Expansivity & compressibility

Coefficient of volume expansion

Suppose we write the specific volume as a function of temperature and pressure, $$v = v(T,P).$$

Because $v=V/n$ ($n$ is a constant, at least for a closed system), $v$ is also a state variable just like $V$. It has an exact differential: $$dv = \left( \frac{\partial v}{\partial T} \right)_P dT + \left( \frac{\partial v}{\partial P} \right)_T dP.$$

The (thermal) coefficient of volume expansion, $\beta$, also called the "expansivity", is the fractional change in volume as temperature changes: $$\beta \equiv \frac{1}{v}\left( \frac{\partial v}{\partial T} \right)_P = \frac{1}{V}\left( \frac{\partial V}{\partial T} \right)_P.$$

For an ideal gas, $v=RT/P$, so $$\beta = \frac{1}{v}\left( \frac{\partial v}{\partial T} \right)_P = \frac{1}{v} \left( \frac{R}{P} \right) = \frac{1}{T}.$$

But for many solids and liquids, $\beta$ is not changing "very much" with temperature, so let's assume it's approximately constant:

Throw a penny ($d = 19.05$ mm and $\beta \approx 5.2 \times 10^{-5} $ K${}^{-1}$) into a kettle at 20 C, and then heat the water in the kettle to boiling, by how much does the diameter of the penny change? [First, calculate the fractional change in the volume...]

• First, argue that $dP\approx 0$...
• Then write an approximate expression for $dV$ that involves $\beta$,
• and integrate $\int dV$.
• Finally, relate the change of volume to the change in diameter.

The contents of the kettle is in contact with the surroundings--It's in mechanical equilibrium with the atmosphere. We'll assume that the pressure is not changing during this process: $dP = 0$. So, $$dV = \left( \frac{\partial V}{\partial T} \right)_P dT = \beta V\ dT \approx \beta V_0\ dT.$$

Integrating to get the change in volume... $$\int_{V_0}^{V} dV' = \beta V_0 \int_{T_0}^{T} dT'.$$ Gives... $$V - V_0 = \Delta V = \beta V_0 \left( T-T_0 \right),$$ Dividing by $V_0$: $$\frac{\Delta V}{V_0} = \beta \left( T-T_0 \right) = 5.2 \times 10^{-5} {\rm K}^{-1} * (80 {\rm K}) = 0.0042.$$

To relate the fractional volume expansion to the fractional linear expansion. Let's assume that all dimensions change by the same fraction, that is: $$\frac{\Delta x}{x} = \frac{\Delta y}{y} = \frac{\Delta z}{z} = \frac{\Delta l}{l}.$$

We'll show in class that... $$ \frac{\Delta V}{V_0} = 3 \frac{\Delta l}{l_0}.$$

So, we'll have... $$\Delta l = \frac{l_0}{3}\frac{\Delta V}{V_0}= \frac{19.05{\rm\ mm}}{3}(0.0042)=0.027{\rm \ mm}=27 \mu.$$

An approximate equation of state

In the penny problem, we ignored the pressure dependence of the volume, but now, let's not.

The isothermal compressability, $\kappa$ is defined as $$\kappa \equiv -\frac{1}{v}\left( \frac{\partial v}{\partial P} \right)_T.$$

Using this definition, and the expansivity to re-write the partial derivatives, we have $$\begineq dv&= \left( \frac{\partial v}{\partial T} \right)_P dT + \left( \frac{\partial v}{\partial P} \right)_T dP \\ &= \beta v\ dT - \kappa v \ dP \approx \beta v_0 \ dT - \kappa v_0 \ dP.\endeq$$

Integrating, $$\begineq \int_{v_0}^{v} dv' &= \beta v_0 \int_{T_0}^{T} dT' -\kappa v_0 \int_{P_0}^{P} dP'\\ \left(v-v_0\right) &= \beta v_0 \left(T-T_0\right) - \kappa v_0 \left(P-P_0\right).\endeq$$

Re-arranging this, gives us a tidy equation of state as $v(T,P)$: $$v = v_0\left( 1+\beta\left(T-T_0\right)-\kappa \left(P-P_0\right) \right).$$

This works well for solids (and some liquids) for small changes of volume, and over temperature ranges where $\beta$ and $\kappa$ are not changing very much w.r.t. temperature.