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I have a pipe, inside of wich there is a constant amount pressure supplied from one end (like a compressor). On the other, the gas (air), is free to leave and experiences decompression.

Knowing the temperature of the gas inside the pipe, as well as the temperature and pressure (1 atmosphere) when it leaves, how would one go about calculating the initial pressure? Would this apply to liquid (water) as well?

I tried to have a go at this question, and found this handy dandy Gay-Lussac's Law. Moved around a bit to my favor it reads: $$ P_1 = P_2 \frac{T_1}{T_2} $$

If for example,

\begin{align} P2 &= 101325\rm\ Pa = \text{1 atmosphere} \\ T2 &= 173.15\rm\ K = -100°C \text{ (My goal is to cool to this temp via decompression.)} \\ T1 &= 298.15\rm\ K = 25°C \end{align}

The result is about $174473.28\rm \ Pa$. This is just over 1.7 atmospheres. So apparently I can make dry ice with my bike pump? That's not right.

NEW HEADLINE He figured out how to make dry ice with a bike pump, chemical supply companies hate him. Learn how, click here!

Or better yet: NEW HEADLINE He figured out how to cool water to absolute zero using a pressure washer (5000psi), scientists hate him. Learn how, click here!

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The problem with Gay-Lussac is that it assumes constant mass and constant volume. Your bike pump can't actually make a direct change in pressure, it can only make a change in volume (which then leads to a change in pressure).

That means you need to use the ideal gas law (and consider the volume change of a mass of air) to get closer to what the temperature might be doing.

Gay-Lussac relationships usually describe what the pressure will be when the temperature changes, not the other way around.

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  • $\begingroup$ But what about like a compressor on the end of the pipe? Just as much air enters the pipe as leaves, keeping the mass and volume constant. $\endgroup$ – user2990508 Nov 5 '16 at 5:47
  • $\begingroup$ You are asking how the temperature of the gas changes as it goes from the compressed section to atmospheric. As that air leaves it expands. That means either a constant mass of air occupies a larger volume, or that a constant volume of space contains a smaller mass of air. Both are not constant during the decompression. That means you can't use that law for that transition. $\endgroup$ – BowlOfRed Nov 5 '16 at 5:53
  • $\begingroup$ Never mind I am just dumb. Thanks though! $\endgroup$ – user2990508 Nov 5 '16 at 5:54

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