Question

As 2 moles of sirenite is heated under a constant pressure of 1 atm it undergoes a solid phase transition, at a temperature of 750 K, from α-sirenite to β-sirenite. The molar specific heat capacity of sirenite varies with thermodynamic temperature, T, such that Cvm = 2.50 x 10-6 T 2 Jmol-1K-1.

a) Calculate the entropy change of the sirenite between the transition temperature of 750 K and room temperature (300 K).

b) By what factor does the number of accessible microstates increase across this temperature range? Express your answer in terms of the original number of microstates.

Answer 1

Basic Thermodynamic relation to start with

$dU=TdS-PdV$

For solid we can assume $dV\approx0$, and by definition $dU=nC_vdT$

Thus,

$dS=n\frac{C_v}{T}dT$

You should have written the mathematical expression properly, but I think you have written the following

$C_v=2.5\times10^{-6}\ T^2$

in which case for n=2,

$\int_0^{\Delta S} dS=5\times10^{-6}\int_{300}^{750} TdT$

$\Delta S\approx1.18\textup{ J/K}$

(b)

The Boltzmann relation for the entropy,

$S=k_B\log\Omega$

Logarithm is on base e. So the change in entropy,

$\Delta S=k_B\log\left (\frac{\Omega_2}{\Omega_1} \right )$

$\frac{\Omega_2}{\Omega_1}=e^{\frac{\Delta S}{k_B}}=e^{9.6\times10^{22}}$