thermodynamics - Work done by isothermal expansion from two different viewpoints

Consider an adiabatic system as follows. It consists of a gas in a container and a piston. Initially, the system is at equilibrium and the gas inside it occupies a volume $V_i$ at a pressure $p_i$ which is equal to the outside pressure. Suddenly, the outside pressure changes and reduces to $p_{atm}$. The piston moves to equalize the pressure and the gas expands isothermally to obtain equilibrium. The gas now occupies a volume $V_f$ at a pressure of $p_f$ which is equal to $p_{atm}$

Now, two textbooks I have define the work done from two different viewpoints.

1: From the viewpoints of the surroundings:

The work done on the system by the surroundings equals $-p_{atm}\Delta V$. Since $p_{atm}$ is pretty much constant for the whole of the whole of the process, we can say that the work done equals:

$$W_1 = -p_{atm}(V_f - V_i)\tag1$$

2: From the viewpoint of the system:

We can write the internal pressure of the system as a function of its volume: $p_{in}(V)$. As during the expansion, the internal pressure changes, the work done by the system equals

$$W_2 = \int_{V_i}^{V_f}p_{in}(V)dV$$

Now, I don't know which definition to use. The work is done by the system (from definition 2) is done on the surroundings. But what about the negative work that the surroundings did on the system? Where did that energy go? Maybe, the two definitions express the same thing: the work done by the system. The negative sign in definition 1 signifies that the work is done by the system. But that would mean that

$$W_1 = W_2$$

We can simplify $W_2$ as follows. Clearly,

$$p_{in}(V) = \frac{p_iV_i}{V}$$

$$W_2 = \int_{V_i}^{V_f}\frac{p_iV_i}{V}dV = p_iV_i \ln{\frac{V_f}{V_i}} \tag2$$

The internal pressure when the volume is equal to $V_f$ is $p_{atm}$

$$\implies p_{atm} = p_{in}(V_f) = \frac{p_iV_i}{V_f}$$

$$\implies V_f = \frac{p_iV_i}{p_{atm}}$$

Putting this into $(1), (2)$ gives us,

$$W_2 = p_iV_i\ln{\frac{p_{i}}{p_{atm}}}$$

$$W_1 = -p_{atm}\left(\frac{p_iV_i}{p_{atm}} - V_i\right) = -V_i(p_{atm} - p_i)$$

Consider some values. Let $p_i = 5 \ \rm{Pa}, V_i = 1 \ \rm{m^3}, p_{atm} = 1 \ \rm{Pa}$.

$$W_1 = -1\cdot(1-5) = 4$$

$$W_2 = 1\cdot1\cdot\ln{\frac{5}{1}} = 1.609$$

Where am I making a mistake?