How can the piston be pushed outside during an isothermal expansion if the internal pressure is less than the external one?

Question

Let us say we have an ideal gas inside a cylinder closed with a piston. We put the cylinder over a heat bath at temperature $T_{1}$ and allow the gas to expand isothermally as bellow (we ignore gravity).

We assume that initially the pressure on both sides of the piston is $P_0$.

Now, since we have an ideal gas, during each step of the process we must have $$PV=NkT,$$ and since we're considering an isothermal expansion, the equation reduces to $$PV=const.$$ This equation tells us that as the gas is expanding, the pressure inside the cylinder $P$ is decreasing, thus we will have $$P<P_{0}.$$ But the external pressure $P_{ext}$ is actually increasing (or, worst case scenario, remains at the same initial value $P_{0}$), thus $$P_{ext}\ge P_{0}.$$

So how can the piston be pushed outside if the internal pressure $P$ is less than the external pressure $P_{ext}$ during each step of the process?

Answer 1

This equation tells us that as the gas is expanding, the pressure inside the cylinder 𝑃 is decreasing...

Yes the pressure inside the cylinder is decreasing, but that is because the pressure outside the cylinder is also decreasing at the same time but the decrease in external pressure is infinitesimally greater than the decrease in internal pressure so that expansion can occur.

So how can the piston be pushed outside if the internal pressure 𝑃 is less than the external pressure 𝑃𝑒𝑥𝑡 during each step of the process?

Because the internal pressure is not less than the external pressure. The internal pressure is infinitesimally greater than the external pressure.

Hope this helps.