# Plate Theory and Extensions - The Column Dead Volume > Page 30

Initially, (V_{M}) will be divided into two parts, that
contained in the pores (V_{p}) and that contained in the interstitial volume
(V_{I}), thus,

V_{M} = V_{I} + V_{p }_{ (30)}

_{ }

The
interstitial volume can also be divided into two parts, the interstitial volume
that is actually moving (V_{I(m)})
and that portion of the interstitial volume close to the points of contact of
the particles that is static (V_{I(s)}).

Hence,
V_{I} = V_{I(m)} + V_{I(}_{S)} (31)

If the mobile
phase is a solvent mixture, the pore contents will not be homogeneous. On
component (that with stronger interactions with the stationary phase) will be
preferentially adsorbed on the surface [10] relative to the other. Thus,
although the bulk of the pore contents (V_{p(1)}), will have the same
composition as the mobile phase, the pore contents close to the surface, (V_{p(2)}),
will have a composition that differs from the bulk mobile phase.

Hence,
V_{p }= V_{p}_{(1)} + V_{p}_{(2)} (32)

Substituting
for (V_{i}) and (V_{p}) from (31) and (32) in (30), a
more informative distribution of the mobile phase becomes apparent,

V_{M} = V_{I(m)} + V_{I(}_{S)} + V_{p}_{(1)} +
V_{p}_{(2)} (33)

The stationary
phase can be apportioned in a similar manner. For a bonded phase, as the
support is porous, some of the pores will be blocked with stationary phase and
the total amount of stationary phase can be divided into that which is
chromatographically *available* (V_{S(A)})
and that which is chromatographically *unavailable* (V_{S(U)}).

It follows that,

V_{S} = V_{S(A)} + V_{S(U)} (34)

Substituting
for (V_{M}) and (V_{S}) from equations (33) and (34) in
equation (25), an expression for the total column volume can be obtained,

V_{c} = V_{I(m)} + V_{I(s)} + V_{p}_{(1)} +
V_{p}_{(2)} + V_{S(A)} + V_{S(U)}
+V_{Si} (35)