Question

Heat of 20 Kcal is
supplied to the system
and 8400 J of external
work done on the system
so that its volume
decreases at constant
pressure. The change in
internal energy is
(J = 4200 J/kcal
30
Ho
ng
pla​

Answer 1

Answer:

ΔQ=20×10

3

cal

=20×4200J

=84×10

3

J

ΔW=−8400J

∴ΔV=ΔQ−ΔW

=84000+8400

=92400

=9.24×10

4

Hence, 4 is the correct answer.

Explanation:

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