A Statistical Mechanical Derivation of the van der Waals Equation of ...

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⎛ ∂Em ⎞ ⎜ ⎟ ⎝ ∂T ⎠ p = C V ⎛ ∂p ⎞ + T⎜ ⎟ ⎝ ∂T ⎠ Vm ⎛ ∂Vm ⎞ ⎜ ⎟ ⎝ ∂T ⎠ p ⎛ ∂Vm ⎞ − p⎜ ⎟ ⎝ ∂T ⎠ p (20) Substituting equation (20) into equation (19) yields C p ⎛ ∂p ⎞ ⎛ ∂Vm ⎞ 3 1 = CV + T⎜ ⎟ ⎜ ⎟ = k B + k B (21) 2 ⎝ ∂T ⎠V ⎝ ∂T ⎠ p 2 a( Vm − b) m 1− 2 k TV B 3 m The constant-pressure heat capacity is a function of temperature and molar volume. Now let’s look at vapor-liquid equilibrium. The internal energy of vaporization is ∆E m = E V m − E L m ⎛ 1 = a⎜ L ⎝Vm 1 − V V m ⎞ ⎟ ⎠ (22) The enthalpy of vaporization is ∆H m = H V m − H L m = ∆E m + p∆V m ⎛ 1 = a⎜ ⎝V L m 1 − V V m ⎞ ⎟ + ⎠ p∆ V L ( V −V ) m m (23) The entropy of vaporization is V V L ⎛V ⎞ m − b ∆S = − = ⎜ ⎟ m S m S m k B ln L (24) ⎝ Vm − b ⎠ The molecular Gibbs free energy of vaporization is ∆G m = G V m − G L m = k B L V L ⎡ ⎛ V ⎞ ⎤ ⎛ ⎞ m − b Vm Vm 1 1 T ⎢ln ⎜ ⎟ + − ⎥ + 2a ⎜ − ⎟ (25) V V L L V ⎢⎣ ⎝Vm − b ⎠ Vm − b Vm − b⎥⎦ ⎝Vm Vm ⎠ Since the molecular Gibbs free energy of vaporization is zero at equilibrium, this gives a condition for vapor-liquid equilibrium, namely, L V L ⎡ ⎛V ⎞ ⎤ ⎛ 1 1 ⎞ m − b Vm Vm k ⎢ln ⎜ ⎟ + − ⎥ + 2 ⎜ ⎟ BT a − = 0 (26) V V L L V ⎢⎣ ⎝Vm − b ⎠ Vm − b Vm − b⎥⎦ ⎝Vm Vm ⎠ The chemical potentials should also be equal at equilibrium 4
∆µ = µ V − µ L = k B L ⎡ ⎛V ⎞ ⎤ ⎛ ⎞ m − b b b 1 1 T ⎢ln ⎜ ⎟ + − ⎥ − 2a ⎜ − ⎟ (27) V V L V L ⎢⎣ ⎝Vm − b ⎠ Vm − b Vm − b⎥⎦ ⎝Vm Vm ⎠ This expression doesn’t look the same as equation (26), but if you get a common denominator in equations (26) and (27) then it turns out that they are identical equations. So at VLE, given the temperature, we have three unknowns: the vapor molar volume, the liquid molar volume, and the vapor pressure. The two molar volumes are found by simultaneously solving equation (27) and the condition of mechanical equilibrium, namely k BT a k BT a ∆p = pV − pL = − − + = 0 (28) V 2 L 2 V − V − L m b V Vm b V m Once the molar volumes are known, the vapor pressure can be obtained by evaluating equation (8) at either molar volume. For a certain application, we may want to know the change in the molar volume as a function of temperature along the two-phase boundary of the phase diagram, also known as the saturation line. In order to obtain an analytical expression for this derivative, we must differentiate both equation (27) and (28) with respect to temperature. We will then obtain two derivatives, one for the liquid and one for the vapor. We will combine the two equations to solve for the two derivatives individually. We begin by differentiating equation (27) with respect to temperature along the saturation line. m ⎡ k BT ⎢ ⎢⎣ V + k B L m ⎡ ⎛V ⎢ln ⎜ ⎢⎣ ⎝V 1 ∂V − b ∂T L m V m L m sat − b ⎞ ⎟ + − b ⎠ V − V V m V m V 1 ∂Vm − b ∂T b − − b V L m sat − b ⎤ ⎛ 1 ⎥ + 2a⎜ − b⎥ ⎜ V ⎦ ⎝Vm V 2 ( ) ∂ L V − b T sat ( V − b) m b 2 ∂V ∂V V m ∂T V m sat + 1 − V L 2 m m b ∂V L m ∂T 2 sat ∂V L m ∂T ⎞ ⎟ = 0 ⎟ ⎠ sat ⎤ ⎥ ⎥⎦ (29) Next we differentiate equation (28) − V V L L ∂Vm k B 2 a ∂Vm k BT ∂Vm k B a ∂Vm + + + − − 2 = 0 2 2 L ∂T V − L 3 m b ∂T m m sat Vm sat (30) V V 3 ( ) ∂ − V − ∂ L V b T V sat m b V T sat ( V − b) m k B T Now we solve these two equations for the two partial derivatives. Take equation (30) and combine like terms. ⎛ ⎜ ⎜ ⎝ k T a ⎞ V L ∂V ⎛ k T a ⎞ m B ∂Vm k B k B + 2 ⎟ + ⎜ − ⎟ = − (31) m ⎠ ⎝ m m ⎠ − B 2 2 3 m V V 3 ( V b) V ⎟ T ⎜ L 2 L L V − ∂ ( V b) V ⎟ ∂T Vm − b Vm − b sat − sat 5
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