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.

  • $\begingroup$ 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. $\endgroup$
    – Arpan
    Commented Aug 4, 2021 at 11:54
  • 1
    $\begingroup$ Note that it applies rather to a perfect gas which is an ideal gas with the constant heat capacity. $\endgroup$
    – Poutnik
    Commented Aug 4, 2021 at 12:12
  • $\begingroup$ Irreversible isothermal expansion also gives $\Delta U=0$ $\endgroup$
    – Jay
    Commented Aug 14, 2021 at 4:17

1 Answer 1


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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.