Question

Please help on this one?

Answer 1

OK, this is pretty interesting.

First get out your parameters.

Pressure of gas, P= 1.5×10^5 Pa

Initial volume, V1=0.0002m³

Final volume, V2=0.0006m³

Heat change, ∆H= +32J (it is positive since it was absorbed.)

Now you can solve.

∆U= ∆H—P∆V

∆U=32J—(1.5×10^5×(0.0006—0.0002))

∆U=32J—(1.5×10^5×0.0004)

∆U=32J—(1.5×10^5×4×10^—4)

∆U=32—60

∆U=—28J

That corresponds to option C.