Why change in entropy in reversible adiabatic process is zero?

The temperature also decreases due to the system doing work on the surrounding. So you have more volume but less energy throughout that volume. Entropy is not only concerned with spatial configurations.

For an ideal gas $PV = n RT$, the expression of entrop can be calculate:

begin{align} dS =& frac{dU}{T} + frac{PdV}{T}; =& n C_v frac{dT}{T} + nRfrac{dV}{V}; end{align}

Carry out the integal from state 1 to state 2: begin{align} S_2 - S_1 =& n C_v left(ln T_2 - ln T_1right) + nR left(ln V_2 - ln V_1right); =& nC_v ln frac{T_2}{T_1} + nRlnfrac{V_2}{V_1}. tag{1} end{align}

In an adiabatic expanion, $V_2 > V_1$, the second term in Eq.(1) is positive; but the temperature becomes lower $T_2 < T_1$, the first term is negative. It is a good practice to show explicitly that these tow terms cancel each other in an adiabatic expansion process.

In classical statistical mechanics the entropy is related to the volume of phase space, that is the space of the positions and momentum. The volume of this space is related to the number of microstates, and it's proportional to the volume of the container, but also to the volume in momentum space. In this case the space volume increases, but the temperature (that is a measure of the mean kinetic energy) decreases, affecting the momentum space.