# What is the internal energy for an isothermal process?

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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

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

#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

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

#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

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

From

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

and also from

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

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

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

or we thus go back to

#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** **isothermal compressibility**

#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

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.