## Essential University Physics: Volume 1 (3rd Edition)

a) The molecule is diatomic, so it has 5 degrees of freedom, making the value for $C_v=\frac{5}{2}R$. Thus, we use the equation for entropy change at constant volume (found in problem 45) to find: $\Delta S = (6.36)(\frac{5}{2}R)ln(\frac{552}{288})=86\ J/K$ b) The molecule is diatomic, so it has 5 degrees of freedom, making the value for $C_p=\frac{7}{2}R$. Thus, we use the equation for entropy change at constant pressure (found in problem 46) to find: $\Delta S = (6.36)(\frac{7}{2}R)ln(\frac{552}{288})=120.40 \ J/K$ c) In the adiabatic process, the change in entropy will be 0.