thermodynamics - Calculation of entropy for an isothermal irreversible expansion

How do I calculate the entropy change for an irreversible isothermal expansion against $p_\text{ext}=p_0$?

For a reversible isothermal expansion we know $\delta S_\text{uni}=0$ so $\delta S_\text{sys}=-\delta S_\text{surr}=nR\ln\frac{V_\mathrm f}{V_\mathrm i}$

However I can't figure out why $\delta S_\text{sys}=nR\ln\frac{V_\mathrm f}{V_\mathrm i}$ remains for irreversible expansion.

I also found out an equation for irreversible expansion $\delta S_\text{uni}=q_\text{rev}/T-q_\text{irrev}/T$ Is it correct? If yes? Can you explain how do we get it?

Edit

Is this correct ?

$\delta S_\text{uni}=\delta S_\text{sys}+\delta_\text{surr}$

$\delta S_\text{sys}=\frac{q_\text{rev}}{T}$

$\delta S_\text{surr}=-\frac{q_\text{irrev}}{T}=p_\text{ext}\delta V/T$