Question

Heat of 30 kCal is supplied to a system
and 4200 J of external work is done on the
system so that its volume decreases at
constant pressure. The change in its
internal energy is (J= 4200 J/kCal)
1) 1.302 x 105)
2) 2.302 x 105 J
3) 3.302 x 105)
4) 4.302 x 105J​

Answer 1

Answer:

ΔQ=30×4200

=126000J

ΔW=−4200J

∴Δu=ΔQ−ΔW

=126000+4200J

=1.302×10

5

J

Hence,

option (A) is correct answer.

Answer 2

1) 1.302 105 J

Explanation:

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