Question
Question: for a small positive coefficient of expansion in case of solid : then comment on Cp and Cv...
for a small positive coefficient of expansion in case of solid : then comment on Cp and Cv
Cp is slightly greater than Cv
Solution
For a solid, the relationship between the molar heat capacities at constant pressure (Cp) and constant volume (Cv) is given by the thermodynamic relation:
Cp−Cv=βTα2VT
where: α is the coefficient of volume expansion. V is the molar volume. T is the absolute temperature. βT is the isothermal compressibility, defined as βT=−V1(∂P∂V)T.
For a typical solid at a positive absolute temperature T>0, the molar volume V>0, and the isothermal compressibility βT>0 (solids are compressible, and increasing pressure decreases volume, so (∂V/∂P)T is negative).
The question states that the solid has a small positive coefficient of expansion. This means α is positive (α>0) and small. Since α is positive, α2 is also positive (α2>0). Given that V>0, T>0, and βT>0, the term βTα2VT is positive. Therefore, Cp−Cv>0, which implies Cp>Cv.
Furthermore, the question states that the coefficient of expansion α is small. The difference Cp−Cv is proportional to α2. If α is small, then α2 is very small. Thus, the difference Cp−Cv is small.
Combining these two points, Cp is greater than Cv, and the difference is small. So, Cp is slightly greater than Cv.
This is consistent with the physical understanding that the difference Cp−C_v arises from the work done during expansion against the external pressure when heat is supplied at constant pressure. For solids, the expansion upon heating is generally small compared to gases, leading to a smaller difference between Cp and Cv. If the coefficient of expansion is explicitly stated to be small, this difference will be particularly small.
The comment on Cp and Cv is that Cp is slightly greater than Cv.