Exercise 3 Solution
From Equation 8c, the condition NA + NB = 0 requires that the viscous flux cancel the non-equimolar flux: Nv = –ND.
The value of the non-equimolar flux (ND = −0.0222 ) computed in Exercise 1 applies here because the diffusive fluxes are not appreciably affected by the pressure gradient. So Nv = −0.0222
and, from Equation 2,
= 0.0222
.
This is integrated to obtain the following expression
which can be used to directly calculate the desired pressure difference. However, computations are facilitated by rewriting the left side using:
where, is the average pressure in the chamber.
The pressure drop is quite small (about 6 cm H2O) relative to the 1 × 105 Pa pressure maintained in the right-hand header, justifying the assumption that the mean pressure is closely approximated by the controlled pressure in the right-hand header.