Question

Calculate ΔS for the isothermal compression of 3.05 mol of Cu(s) from 1.00 bar to 1370. bar at 298 K. α=0.492×10−4K−1,κT=0.78×10−6 bar−1, and the density is 8.92 g⋅cm−3.

Answer 1

Explanation:

For compression, equation for entropy change ($\Delta S$) is as follows.

   $\Delta S = \int_{T_{i}}^{T_{f}} \frac{C_{p}}{T} dT - \int_{P_{i}}^{P_{f}}V \alpha dP$

              = $nC_{p,m} ln \frac{T_{f}}{T_{i}} - nV_{m,t} \alpha (P_{f} - P_{i})$  

Hence, for an isothermal process $T_{i} = T_{f}$

Now, we will put the given values into the above formula as follows.

 $\Delta S = nC_{p,m} ln \frac{T_{f}}{T_{i}} - nV_{m,t} \alpha (P_{f} - P_{i})$

           = $nC_{p,m} ln(1) - [3.05 mol \times \frac{63.546 g/mol}{8.92 g/cm^{3}} \times \frac{1 m^{3}}{10^{6} cm^{3}} \times 0.492 \times 10^{-4} (1370 - 1)bar \times \frac{10^{5} Pa}{1 bar})$            = -0.146 J/K

Thus, we can conclude that value of $\Delta S$ for the isothermal compression is -0.146 J/K.