Question 6.11.5: 1. Determine the expression of the internal energy U(T, V) o...

1. Determine the expression of the internal energy U(T, V) of the van der Waals gas, assuming that the specific heat at constant volume C_V is independent of temperature.
2. Show that during a Joule expansion, where the volume of gas increases while the system is isolated, the temperature T of the gas decreases.

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1. Using Maxwell relation (4.71) and the van der Waals equation of state (6.63), we obtain the latent entropy of expansion,

\frac{\partial p}{\partial T} = \frac{\partial S}{\partial V}.

\Bigl(p+\frac{N^2\alpha }{V^2} \Bigr)(V-N b) = N R T.

\frac{\partial S}{\partial V}=\frac{\partial p}{\partial T}=\frac{N R}{V-N b}.

Furthermore, taking into account the expression (5.11) for the entropy capacity, the entropy differential can be written,

K_V(T,V)=\frac{C_V(T,V)}{T} =\frac{\partial S(T ,V)}{\partial T}.

dS=\frac{\partial S}{\partial T} dT+\frac{\partial S}{\partial V} dV=\frac{C_V}{ T} dT+\frac{NR}{V-Nb} dV.

Using the internal energy differential,

dU=\frac{\partial U}{\partial T} \mid _V  dT+ \frac{\partial U}{\partial V} \mid _T dV.

the entropy differential (6.15) is written,

dS=\frac{1}{T} dU+\frac{p}{T} dV.

dS =\frac{dU}{T}+=\frac{p}{T}dV =\frac{1}{T} \frac{\partial U}{\partial T} \mid _V dT +\biggl(\frac{1}{T} \frac{d U}{d V}\mid _T+\frac{p}{T} \biggr)dV.

Equating the equivalent terms in both expressions of the differential dS, and taking into account the van der Waals equation of state (6.63), we find,

\Bigl(p+\frac{N^2\alpha }{V^2} \Bigr)(V-Nb)= NR T.

\frac{\partial U}{\partial T} \mid _V=C_V

\frac{d U}{d V}\mid _T=\frac{NRT}{V-Nb}-p=\frac{N^2α}{V^2}.

Substituting these expressions in the internal energy differential, dU becomes,

dU= C_V dT + \frac{N^2α}{V^2}dV.

Now, we show that the specific heat at constant volume C_V of the van der Waals gas is independent of the volume. Taking into account the first principle (1.38), the infinitesimal heat (5.4) provided to the system and the infinitesimal work (2.41) performed on the gas during a reversible process, we can rewrite the differential of the internal energy as,

dU = δW + δQ      (closed system)

δQ = TdS(T, V) ≡ C_V (T, V) dT + L_V (T, V) dV.

δW = P_W dt = −p (S, V) dV     (reversible process)

dU = δQ + δW = C_V dT + (L_V − p) dV

The Schwarz theorem applied to the internal energy U\bigl(S(T,V),V\bigr) reads

\frac{\partial }{\partial V} \biggl(\frac{\partial U}{\partial T}\mid _V \biggr)=\frac{\partial }{\partial T} \biggl(\frac{\partial U}{\partial V}\mid _T \biggr).

and gives the following Maxwell relation,

\frac{\partial  C_V }{\partial V}=\frac{\partial }{\partial T}(L_V-p).

Taking into account expression (5.10) for the coefficient L_V and the van der Waals equation of state (6.63), we obtain,

L_V(T,V)=T\frac{\partial p(T,V)}{\partial T}.

\Bigl(p+\frac{N^2\alpha }{V^2} \Bigr)(V-Nb)= NR T.

L_V -p =T \frac{\partial p(T,V)}{\partial T}-p = \frac{NR}{V-Nb}-p=\frac{N^2 α}{V^2}.

Thus, the Maxwell relation reduces to,

\frac{\partial C_V}{\partial V} = \frac{\partial }{\partial T}\Bigl(\frac{N^2\alpha }{V^2} \Bigr)=0.

This shows that the specific heat at constant volume C_V of the van der Waals gas is independent of volume. Since the van der Waals gas tends towards the ideal gas when the volume of the system is sufficiently large, the specific heat at constant volume of the van der Waals gas, which is independent of volume, has to be equal to that of the ideal gas, i.e. C_V = cNR [64]. Thus, we find that C_V is independent of temperature. Finally, we conclude by integrating the differential dU to obtain the following expression for the internal energy of the van der Waals gas,

U(T, V) = cNRT-\frac{N^2\alpha }{V}.

2. In a Joule expansion, we note the initial and final volumes V_i and V_f, respectively, and the initial and final temperatures T_i and T_m, respectively. The internal energy of an isolated system remains constant, i.e. U(T_i, V_i) = U(T_m, V_f). Therefore, during a Joule expansion, the temperature variation is related to the volume variation by,

cNR(T_m − T_i) − N^2 α \Bigl(\frac{1 }{V_f}-\frac{1 }{V_i} \Bigr) =0.

which can be recast in the form

T_m − T_i =\frac{Nα }{c R} \Bigl(\frac{V_i-V_f }{V_iV_f} \Bigr).

During a Joule expansion, the volume increases, i.e. V_f > V_i, which implies that the temperature decreases, i.e. T_m < T_i. We note that the introduction of an attractive force, characterised by the parameter a, is what accounts for the cooling. For a real gas, this represents a temperature variation of about 0.01% [41].

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