The genie is out of the bottle 1

Let us consider this process in two steps.

First we expand the gas from its original volume, $V_i$ , to a final volume, $V_f$ , so that the pressure changes from $p_i=200atm$ to $p_f=1atm$ . In an isothermal process $p_iV_1=p_fV_f$ and therefore $V_f/V_i=200$ .

The entropy change is $\Delta S= \int_i^f dQ/T$ , where $dQ$ is the heat transfer. The total energy of the gas is the same (for ideal gas the energy depends only on the temperature). Therefore the heat transfer must be equal to the work done by the gas, $dQ=pdV$ . According to the ideal gas law $pV=nRT$ , where $n=1kg/(4.002 g/mol)= 249.87 mol$ is the amount of He gas in the container. We get $dQ=pdV=\frac{nRT}{V} dV$ and in the integration the temperature $T$ drops out. We get

$\Delta S= nR \ln \frac {V_f}{V_i}= nR \ln \frac{p_i}{p_f}$ ,

This as a general expression for the entropy change of the ideal gas in an isothermal process. With our values we get $\Delta S=1.1 \times 10^4 J/K$ .

In the second part we let most of the gas out to the atmosphere, but keep the original volume. In this process the gas mixing with the atmosphere generates a lot more entropy, but we do not care about that. The gas we keep will have the entropy that corresponds to the volume (or mole) fraction of the total amount of gas:

$\Delta S_{bottle} = \frac{V_i}{V_f} \Delta S= nR \frac{p_f}{p_i} \ln\frac{p_i}{p_f}$

That comes out to be $\Delta S_{bottle} = 55.0 J/K$ .

Note: Because the pressure of the He gas in the final state is equal to the atmospheric pressure, we are tempted to think that the system reached an equilibrium state. This is far from being true. Once the gas flow stops the diffusion of atoms and molecules keep going on (as long the valve is leaking). In the true equilibrium all the He gas will be in the atmosphere, eventually leaving the Earth, because the thermal velocity is larger than the escape velocity required to break free from the gravity of Earth. The bottle will have only air in it.