When work is done on a piston to compress the air quasi-statically but an amount of heat equal to the work done on the air is also removed from air, how is the work done isothermal in this situation? Please explain.
Thermodynamics – Understanding Isothermal Process and Work Done
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Best Answer
For an ideal gas, internal energy is only a function of temperature: $$U = c_v T.$$ Furthermore, the change in internal energy is equal to the work done on the system plus the heat absorbed from the surroundings: $$dU = \delta W + \delta Q.$$ Therefore, if $\delta Q = - \delta W$, the internal energy remains constant, which means the temperature remains constant as well.