Work done by gas in isothermal process
The compression of the air outside the cylinder by the $\ce{N2}$ escaping from the cylinder takes place reversibly (as far as the air is concerned), but the process of $\ce{N2}$ leaking out of the cylinder through a valve or small orifice is definitely irreversible because of the Joule–Thompson throttling that takes place as the $\ce{N2}$ flows from a high pressure inside the cylinder to a much lower pressure outside the cylinder.  So to calculate the work done by the $\ce{N2}$, one cannot treat the expansion of the nitrogen as reversible.  Instead, one must focus on the reversible work required to compress the air.

Assuming that the $\ce{N2}$ does not diffuse into the air (i.e., the boundary between the air and the $\ce{N2}$ remains well-defined), the final volume occupied by the $\ce{N2}$ is $11 × 20/1.19 = 184.9$ liters.  That means that the volume of $\ce{N2}$ increased by $184.9 - 20 = 164.9$ liters.  This is the volume of $\ce{N2}$ outside the cylinder (in the jar) in the final state of the system.  If $V_0$ represents the volume in the jar outside the cylinder (and also the initial volume of air), the final volume of air is $V_0 - 164.9$.  So, the ratio of the initial volume of air to the final volume of air is equal to the pressure ratio of the air, and given by:

$$\frac{V_0}{V_0 - 164.9} = 1.19$$

Solving for $V_0$ then yields $V_0 = \pu{1033 L}$.  

The work done by the air on the $\ce{N2}$ is given by

$$
\begin{align}
W_\mathrm{a} &amp;= n_\mathrm{a}RT\ln{\left(\frac{1033 - 164.9}{1033}\right)} \\
  &amp;= P_0V_0\ln{\left(\frac{1033-164.9}{1033}\right)} \\
  &amp;= 1\cdot 1033\ln{\left(\frac{1033-164.9}{1033}\right)} \\
  &amp;= \pu{-180 L atm}
\end{align}
$$

This is equal in magnitude and opposite in sign to the work done by the $\ce{N2}$ on the air.  So the work done by the $\ce{N2}$ on the air is 180 liter-atm.