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I want to study the optical properties of a molecule under pressure. The method that I am planning to use to apply the pressure on my sample is the following:

A silicone rubber tube is connected to the mouth of a quartz cuvette. The cuvette and the tube are completely filled with the sample which is dissolved in chloroform, and then the tube is capped. The sample is placed in a chamber filled with hexane. Pressure is added into the chamber and transmitted via the hexane, which applies pressure on the sample by compressing the rubber tubing.

This is the diagram:

Simplified diagram of the setup

I want to know what is the amount of pressure that I need to exert on the hexane in order to compress the tubing to say... 10 - 20% of its original volume. And, when I do this, would the pressure inside the cuvette be the same as the pressure outside the cuvette, which I can directly measure with a barometer?

Any information on what type of textbooks would have this information would also be very useful!

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    $\begingroup$ Max pressure will be either the pump reaching its limit or the container failng. $\endgroup$ – user207455 Jun 16 '19 at 13:32
  • $\begingroup$ @SolarMike I am expecting the cuvette to fail after the tube is fully compressed. I want to begin by compressing the tubing to 10 - 20% of its original size, and hopefully the cuvette won't fail by this point. I am having a problem calculating at what pressure this will happen, and what the pressure will be on the sample. $\endgroup$ – Max Jun 16 '19 at 13:52
  • $\begingroup$ 10% to 20% is a different question, edit your original question to make it clear what you really mean to ask. When I used the term "container" I meant the outer one, but the cuvette is also relevant. $\endgroup$ – user207455 Jun 16 '19 at 13:55
  • $\begingroup$ @SolarMike Ah, ok! Thank you. The outside container is a very sturdy metallic box. We have already used for other experiments in which a piston was used to transmit the pressure. I have edited the question to include the 10% - 20%. $\endgroup$ – Max Jun 16 '19 at 14:11
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Let the volume of the cuvette be $V_C$ and the original volume of the compressible tube be $V_T$. If the cuvette is so rigid that its volume doesn't change under pressure and the volume of fluid in the compressible tube is reduced to 15% of the original volume, then the fluid volume ratio is $$\frac{(V_C+0.15V_T)}{V_C+V_T}=1-\frac{0.15V_T}{V_C+V_T}=\exp{[-\beta (P-P_0)]}$$where $\beta$ is the bulk compressibility of the fluid. If the volume change can be brought about by a relatively modest pressure increase, this equation approximately reduces to: $$(P-P_0)=\frac{1}{\beta}\frac{0.15V_T}{(V_C+V_T)}$$

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  • $\begingroup$ Thank you! This was very helpful. $\endgroup$ – Max Jun 22 '19 at 16:29

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