For a homework problem, I've been asked to show that the work done by increasing pressure isothermally on a solid of mass m is: $$W \approx -\frac{m\bar{\kappa}}{2\bar{\rho}}(P_f^2 – P_i^2)$$
where $\bar{\kappa}$ and $\bar{\rho}$ are, respectively, the mean compressibility and density (unchanged from $P_i$ to $P_f$). Starting with the definition of both the compressibility ($\kappa = -\frac{1}{V}(\frac{\partial V}{\partial P})_T$) and the density ($\rho = \frac{m}{V}$), you can work backwards to get: $$W \approx -\frac{m}{2}\left(-\frac{1}{V} \left(\frac{\partial V}{\partial P}\right)_T \right)\left(\frac{V}{m}\right)(P_f^2 – P_i^2)$$
$$W \approx\frac{1}{2}\left(\frac{\partial V}{\partial P}\right)_T (P_f^2 – P_i^2)$$
The part about it being a solid confuses me because I'm not sure if the ideal gas law can be utilized. Is it enough to simply use the definition of work to get you the rest of the way from there:
$$W = \int P\,dV = \int \frac{nRT}{V}\,dV$$
I think the multivariable calculus is getting in the way. I'm not sure how to interpret the partial derivative.
Can someone help me connect the dots?
Best Answer
In general, you could write V as a function of p and T, so:
$$dV = \frac{\partial V}{\partial p} dp + \frac{\partial V}{\partial T} dT$$
Because the transformation is isothermal, you have $dT = 0$, so:
$$dV = (\frac{\partial V}{\partial p})_T ~ dp $$
The work is $W = -\int_i^f p dV = - \int_i^f p (\frac{\partial V}{\partial p})_T ~ dp$
Suppose that $(\frac{\partial V}{\partial p})_T, \kappa, \rho$, are constant, we would have :
$$W = - (\frac{\partial V}{\partial p})_T \int_i^f p ~ dp = - \frac{1}{2} (\frac{\partial V}{\partial p})_T (p_f^2 - p_i^2) = -\frac{m{ \kappa}}{2{\rho}} (p_f^2 - p_i^2)$$
If the quantities $(\frac{\partial V}{\partial p})_T, \kappa, \rho$ are varying slowly with p, an acceptable approximation of W is : $$W \approx -\frac{m{\bar \kappa}}{2{\bar \rho}} (p_f^2 - p_i^2)$$, where $\bar \kappa$ and $\bar \rho$ are mean quantities (based on the interval variation of the pressure).