¿How does pressure decrease in an isothermal process when heat is transfered? Let's suppose there is a gas(ideal) inside a cilinder with a frictionless, massless piston. The gas is under a pressure $P_{1}$ and has volume $V_{1}$ at a temperature $T$. We carry out a isothermal process (I have no idea how to conduct one, but i'm assuming I do). Here's a picture. Heat is transfered into the system.

Here's my question. What I understand is that for the piston to start moving, more force is required to defeat the initial pressure. Therefore, the pressure increases so does the volume. I kwow I'm wrong because according to the state equation $PV=RTN$ pressure should decrease. I hope you could provide me a deepest insight about what is happening at the molecular level to understand this process.
 A: In the picture you've just drawn, pressure is constant for the entire process. Why? Well, the forces on the piston are $PA$ and $-mg$, and they have to sum to zero for the piston not to accelerate off either up or down. $A$, $m$, and $g$ don't change, so $P$ doesn't change either. Then the only way for V to change is for T to increase. So you haven't drawn an isothermal process.
But let's pretend you did draw an isothermal process. Then $T$ is constant, so either $P$ decreases and $V$ increases or vice-versa. Let's consider what has to happen to increase $V$, decrease $P$, and keep $T$ constant.
First: if we're going to increase $V$, the gas is going to do work on the environment. So, we need to supply some heat $Q$ which is exactly equal to the work done. So we're going to heat this container during this process, and carefully control the heat to keep $T$ constant. Alternatively, we're going to perform this process VERY SLOWLY, and allow the gas time to gain heat from the environment. Second, we need the gas to expand. How do we do that? We have to decrease the external force on the piston. We either take some weight off the top of the piston (decrease m in your picture), or grab the piston and pull up. In general, you're correct in saying that the pressure of $P$ is not going to make the gas expand without us doing something.
What I'm trying to get across is that these processes don't happen spontaneously. The ideal gas law has a lot of constants that can vary in a lot of different ways; if you want a specific process to happen (e.g. adiabatic, isothermal, isobaric, etc) you have to do something VERY SPECIFIC to the gas. In this case, you need to simultaneously provide heat AND pull on the the piston to create an isothermal process. You asked why the container would expand if $P$ was decreasing; the answer is, $P$ decreased because YOU did something to expand the container.
A: Initially in isothermal process that happens at temperature $T$, system's pressure is in mechanical equilibirium with external pressure of weight so no movement of piston. As heat is added that changes enthalpy of system which converted in work done and as heat is allowed to expand, there is no change in internal energy so, $P_iV_i=P_fV_f$.
This means that pressure of system changes in expansion or work done in isothermal process. So how does system's pressure is decreasing then at mechanical equilibirium, $P_f=P_{ext}$. But adding of little heat changes equilibrium by $P_f+\delta P=P_{ext}$ or $P_f=P_i-\delta P$, where $P_{ext}=P_i$. Each final value becomes initial value for next step.
This can be achieved by gradually removing weight or external pressure from the piston as surrounding. So with each step, internal pressure comes at mechanical equilibirium with external pressure or system is at equilibirium with surrounding. This equilibirium are quasi-static and make this process as reversible.
Pressure at each equilibirium by system is given by, $P=\frac{nRT}{V}$ at the value of volume. As work done at given pressure is, $dW=PdV=\frac{nRTdV}{V}$. Integrating it upto final volume from initial volume gives total work done in isothermal expansion as, $W=nRT\ln \frac{V_f}{V_i}$
