Calculate the reversible work done in compressing 
of mercury at a constant temperature of 32(°F) from 1(atm) to 3000(atm). The isothermal compressibility of mercury at 32(°F) is:
By definition, 
, and we know that we can find the work done on the sample by integrating –PdV.
From our definition of ? above, we see that at constant T, we have dV = -?VdP.
We can use this to first find V as a function of P, and then substitute that back in to do the integral of –PdV. To find the volume, we integrate the above relationship from the initial volume (Vo) at the initial pressure (Po) to an arbitrary pressure P, at which we have the volume V given by:
In these expressions, the units of V are arbitrary, but the units of P must be atm, since that is the units in which the numerical value of ? is given. If we only want to find the work required, we don’t actually have to solve for the final volume.
Rather, we can simply substitute dV = -?VdP into our expression for work, dW = -PdV = ?PVdP, which finally gives us
we can put into this 
and 
atm, and then integrate it from P = 1 atm to P = 3000 atm, to get
In the second expression above, we have factored out 
, both of which we can take to be negligibly different from 1, and can therefore ignore. Doing the last integral with a little help from Maple and/or an integral table gives us W = 16.53. We started out with P in atm and V in ft3, so this has units of W = 16.53 atm ft3, which is not a particularly common unit for energy. We can find the conversion factor to get to more reasonable units by using the values of R (the ideal gas constant) as as R = 0.7302 ft3 atm lbmol-1 R-1 = 1.986 Btu lbmol-1 R-1, so 0.7302 ft3 atm =1.986 Btu, or 1 ft3 atm = 2.720 Btu. So, we can multiply our result by 2.72 to get W = 45.0 Btu.
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