Thermodynamics : an empty container surrounded by an atmospheric gas I have a small exercices of thermodynamics if anyone can help me :
We start with an empty container of volume $V$. The walls of this container are adiabatic and will not change over time surrounded by a gas of pressure $P_0$ and of temperature $T_0$.

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*Define and give the characteristics of the system we'll study

*Write the internal energy inside the box using the characteristic of the exterior gas when the equilibrium is reached

*What is the internal energy if the box is closed when the mechanical equilibrium is reached but not the thermal one ?

I know the system is the volume of the box + the matter inside the atmosphere which will go inside the box.
Idk how to characterize it and how I can write just the internal energy and not its variations.
Edit : by applying the first principle in open system, supposing that work and thermal energy are equal to 0 :
$u_2 = P_0(v+v_0) - P_f \times v$
with $m$ the mass of the matter of the system, $v = \frac{V}{m}$, $v_0$ the volume occupied by the gas outside the box at initial time.
I think this relation is wrong because next question ask for internal energy when the box is closed when you reach mechanical equilibrium but not thermal one (my relation doesn't need temperature at all ??!!).
 A: Since the container is empty, the pressure inside the container is 0. There is no material/mass inside the system whose thermodynamical quantities can be considered. Please check the question again.
A: There are two ways of doing this.  One is using the closed system version of the 1st law, and the other is using the open system version.  I'll show you how it is done using the open system version.  Let n be the number of moles of gas which eventually enter the container.  Applying the open system version to the situation yields:  $$\Delta U=nu=h_0n$$where $\Delta U$ is the change in internal energy within the container during the process, and the enthalpy per mole entering the container during the change is $$h_0=u_0+P_0v_0=u_0+RT_0$$If we combine these two equations, we obtain $$u=u_0+RT_0$$or$$u-u_0=C_v(T-T_0)=RT_0$$The rest is easy.
I leave it to you to show how the closed system version of the 1st law delivers this same result.  In this case, you would choose as your closed system all the gas that eventually ends up in the container.
