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A motor tyre has a pressure of $3$ atm at a temperature of $27^\circ C$ If the tyre suddenly bursts what is the resulting temperature?
First of all, I believe this is not a quasi-static process and by "suddenly" i dont think there is any equillibrium maintained. This question is given as a homework problem under the section of Applications of the First Law of Thermodynamics. There was a solution on some other website to a very similar problem but they use the $PV^{\gamma}=constant$ equation and also the equation for ideal gas. But I dont see how they are able to use it since this is not a quasi-static process.

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  • $\begingroup$ Well, the question arises whether the process is quasistatic (i.e. reversible) or not? What do you think? Based on the formula you've given in the problem, what kind of process does it look to you? And, does that process necessarily have to be a quasistatic one? $\endgroup$ Jun 22, 2020 at 13:40
  • $\begingroup$ based on the formula it is an adiabatic process, and to apply the equation which i meant i thought the process needs to be necessarily quasistatic. $\endgroup$
    – 1500kook12
    Jun 22, 2020 at 14:11
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    $\begingroup$ Well, if the process is happening suddenly, then it's not quasistatic, is it? The reason is that, adiabatic processes can be irreversible as well, as a sudden 'bursting' of tires involves no heat exchange. This is an example of an irreversible adiabatic process. $\endgroup$ Jun 22, 2020 at 14:58
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    $\begingroup$ See this for the difference between a reversible and a quasistatic process: Quasistatic vs Reversible. Also, read these sites to understand what I am talking about:1. 2. $\endgroup$ Jun 22, 2020 at 17:08
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    $\begingroup$ thank you, i finally did get what you mean and this problem should be assumed as a quasi-static irreversible adiabatic process. $\endgroup$
    – 1500kook12
    Jun 23, 2020 at 10:12

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In my judgment, your assessment is correct. The subsequent expansion of the air is going to be irreversible. To a first approximation, the expansion takes place against constant atmospheric pressure, and the behavior is essentially the same as if the air were contained in an insulated cylinder with a massless piston. Of course, that part that ends up outside the carcass will, after a while, come to equilibrium with the surrounding air. But the air remaining inside the carcass of the tire will heat up much more slowly, and so, at least for the purposes of this problem, the expansion can be considered adiabatic (and, as indicated previously, against constant atmospheric pressure).

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  • $\begingroup$ heat up? i got a lower temperature value on applying the $PV^{\gamma}=constant$ equation. $\endgroup$
    – 1500kook12
    Jun 22, 2020 at 13:48
  • $\begingroup$ and are we assuming the gas to be ideal gas too? $\endgroup$
    – 1500kook12
    Jun 22, 2020 at 14:10
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    $\begingroup$ First it cools down, but then it very slowly reheats after after heat is exchanges through the carcass with the outside air (the latter not part of the process we are looking at). And, for air, assuming ideal gas behavior up to 3 atm is certainly a valid approximation. $\endgroup$ Jun 22, 2020 at 14:48

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