Given,
V1=30dm3V2=45dm3P=80kPa
Using, first law of thermodynamics,
ΔU=Q+W
Q is the heat given or lost
ΔU is the change in internal energy
W is the work done
ΔU=0 : Process is isothermal
Q=−W
W=−PextΔV: Irreversible
W=−Pext(V2−V1)
=−80×103(45−30)×10−3
=−1200J
Hence, the amount of heat transferred is 1200J