# Why is change in enthalpy zero for isothermal processes?

##### 1 Answer

*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

#((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

#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,

#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.