# What is the enthalpy change for an isothermal process?

##### 1 Answer

#DeltaH = int_(P_1)^(P_2) ((delH)/(delP))_TdP = int_(P_1)^(P_2) V - T((delV)/(delT))_PdP#

Now decide what gas law to use, or what

Well, from the total differential at constant temperature,

#dH = cancel(((delH)/(delT))_PdT)^(0) + ((delH)/(delP))_TdP# ,

so by definition of integrals and derivatives,

#DeltaH = int_(P_1)^(P_2) ((delH)/(delP))_TdP# #" "bb((1))#

The natural variables are

#dG = -SdT + VdP# #" "bb((2))#

This is also related, obviously, by the well-known isothermal Gibbs relation

#dG = dH - TdS# #" "bb((3))#

Differentiating

#((delG)/(delP))_T = ((delH)/(delP))_T - T((delS)/(delP))_T#

From

#((delG)/(delP))_T = V#

and also from

#((delS)/(delP))_T = -((delV)/(delT))_P#

since the Gibbs' free energy is a state function and its cross-derivatives must be equal. Thus from

#V = ((delH)/(delP))_T + T((delV)/(delT))_P#

or we thus go back to

#barul|stackrel(" ")(" "DeltaH = int_(P_1)^(P_2) ((delH)/(delP))_TdP = int_(P_1)^(P_2)V - T((delV)/(delT))_PdP" ")|#

*And what remains is to distinguish between the last term for gases, liquids and solids...*

**GASES**

Use whatever gas law you want to find it. If for whatever reason your gas is ideal, then

#((delV)/(delT))_P = (nR)/P#

and that just means

#((delH)/(delP))_T = V - (nRT)/P#

#= V - V = 0# which says that

ideal gases have changes in enthalpy as a function of only temperature.One would get

#color(blue)(DeltaH = int_(P_1)^(P_2) 0 dP = 0)# .Not very interesting.

Of course, if your gas is **not** ideal, this isn't necessarily true.

**LIQUIDS AND SOLIDS**

These data are tabulated as **coefficients of volumetric thermal expansion**

#alpha = 1/V((delV)/(delT))_P# at VARIOUS temperatures for VARIOUS condensed phases. Some examples at

#20^@ "C"# :

#alpha_(H_2O) = 2.07 xx 10^(-4) "K"^(-1)# #alpha_(Au) = 4.2 xx 10^(-5) "K"^(-1)# (because that's REAL useful, right?)#alpha_(EtOH) = 7.50 xx 10^(-4) "K"^(-1)# #alpha_(Pb) = 8.7 xx 10^(-5) "K"^(-1)#

In that case,

#((delH)/(delP))_T = V - TValpha#

#= V(1 - Talpha)#

Thus,

#color(blue)(DeltaH = int_(P_1)^(P_2) V(1 - Talpha)dP ~~ V(1 - Talpha)DeltaP)#

since liquids and solids are very incompressible and require a large change in pressure.