 Number-based dissociation equilibrium ratio

For ideal-gas single-particle functions $Z_1^{(i)}=C_{i}V{T}^{3/2}$ and reaction $A\rightleftharpoons 2B$, equilibrium gives $K_N=N_B^2/N_A=(C_B^2/C_A)V{T}^{3/2}$. This particle-number ratio contains a volume factor; the concentration ratio $n_B^2/n_A=K_N/V$ does not.