(Solved): diffusion relative to volume average velocityplease do A and B. thanks! 1-2. Difinsion Relative to V ...

1-2. Difinsion Relative to Volume-Average Velocity A reference frame sometimes chosen for diffusion is the volume-average velocity \( v^{(v)} \). For a binary mixture this velocity is given by \[ \left(v(v)=\mathrm{N}_{A} \bar{V}_{A}+\mathrm{N}_{B} \bar{V}_{B},\right. \] where \( \bar{V}_{A} \) and \( \bar{V}_{B} \) are the partial molar volumes of species \( A \) and \( B \), respectively. The total molar concentration is related to the partial molar volumes and mole fractions by \[ C=\left(x_{A} \bar{V}_{A}+x_{B} \bar{V}_{B}\right)^{-1} \text {. } \] The molar flux of \( A \) relative to the volume-average velocity. \( J_{A}^{(v)} \), is defined as \[ \mathbf{J}_{A}^{(v)}=\mathbf{N}_{A}-C_{A} v^{(v)} \] (a) Show that \[ \mathbf{J}_{A}{ }^{(v)} \bar{V}_{A}+\mathbf{J}_{B}^{(v)} \bar{V}_{B}=\mathbf{0} \] (b) Show that \[ \mathbf{J}_{A}^{(v)}=C \bar{V}_{B} \mathbf{J}_{A}^{(M)} \]