What is isothermal process?

An isothermal process has no change in temperature by the time the process is over. No matter how the temperature changes, as long as the final state is the same as the initial state, it is isothermal.

If ever you want to calculate #DeltaU# or #DeltaH# for such processes...

https://socratic.org/questions/what-is-the-internal-energy-for-an-isothermal-process
https://socratic.org/questions/what-is-the-enthalpy-change-for-an-isothermal-process

The main consequences of this are:

• The change in internal energy #DeltaU# of an ideal gas is zero.
• The change in enthalpy #DeltaH# of an ideal gas is zero.

Real gases would have significantly smaller #DeltaU# and #DeltaH# (compared to #DeltaT ne 0#), but they would not be zero.

For the internal energy we have:

#DeltaU = int_(V_1)^(V_2) ((delU)/(delV))_TdV = int_(V_1)^(V_2) [-P + T((delP)/(delT))_V]dV#

For ideal gases, #P = (nRT)/V# and #((delP)/(delT))_V = (nR)/V#, so

#color(blue)(DeltaU = int_(V_1)^(V_2) [-(nRT)/V + T((nR)/V)]dV = 0)#

For the enthalpy we have:

#DeltaH = int_(P_1)^(P_2) ((delH)/(delP))_TdP = int_(P_1)^(P_2) [V - T((delV)/(delT))_P]dP#

For ideal gases, #V = (nRT)/P# and #((delV)/(delT))_P = (nR)/P#, so

#color(blue)(DeltaH = int_(P_1)^(P_2) [(nRT)/P - T((nR)/P)]dP = 0)#

For real gases one would plug in either the van der Waals equation of state, Redlich-Kwong, Peng-Robinson, or whatever equation of state suits your gas.