# How do I calculate the new pressure of liquid water using its isothermal compressibility kappa and expansion coefficient alpha?

## I'm given that the isothermal compressibility $\kappa = 4.7 \times {10}^{- 5} {\text{atm}}^{- 1}$ and expansion coefficient $\alpha = 1.7 \times {10}^{- 4} {\text{K}}^{- 1}$ at ${17}^{\circ} \text{C}$. The rest of the question states: closed rigid container filled with liquid water at an initial temperature of ${14}^{\circ} \text{C}$ and initial pressure of $\text{1 atm}$. The temperature is changed to ${20}^{\circ} \text{C}$, and the new pressure is to be found. Assume that $\alpha$ and $\kappa$ are negligibly different whether it's ${17}^{\circ} \text{C}$ or ${14}^{\circ} \text{C}$ or ${20}^{\circ} \text{C}$. What is this new pressure? I know it is $\text{23 atm}$, but I don't know how to do the problem. Here are the definitions you need to know, and the calculus you may need to know: $\alpha = \frac{1}{V} {\left(\frac{\partial V}{\partial T}\right)}_{P} \implies \alpha {\mathrm{dT}}_{P} = \frac{1}{V} {\mathrm{dV}}_{P}$ $\kappa = - \frac{1}{V} {\left(\frac{\partial V}{\partial P}\right)}_{T} \implies \kappa {\mathrm{dP}}_{T} = - \frac{1}{V} {\mathrm{dV}}_{T}$ ${\left(\frac{\partial V}{\partial P}\right)}_{T}$ is the partial derivative of the volume with respect to the pressure at a constant temperature. ${\left(\frac{\partial V}{\partial T}\right)}_{P}$ is the partial derivative of the volume with respect to the temperature at a constant pressure. Note that $\frac{1}{V} {\mathrm{dV}}_{P}$ and $- \frac{1}{V} {\mathrm{dV}}_{T}$ are under different conditions so they are not equivalent. ${\text{_}}_{P}$ is constant pressure, and ${\text{_}}_{T}$ is constant temperature. Also, you cannot use the ideal gas law, because it fails on liquids.

Aug 24, 2016

AHA! Actually, right after I wrote this question out, I figured it out.

The logic behind this is to express the change in pressure with respect to temperature at a constant volume in terms of $\alpha$ and $\kappa$, since:

• The container is rigid and full of liquid water. i.e. constant volume.
• The water changes pressure due to the change in temperature. i.e. (dP)/(dT) = ???

Starting from the total derivative of the differential volume

$\mathrm{dV} = {\left(\frac{\partial V}{\partial T}\right)}_{P} \mathrm{dT} + {\left(\frac{\partial V}{\partial P}\right)}_{T} \mathrm{dP}$,

we can divide by the partial differential temperature at a constant volume, $\partial {T}_{V}$, to find ${\left(\frac{\partial P}{\partial T}\right)}_{V}$.

${\cancel{{\left(\frac{\partial V}{\partial T}\right)}_{V}}}^{0} = {\left(\frac{\partial V}{\partial T}\right)}_{P} {\cancel{{\left(\frac{\partial T}{\partial T}\right)}_{V}}}^{1} + {\left(\frac{\partial V}{\partial P}\right)}_{T} \stackrel{\text{goal}}{\overbrace{{\left(\frac{\partial P}{\partial T}\right)}_{V}}}$

The term that cancels to $0$ is because the change in volume must be $0$ at a constant volume, and the term that cancels to $1$ cancels because $\frac{\mathrm{dT}}{\mathrm{dT}} = 1$, and $1$ at a constant volume is still $1$.

Thus, our big equation becomes:

${\left(\frac{\partial P}{\partial T}\right)}_{V} = \frac{- {\left(\frac{\partial V}{\partial T}\right)}_{P}}{\frac{\partial V}{\partial P}} _ T$

If we note that the definitions of $\alpha$ and $\kappa$ say $\alpha = \frac{1}{V} {\left(\frac{\partial V}{\partial T}\right)}_{P}$ and $\kappa = - \frac{1}{V} {\left(\frac{\partial V}{\partial P}\right)}_{T}$, then:

${\left(\frac{\partial P}{\partial T}\right)}_{V} = \frac{\cancel{- V} \alpha}{\cancel{- V} \kappa} = \frac{\alpha}{\kappa}$

Now, we can do this trick where we multiply out a partial differential temperature at a constant volume to turn it into an exact differential, ${\mathrm{dT}}_{V}$, and then integrate.

${\mathrm{dP}}_{V} = \frac{\alpha}{\kappa} {\mathrm{dT}}_{V}$

${\int}_{{P}_{1}}^{{P}_{2}} {\mathrm{dP}}_{V} = \frac{\alpha}{\kappa} {\int}_{{T}_{1}}^{{T}_{2}} {\mathrm{dT}}_{V}$

${P}_{2} - {P}_{1} = \frac{\alpha}{\kappa} \left({T}_{2} - {T}_{1}\right)$

So the final expression is:

$\boldsymbol{{P}_{2} = {P}_{1} + \frac{\alpha}{\kappa} \left({T}_{2} - {T}_{1}\right)}$

And if we use this to evaluate the final pressure:

color(blue)(P_2) = "1 atm" + (1.7xx10^(-4) "K"^(-1))/(4.7xx10^(-5) "atm"^(-1))("6 K")

$= 22.7 \approx \textcolor{b l u e}{\text{23 atm}}$