What is the internal energy for an isothermal process?

This is here to supplement the question about enthalpy for an isothermal process:
https://socratic.org/questions/what-is-the-enthalpy-change-for-an-isothermal-process

1 Answer
Truong-Son N.
Apr 6, 2018

#DeltaU = int_(V_1)^(V_2) ((delU)/(delV))_TdV = int_(V_1)^(V_2) -P + T((delP)/(delT))_VdV#

Now decide what gas law to use, or what #alpha# and #kappa# corresponds to your substance.

Well, from the total differential at constant temperature,

#dU = cancel(((delU)/(delT))_VdT)^(0) + ((delU)/(delV))_TdV#,

so by definition of integrals and derivatives,

#DeltaU = int_(V_1)^(V_2) ((delU)/(delV))_TdV# #" "bb((1))#

The natural variables are #T# and #V#, which are given in the Helmholtz free energy Maxwell relation.

#dA = -SdT - PdV##" "bb((2))#

This is also related by the isothermal Helmholtz relation (similar to the isothermal Gibbs' relation):

#dA = dU - TdS##" "bb((3))#

Differentiating #(3)# at constant temperature,

#((delA)/(delV))_T = ((delU)/(delV))_T - T((delS)/(delV))_T#

From #(2)#,

#((delA)/(delV))_T = -P#

and also from #(2)#,

#((delS)/(delV))_T = ((delP)/(delT))_V#

since the Helmholtz free energy is a state function and its cross-derivatives must be equal. Thus from #(3)# we get

#-P = ((delU)/(delV))_T - T((delP)/(delT))_V#

or we thus go back to #(1)# to get:

#barul|stackrel(" ")(" "DeltaU = int_(V_1)^(V_2) ((delU)/(delV))_TdV = int_(V_1)^(V_2)-P + T((delP)/(delT))_VdV" ")|#

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

#((delP)/(delT))_V = (nR)/V#

and that just means

#((delU)/(delV))_T = -P + (nRT)/V#

#= -P + P = 0#

which says that ideal gases have changes in internal energy as a function of only temperature. One would get

#color(blue)(DeltaU = int_(V_1)^(V_2) 0 dV = 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# and isothermal compressibility #kappa#,

#alpha = 1/V((delV)/(delT))_P#

#kappa = -1/V((delV)/(delP))_T#

#alpha/kappa = [ . . . ] = ((delP)/(delT))_V#

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)#

  • #kappa_(H_2O) = 4.60 xx 10^(-5) "bar"^(-1)#

  • #kappa_(Au) = 5.77 xx 10^(-7) "bar"^(-1)#
  • #kappa_(EtOH) = 1.10 xx 10^(-4) "bar"^(-1)#
  • #kappa_(Pb) = 2.33 xx 10^(-6) "bar"^(-1)#

In that case,

#((delU)/(delV))_T = -P + (Talpha)/kappa#

Thus,

#color(blue)(DeltaU) = int_(V_1)^(V_2) -P + (Talpha)/kappadV#

#= color(blue)((alphaTDeltaV)/kappa - int_(V_1)^(V_2) P(V)dV)#

Next, the pressure will vary a lot if the volume varies a little for a solid or liquid...

Hence, to find #DeltaU#, we first need to measure the temperature of the system, and reference #alpha# and #kappa#.

Then, we'll need to measure how the volume changes (to high precision!) with the change in pressure to empirically find the pressure as a function of the volume for the given solid or liquid.

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