Why is change in enthalpy zero for isothermal processes?

The CHANGE in enthalpy is zero for isothermal processes consisting of ONLY ideal gases.

For ideal gases, enthalpy is a function of only temperature. Isothermal processes are by definition at constant temperature. Thus, in any isothermal process involving only ideal gases, the change in enthalpy is zero.

The following is a proof that this is true.

From the Maxwell Relation for the enthalpy for a reversible process in a thermodynamically-closed system,

`dH = TdS + VdP`, `" "bb((1))`

where `T`, `S`, `V`, and `P` are temperature, entropy, volume, and pressure, respectively.

If we modify `(1)` by infinitesimally varying the pressure at constant temperature, we get:

`((delH)/(delP))_T = T((delS)/(delcolor(red)(P)))_(color(red)(T)) + Vcancel(((delP)/(delP))_T)^(1)` `" "bb((2))`

Now, examine the entropy term, which changes due to the change in pressure at constant temperature.

The Gibbs' free energy is a function of temperature and pressure from its Maxwell Relation for a reversible process in a thermodynamically-closed system:

`dG = -SdT + VdP` `" "bb((3))`

Since the Gibbs' free energy (as with any thermodynamic function) is a state function, its cross-derivatives are equal

`((delS)/(delP))_T = -((delV)/(delT))_P`, `" "bb((4))`.

Utilizing `(4)` in `(2)`, we get:

`color(green)(bar(|ul(" "((delH)/(delP))_T = -T((delV)/(delT))_P + V" ")|))` `" "bb((5))`

This relation, which is entirely general, describes the variation of the enthalpy due to a change in pressure in an isothermal process.

The ideality assumption comes in when we use the ideal gas law, `bb(PV = nRT)`.

Thus, `V = (nRT)/P`, and `(5)` becomes:

`color(blue)(((delH^"id")/(delP))_T) = -T(del)/(delT)[(nRT)/P]_P + (nRT)/P`

`= -(nRT)/P cancel((d)/(dT)[T]_P)^(1) + (nRT)/P`

`= color(blue)(0)`

Thus, we have shown that for ideal gases at constant temperature, their enthalpy does not change. In other words, we've shown that for ideal gases, the enthalpy is only a function of temperature.