High School What Happens to Volume Change Ratio in Caesium When Pressure Exceeds 1.6 GPa?

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The discussion centers on the behavior of Caesium under pressure, particularly regarding its low bulk modulus of 1.6 GPa. When pressure exceeds this value, the volume change ratio cannot exceed one, leading to confusion about the implications of the bulk modulus formula. It is clarified that the bulk modulus is defined in terms of the rate of fractional change in volume with pressure, not absolute volume changes. The correct mathematical representation indicates that volume approaches zero asymptotically but never actually reaches it, preventing the sample from being crushed completely. This highlights the importance of understanding the correct application of bulk modulus in solids and liquids.
Fitz Watson
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Today, while studying about bulk Modulus, I encountered a doubt. Please consider this thought experiment.

I'm considering Caesium as an example as it seems to have a quite low Bulk Modulus (comparatively) of 1.6 GPa.
Let's say I apply a pressure of X GPa.
Volume change ratio can be given by X/1.6

What happens if I increase X to more than 1.6 GPa? Practically speaking, its impossible to have Volume change ratio of > 1. So, where am I going wrong?
 
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Fitz Watson said:
What happens if I increase X to more than 1.6 GPa?
Of course you can't expect to squeeze the sample down to nothing. :wink:
The expression for Bulk Modulus K is not what you are implying though. Bulk modulus is useful for describing the behaviour of solids and liquids.
e782c2d3008ff51d03795ad928b9c6b1b1f99515

Is the way it's defined - in terms of the rate of fractional change in volume with pressure and not the change of volume with pressure. Does that help with your confusion?
Gases follow the 'Gas Laws' because there is a lot of space between the molecules so Boyle's Law PV=Constant can be used over a huge range of pressures and volumes.
 
sophiecentaur said:
Of course you can't expect to squeeze the sample down to nothing. :wink:
The expression for Bulk Modulus K is not what you are implying though. Bulk modulus is useful for describing the behaviour of solids and liquids.
e782c2d3008ff51d03795ad928b9c6b1b1f99515

Is the way it's defined - in terms of the rate of fractional change in volume with pressure and not the change of volume with pressure. Does that help with your confusion?
Gases follow the 'Gas Laws' because there is a lot of space between the molecules so Boyle's Law PV=Constant can be used over a huge range of pressures and volumes.

Is this formula wrong?
Bulk Modulus = Pressure/Volume Strain?
I haven't gone into the differential form of it yet, so that's why I'm using this one
 
If you apply the nominal 1.6 GPa pressure, you'll compress the substance by a factor of e. To make a long story short, that's because ##\lim_{n \to \infty}(1-\frac{1}{n})^n## = 1/e.
 
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Fitz Watson said:
Is this formula wrong?
I don't recognise it. You yourself have pointed out a problem with it as it suggests that you could crush a sample out of existence. As @jbriggs444 points out, the proper formula is consistent with experience - always a good thing. :wink:
 
If one assumes that the bulk compressibility is approximately constant, then the solution to the (correct) differential equation posted by @sophiecentaur in post #4 is not $$V=V_0\left[1-\frac{P}{K}\right]$$The correct solution to this equation is $$V=V_0\exp{\left(-\frac{P}{K}\right)}$$So, no matter how high P gets, V never goes to zero.
 
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I do not have a good working knowledge of physics yet. I tried to piece this together but after researching this, I couldn’t figure out the correct laws of physics to combine to develop a formula to answer this question. Ex. 1 - A moving object impacts a static object at a constant velocity. Ex. 2 - A moving object impacts a static object at the same velocity but is accelerating at the moment of impact. Assuming the mass of the objects is the same and the velocity at the moment of impact...

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