In order to look at why supersaturation occurs we must first introduce some equations of the first law of thermodynamics. As stated earlier, this is the conservation of energy; it is often formalised as:

*ΔQ= ΔU + ΔW*

Where:

Δ is the Greek letter delta symbolising a change of; Q is the heat transferred into or out of the system (Heat transferred into is positive)

U is the internal energy of system

W is work done on or by the system, where work done by is positive and work done on is negative

It is also known that work= force multiplied by the perpendicular distance:

*W=F.d*

In a gas P=F/A, which can be re written as:

*F=PA*

P being the pressure and A being the cross sectional area of the gas (i.e. it’s container). If we combine the above two equations we now have:

W=PA.d

which gives

*W=PV*

as an area multiplied by a distance gives a volume. Therefore if we get an increase in the volume then work is done by the gas. We can then also find that from the kinetic theory of gases that the total kinetic energy of an ideal gas is related to the absolute temperature of the gas, and the internal energy of the gas is therefore also linked to the absolute temperature of the gas.

*KE=3/2NkT*

Where:

k is the Boltzmann constant

N is the number of molecules

T is the absolute temperature

Our original first law of thermodynamics can now be re-written as:

*∆Q=∆ 3/2NkT+∆PV*

As the chamber expands adiabatically there is no heat transfer, so ΔQ=0. However the volume has increased, so the system must have done work. Subsequently ΔW is positive. This means that to conserve energy the internal energy must be negative and of the same magnitude as the work done. Subsequently there must be a temperature drop, since U is related to T as shown above. It is this that causes supersaturation within the chamber.

-- NigelWatson - 04 Jul 2012

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Topic revision: r1 - 04 Jul 2012 - 15:00:30 - NigelWatson

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