Low Pass filter with a resistor
If for a moment, you remove C1 from the circuit you are left with two resistors in series, which when you measure the voltage across R2, it is a voltage divider.

The voltage over R2 will be 
\begin{equation} 
V_{R2} = V1 * \frac{R2}{(R1+R2)}
\end{equation} , so increasing R2 will increase the voltage drop across R2.

Does this answer your question?