Question

4) 7.5 kesi
12. Heat of 20 Kcal is supplied to the
system and 8400 J of external work is done
on the system so that its volume decreases
at constant pressure. The change in
internal enrgy
(J = 4200 J/kcal)
1) 9.24 x 104 ) 2) 7,56 x 10
3) 8.4 x 104 ) 4) 10.5 x 104 )
44
DOCT​

Answer 1

Explanation:

ΔQ=20×10

3

cal

=20×4200J

=84×10

3

J

ΔW=−8400J

∴ΔV=ΔQ−ΔW

=84000+8400

=92400

=9.24×10

4