A rigid box has a volume of $V_1$ and is thermally isolated so that it exchanges no heat with its environment. Initially the box is empty, and the air around it has a pressure of $P_0$ and a temperature of $T_0$. The air can be treated as an ideal gas that has an adiabatic constant of $\frac{C_p}{C_v} = 1$.

We remove the lid and the box quickly becomes filled with air, when the air no longer flows into the box we close it again.

This is the confusing part : "let $V_0$ be the volume occupied by the air immediately when we opened the box (by removing the lid)"

The problem with this is that it asks that you represents all the states and their $P,V,T$ respectively. It says that there are 3 states, how is that possible? There are only two: when the box is full and when we close the box. The $V_0$ part really confused me, is that supposed to be a state too?

The problem asks why we can consider this transformation an adiabatic process, then asks you to prove that the final temperature is equal to the temperature of the air outside.

  • $\begingroup$ Our definition of "homework-like" explicitly includes "i got stuck in this exercise im self studying" if the value of the question is primarily pedagogical. See meta.physics.stackexchange.com/questions/714. $\endgroup$ – dmckee --- ex-moderator kitten Jun 3 '15 at 13:52
  • $\begingroup$ ok how can i add homework tag ? i was selfstudying why should i lie anyways $\endgroup$ – user31731 Jun 3 '15 at 13:53
  • 2
    $\begingroup$ I made some pretty extensive edits attempting to clean up your spelling, grammar, and formatting. I tried not to change the content of your question. Not to be scold-y, but in the future, consider that the more effort you put into presentation, the less effort you ask readers to make in an attempt to understand you. $\endgroup$ – zeldredge Jun 3 '15 at 14:06
  • $\begingroup$ The 3 states are before box open, when box is open, and when box is closed (again). $\endgroup$ – Jiminion Jun 3 '15 at 14:11

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