An ideal gas is undergoing an isothermal expansion against a constant external pressure. Which of the following statements is true?

An ideal gas is undergoing an isothermal expansion against a constant external pressure. Which of the following statements is true?

1) This process is reversible.

2) This process is irreversible.

3) This process is exothermic.

4) The system pressure is always in equilibrium with the external pressure.

5) The information given is insufficient to describe this process.

(source)

I don't get how the answer can be 2, I thought it would be 5. Since it's expansion, $$\Delta S$$ of the system must be > 0. Now since it's an expansion, work must be < 0. If work is less than 0, q > 0 due to the process being isothermal. So that means $$\Delta S$$ of the surroundings < 0 since it is -q/T. That means $$\Delta S$$ of the universe could be > 0, < 0 (non spontaneous) and = 0 depending on the magnitude of $$\Delta S$$ of the system and surroundings. As a result, I choose 5 since we don't know the magnitudes of both $$\Delta S$$.

Why would it be 2?

If the gas is ideal, for an isothermal expansion

$$p=\frac{nRT}{V}=\frac{\textrm{constant}}{V}$$

As the volume of the system changes, so must the pressure.

Now a condition of reversibility is mechanical balance between the system and surroundings, that is, $$p=p_{\textrm{ext}}$$. However, if a process is performed at constant external pressure and temperature, then as shown above the condition of mechanical balance cannot be satisfied, since $$p$$ will change during the expansion whereas $$p_{\textrm{ext}}$$ is constant. Therefore the process must be irreversible.

If the temperature and pressure are constant, then the change in entropy of the system is $$\Delta S_{syst}=nR\ln{(V_2/V_1)}$$For constant temperature, the change in internal energy is zero, so the heat transfer from the surroundings is $$Q=P_{ext}(V_2-V_1)=\frac{nRT}{V_2}(V_2-V_1)=nRT\left(1-\frac{V_1}{V_2}\right)$$It then follows that the change in entropy of the surroundings is $$\Delta S_{surr}=-\frac{Q}{T}=-nR\left(1-\frac{V_1}{V_2}\right)$$So the total entropy change of the system and surroundings in this process is: $$\Delta S=-nR\left[\ln{(V_1/V_2)}+\left(1-\frac{V_1}{V_2}\right)\right]$$ This entropy change is positive for all non-unity values of the volume ratio. The process is therefore irreversible.