Question 4.8.3: Consider a cylinder closed by two sliding pistons separated ...

Consider a cylinder closed by two sliding pistons separated by a permeable fixed wall. The cylinder contains N moles of an ideal gas passing through the wall under the effect of pistons 1 and 2 that keep the pressures p_1 and p_2 constant in the subsystems 1 and 2 on both sides of the wall (Fig. 4.9). The device is an adiabatic closed system.

1. Show that the enthalpy H is conserved if the external pressures exerted by the pistons are equal to the pressures in the corresponding subsystems at all times, i.e. p^{ext}_ 1 = p_1 and p^{ext}_ 2 = p_2. This is called the Joule–Thomson expansion.
2. For an arbitrary gas and an infinitesimal pressure difference d_p, show that the Joule– Thomson coefficient, defined as the partial derivative of temperature T with respect to the pressure p, is given by,

\frac{\partial T}{\partial p} =\frac{(T_\alpha-1)V }{C_p}.

where α is the thermal expansion coefficient, defined by,

\alpha=\frac{1}{V}\frac{\partial V}{\partial T}.

and C_p, the specific heat at constant pressure (see Chapter 5), defined as,

C_p=\frac{\partial H}{\partial T} \mid _p=T\frac{\partial S}{\partial T}.

The last equality is a consequence of the definitions (4.33) and (4.77). In these equations, the volume V and the entropy S are functions of the state variables temperature T and pressure p.

T(S,p,\left\{N_A\right\} )=\frac{\partial H(S,p,\left\{N_A\right\} ) }{\partial S }.

\frac{df}{dy}\mid _z ≡\frac{\partial f\Bigl(x(y,z),y\Bigr) }{\partial x (y,z)} \frac{\partial x(y,z) }{\partial y }+ \frac{\partial f\Bigl(x(y,z),y\Bigr) }{\partial y}.

\frac{df}{dz}\mid _y ≡\frac{\partial f\Bigl(x(y,z),y\Bigr) }{\partial x (y,z)} \frac{\partial x(y,z) }{\partial z }.

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1. According to the first law (1.38) applied to the case of an adiabatic closed system and in view of definition (2.27) for an infinitesimal work, we have,

dU=δW+δQ    (closed system)

δW=P_Wdt=-p(V)dV    (reversible adiabatic process)

dU=dU_1+dU_2=δW_1+δW_2=-p_1dV_1-p_2dV_2.

Since the pressures p_1 and p_2are kept constant,

d (U_1 + p_1 V_1) + d (U_2 + p_2 V_2) = 0.

The definition of the Legendre transform (4.29) implies that,

H = U + pV

dH_1 = d (U_1 + p_1 V_1).     and  dH_2 = d (U_2 + p_2 V_2).

Thus, the total enthalpy H is constant, i.e.

dH = dH_1 + dH_2 = 0.

2. The differential of the entropy S (T, p) is written,

dS=\frac{\partial S}{\partial T}dT+\frac{\partial S}{\partial p}dp.

Substituting for dS this expression in the enthalpy differential (4.32), we obtain,

dH = TdS + Vdp + \sum\limits_{A=1}^{r}{\mu _AdN_A}.

dH = T\frac{\partial S}{\partial T}dT+\biggl(T\frac{\partial S}{\partial p}+V \biggr)dp=0.

The Maxwell relation (4.75) and the definition of the thermal expansion coefficient α imply that,

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

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

According to the definition of the specific heat at constant pressure C_p, the enthalpy differential yields,

C_p dT + (1 − α T) Vdp = 0.

Thus, we obtain the expression of the Joule–Thomson coefficient,

\frac{\partial T}{\partial p}=\frac{(αT-1)V}{C_p}.

In the particular case of an ideal gas that satisfies the equation of state p V = NR T, the Joule–Thomson coefficient vanishes.

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