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.
