# Expansion of a gas under isothermal and reversible conditions

What is the reason for the statement - "When the expansion of an ideal gas is carried out under isothermal and reversible conditions, the internal energy does not change, i.e., $$\Delta$$U=0?"

A relation with the equation ($$\Delta$$H=$$\Delta$$U-P$$\Delta$$V) or ($$\Delta$$U = q + w) would be helpful

U = Internal Energy; H = Enthalpy; V = Volume; q = Heat gained by the system; w = Work done by the system.

• For ideal gas change in internal energy(Delta U) is given by U2-U1=nCv(T2-T1). In isothermal process T1=T2. Hence U2=U1. Aug 4, 2021 at 11:54
• Note that it applies rather to a perfect gas which is an ideal gas with the constant heat capacity. Aug 4, 2021 at 12:12
• Irreversible isothermal expansion also gives $\Delta U=0$
– Jay
Aug 14, 2021 at 4:17

Internal energy is an equilibrium physical property of an ideal gas (or any other substance), independent of any process. So it can't depend directly on q or w. Historically, it was observed experimentally that, for gases that approached ideal gas behavior, U was a function only of temperature (not pressure or volume) and that $$\Delta U$$ depended only on the temperatures of the two end states. Later, after the 2nd law of thermodynamics was developed, it was shown mathematically that, for a substance that obeys the equation of state PV=nRT, U=U(T).