Question

A substance at 4$^\circ$C has a thermal expansion coefficient $\beta = \dfrac{1}{v}\left(\dfrac{\partial v}{\partial T}\right)_P = 0$ K$^{-1}$, an isothermal compressibility, $\kappa_T = -\dfrac{1}{v}\left(\dfrac{\partial v}{\partial P}\right)_T = 5\times10^{-4}$ Pa$^{-1}$ and a molar volume $v = 18\times10^{-6}$ m$^3$ mol$^{-1}$. If $s$ is the molar entropy, then at 4$^\circ$C, the quantity $\left[v\left(\dfrac{\partial s}{\partial v}\right)_T\right]$ evaluated for the substance is \(\underline{\hspace{2cm}}\) J mol$^{-1}$ K$^{-1}$ (rounded off to the nearest integer).

Hint:

Use Maxwell relations and the identity $(\partial P/\partial T)_V = \beta / \kappa_T$ to connect measurable properties like $\beta$ and $\kappa_T$ with entropy derivatives.

Answer 1

Correct Answer: 0

From thermodynamics, using Helmholtz free energy $A = U - TS$ with differential 
$dA = -S\,dT - P\,dV$, we obtain the Maxwell relation 
\[ \left(\dfrac{\partial S}{\partial V}\right)_T = \left(\dfrac{\partial P}{\partial T}\right)_V . \] We therefore need $\left(\dfrac{\partial P}{\partial T}\right)_V$. 
The volume as a function of $T$ and $P$ gives 
\[ \frac{1}{v}dv = \beta\,dT - \kappa_T\,dP . \] For constant $v$ (i.e. at constant volume), $dv = 0$, hence 
\[ 0 = \beta\,dT - \kappa_T\,dP \Rightarrow \left(\dfrac{\partial P}{\partial T}\right)_V = \dfrac{\beta}{\kappa_T} . \] Thus 
\[ \left(\dfrac{\partial s}{\partial v}\right)_T = \left(\dfrac{\partial P}{\partial T}\right)_V = \dfrac{\beta}{\kappa_T} . \] The required quantity is 
\[ v\left(\dfrac{\partial s}{\partial v}\right)_T = v\,\dfrac{\beta}{\kappa_T}. \] At 4$^\circ$C, $\beta = 0$ K$^{-1}$, so 
\[ v\left(\dfrac{\partial s}{\partial v}\right)_T = v\cdot\dfrac{0}{\kappa_T} = 0 \;\text{J mol}^{-1}\text{K}^{-1}. \] Hence, the value rounded to the nearest integer is $0$.