An ideal gas is expanded from volume V₀ to 2V₀ under three different processes. Process 2 is isothermal and process 3 is adiabatic. Let ${\mathrm{\Delta U}}_1,{\mathrm{\Delta U}}_2\text{ and }{\mathrm{\Delta U}}_3$ be the change in internal energy of the gas in these three processes. Then

Process 2 is an isothermal process
Hence, ${\mathrm{\Delta U}}_2=0$
Process 1 is an isobaric (P = constant) expansion. Hence, temperature of the gas will
increase
Or ${\mathrm{\Delta U}}_1$positive
Process 3 is an adiabatic expansion. Hence, temperature will decrease
Or ${\mathrm{\Delta U}}_3=$negative
Therefore , $,{\mathrm{\Delta U}}_1>{\mathrm{\Delta U}}_2>{\mathrm{\Delta U}}_3$.