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

[Fitz Watson]

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?

 

[sophiecentaur]

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.
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.

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.

 

[Fitz Watson]

Of course you can't expect to squeeze the sample down to nothing.
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.

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

 

[jbriggs444]

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.

 

[sophiecentaur]

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.

 

[Chestermiller]

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.