From the graph of an ideal gas given in the question, the process is an isothermal expansion because volume of the ideal gas increases. Given, VB>VA and $p_{B}∵ΔT=0 (isothermal process) ∴ Change in internal energy, ΔU=0(∵ΔU∝ΔT) Since we know, in isothermal process ΔQ=ΔW=n.RTln(
Vf
Vp
) orΔQ=ΔW=nRTln(
VB
VA
) Thus, ΔQ=+ ve and ΔW=+ve because VB>VA′ (from the graph)