# Calculation of water bulk modulus using equation of state

I simply want to calculate the bulk modulus of water at 50C and increasing pressures. I think I am correctly calculating the new specific volume from the original conditions at (25C and 1atm) to 50C and higher pressures. I am rightly getting a decrease in specific volume with increasing pressure at constant temperature (Column7). Here is the spreadsheet: ${V}^{'}={V}_{o}e^{{\beta}(T-25)-\kappa\Delta P}$

where:

${V}^{'}$ is column 7

${V}_{o}$ is column 1, the specific volume of water at 1 atm and 25C

$T$ is in Celsius

$P$ is in atm.

I used the above cross plot to graphically solve the slope $(\frac{\partial v} {\partial P})_{T}$ and input it into Column 8:

Then to calculate the new compressibility at 50C (${\kappa}$) Column 9:

${\kappa}=-\frac{1} {V}(\frac{\partial v} {\partial P})_{T}$

which gives me the new compressibiliy Column 9. Then I I just take the reciprocal and convert the units to GPa.

Oops, bulk modulus (Column 10) should be increasing with pressure at a constant temperature, not decreasing. I know that since dividing by an ever decreasing specific volume as pressure increases will give me a larger compressibility (Column 9) and a decreasing Bulk Modulus (Column 10). But everyone knows increasing pressure should have the opposite effect. Where did I go wrong?

• Can you clearly write out the equations you think you should be using? It's either a wrong equation or a typo in the spreadsheet -- the former we can help with, the latter, not so much. Jan 28 '15 at 17:34
• Check your equations, is it just a typo that your partial derivatives are with respect to $T$ and not to $P$? Jan 29 '15 at 0:06
• The partial differential is correct now. It is with respect to P at constant temperature.
– JimK
Jan 29 '15 at 0:52

This basically means you need to use the original mass or amount of matter. As there is no mass or amount of material (ie. $kJ/mol$ or $kJ/kg$) used in these equations, the connection to some amount of material must be made through original Volume. But it seems that you are using the Specific Volume instead. This means you calculate with a smaller amount of material, which obviously leads to a decrease in Bulk Modulus. Its dimensional form (SI pascal) is basically $M^1L^−1T^−2$ Which pretty much means you have too small $M$ (mass) the $L$ is too big (some length) or the $T$ Temperature is too big.