV=(P_(2)V_(2)-P_(1)V_(1))`
So, `Delta H=100+(100xx99-1xx100)=9900 ` bar `mL=990J`

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

...