# Exercise 3 Solution

From Equation 8c, the condition *N*_{A}* + N*_{B}* = *0 requires that the viscous flux cancel the non-equimolar flux: *N*^{v}* = **–**N*^{D}*.*

The value of the non-equimolar flux (*N*^{D} = −0.0222 ) computed in Exercise 1 applies here because the diffusive fluxes are not appreciably affected by the pressure gradient. So *N*^{v} = −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 H_{2}O) relative to the 1 × 10^{5} 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.