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,
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.
