Question

Heat capacity can be defined in two ways: 27.) ДТ )v-const. AT P=const. Use conservation of energy and the equipartition theorem to calculate expressions for both of these for solids, monatomic gases, and diatomic gases. Use the table of Cp values, provided, to check that you ca approximately, account for most of these.

Answer 1

Given that Heat capacity can be defined in two ways CV = (Q/Delta T)_v = constant C_P = (Q/Delta T) P = constant From the equi partition theorem every degree of freedom has internal energy KBT/2 calculate the expression for solid Degree of freedom for solid is 6 U = 6 K_BT/2 = 3 K_BT u = 3 K_BT C_v = dU/dT = 3 K_B C_v = 3 K_B And molar specific heat is C_v = 3R and Cp = Cv + R Cp = 3R + R cp = 4R Calculate the expression for monatomic gases Degree of freedom f = 3 Cv = fK_B/2 implies fR/2 per mole C_v = 3R/2 [therefore Cp = C_v + R] Cp = 5R/2 Calculate the expression for diatomic gasses diatomic gasses implies Degrees of freedom = 5 Three translations + two Rotations C_v = 5R/2 cp = 7R/2