Work done by gas expanding into a cylinder+piston system

[LCSphysicist]

Homework Statement

A thin-walled metal container of volume V contains a gas at high pressure.
Connected to the container is a capillary tube and stopcock. When the stopcock is
opened slightly, the gas leaks slowly into a cylinder equipped with a nonleaking,
frictionless piston, where the pressure remains constant at the atmospheric value $P_0$.
(a) Show that, after as much gas as possible has leaked out, an amount of work $P_0(V_0-V)$
has been done, where $V_0$ is the volume of the gas at atmospheric pressure and
temperature.
(b) How much work would be done if the gas leaked directly into the atmosphere?

Relevant Equations

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I will summarize briefly my reasoning for both letters, since the answer is immediately after that:

A) The work is quasi-static, and the pressure is approximatelly constant and equal to the atmospheric pressure, so the works is $$W = -p\int dV = -p_{0} (V_{0}-V)$$
B) The work is fast, fast enough that no heat flows thought the gas, so that the work is $$W = -\int p dV = -\int_{V_i}^{V_f} p_i (V_i)^{\gamma} dV /V^{\gamma}$$ $$ P_i V_i^{\gamma} / P_o = V_f ^{\gamma}$$

What do you think? I am almost sure it is wrong, so i am posting here to get any help on letter b

 

[Chestermiller]

If your system is the gas, why do you feel that the answer to B is different from A?

 

[LCSphysicist]

If your system is the gas, why do you feel that the answer to B is different from A?

As i said, because the first one "leaks slowly", and nothing has been said about the second one, so i suppose it is more "explosive".

I have questioned myself also if the work is the same, but i was not able to argue why it should be. Even so the final points on (PV space) is the same, the path could be different, resulting in different work.

 

[Chestermiller]

The question implies that the gas is still leaking slowly in case B, but into the atmosphere instead of into a cylinder and piston.