Show that for a van der Waals gas, #((delC_V)/(delV))_T = 0#, where #C_V = ((delU)/(delT))_V#?

Question

Caleb Adams · Accepted Answer

The constant-volume heat capacity was, by definition,

#C_V = ((delU)/(delT))_V#,#" "" "bb((1))#

where #U# is the internal energy, and #T# and #V# are temperature and volume, respectively, as defined in the ideal gas law and other gas laws.

To show that for a van der Waals gas, the constant-volume heat capacity does not change due to a change in volume at a constant temperature, i.e. #((delC_V)/(delV))_T = 0#, we first write out the van der Waals (vdW) equation of state:

#bb(P = (RT)/(barV - b) - a/(barV^2) = (nRT)/(V - nb) - (an^2)/(V^2))#, #" "" "bb((2))#

where #barV = V/n# is the molar volume, #R# is the universal gas constant, #a# and #b# are the vdW constants for intermolecular forces of attraction and excluded volume (respectively), and #P# is the pressure.

This means we'll need to have an expression for #dU# that includes #P# somehow, or even a partial derivative.

For the internal energy of a closed system going through a reversible process, we can apply the Maxwell relation:

#dU = TdS - PdV#,

If we divide by #(delV)_T# for the entire equation, we get:

#((delU)/(delV))_T = T((delS)/(delV))_T - Pcancel(((delV)/(delV))_T)^(1)# #" "" "bb((3))#

This will be the main expression we'll work with. Note that #T# and #V# are the natural variables of the Helmholtz free energy, #A#, whose Maxwell relation is:

#dcolor(red)(A) = -Scolor(red)(dT) - Pcolor(red)(dV)#

Using the fact that the Helmholtz free energy is a state function (just like #G#, #H#, #S#, etc), there is a cyclic relationship we can use:

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

Plugging into #bb((3))# gives:

#((delU)/(delV))_T = T((delP)/(delT))_V - P# #" "" "bb((4))#

which is something we can relate back to the vdW equation of state, using #((delP)/(delT))_V#. We're almost there.

Given that #U# is a state function (and this is very important!), the order of partial differentiation does not matter:

#(del)/(delV)[((delU)/(delT))_V]_T = (del)/(delT)[((delU)/(delV))_T]_V#

or

#((del^2U)/(delVdelT))_(V,T) = ((del^2U)/(delTdelV))_(T,V)#

This will become relevant if we take the partial derivative of #C_V# with respect to #V# at constant #T# and plug in the definition of #C_V# from #bb((1))#:

#((delC_V)/(delV))_T = (del)/(delV)[C_V]_T#

#= (del)/(delV)[((delU)/(delT))_V]_T#

#= (del)/(delT)[((delU)/(delV))_T]_V##" "" "bb((5))#

Now consider the right side of #bb((5))# and plug in #bb((4))#:

#((delC_V)/(delV))_T#

#= (del)/(delT)[T((delP)/(delT))_V - P]_V#

#= (del)/(delT)[T((delP)/(delT))_V]_V - ((delP)/(delT))_V# #" "" "bb((6))#

Now, we should figure out what #((delP)/(delT))_V# is so we can use it! Plugging in the vdW equation of state from #bb((2))#:

#((delP)/(delT))_V = (del)/(delT)[(nRT)/(V - nb) - (an^2)/(V^2)]_V#

#= (nR)/(V - nb)#

Plugging this back into #bb((6))#, we get:

#color(blue)(((delC_V)/(delV))_T) = (del)/(delT)[(nRT)/(V - nb)]_V - (nR)/(V - nb)#

#= (nR)/(V - nb) - (nR)/(V - nb)#

#= color(blue)(0)#

Axel Alvarado · Answer

undefined To show that for a van der Waals gas, $\left(\frac{\partial C_V}{\partial V}\right)_T = 0$, where $C_V = \left(\frac{\partial U}{\partial T}\right)_V$, we start with the expression for $C_V$ and use the Maxwell relation.

Given $C_V = \left(\frac{\partial U}{\partial T}\right)_V$, where $U$ is the internal energy and $T$ is the temperature, we can use the Maxwell relation:

$$ \left(\frac{\partial C_V}{\partial V}\right)_T = \left(\frac{\partial}{\partial V}\left(\frac{\partial U}{\partial T}\right)_V\right)_T $$

By the exactness of the differential $dU$, we can interchange the order of differentiation:

$$ \left(\frac{\partial}{\partial V}\left(\frac{\partial U}{\partial T}\right)_V\right)_T = \left(\frac{\partial}{\partial T}\left(\frac{\partial U}{\partial V}\right)_T\right)_V $$

Using the Maxwell relation for $ \left(\frac{\partial U}{\partial V}\right)_T $, we have:

$$ \left(\frac{\partial}{\partial T}\left(\frac{\partial U}{\partial V}\right)_T\right)_V = \left(\frac{\partial}{\partial T}\left(P - T\left(\frac{\partial P}{\partial T}\right)_V + \frac{a}{V^2}\right)\right)_V $$

Expanding and differentiating, we get:

$$ \left(\frac{\partial}{\partial T}\left(P - T\left(\frac{\partial P}{\partial T}\right)_V + \frac{a}{V^2}\right)\right)_V = \left(\frac{\partial P}{\partial T}\right)_V - T\left(\frac{\partial^2 P}{\partial T^2}\right)_V $$

Now, for a van der Waals gas, $ \left(\frac{\partial P}{\partial T}\right)_V = 0 $ and $ \left(\frac{\partial^2 P}{\partial T^2}\right)_V = 0 $, thus:

$$ \left(\frac{\partial C_V}{\partial V}\right)_T = 0 $$