As per second law of thermodynamics a process taken place spontaneously if and only if the entropy of the universe increases due to the process. Change in entropy is given by Delta S = (Q_(rev))/(T) A gas C_(V) = (0.2 T) Cal K^(-1). What is the change in its entropy when one mole of it is heated from 27^(@)C to 127^(@)C at constant volume ?

As per second law of thermodynamics a process taken place spontaneousl

1.	As per second law of thermodynamics a process taken place spontaneously if and only if the entropy of the universe increases due to the process. Change in entropy is given by Delta S = (Q_(rev))/(T) A gas C_(V) = (0.2 T) Cal K^(-1). What is the change in its entropy when one mole of it is heated from 27^(@)C to 127^(@)C at constant volume ?
Answer» `20 "cal "K^(-1) mol^(-1)` `15 "cal "K^(-1) mol^(-1)` `35 "cal "K^(-1) mol^(-1)` `25 "cal "K^(-1) mol^(-1)` Solution :`Delta S = UNDERSET(T_(1))OVERSET(T_(2))int (N C_(V) DT)/(T) + underset(V_(1))overset(V_(2))int (pdv)/(T)` For the GIVEN condition `Delta S= underset(T_(1))overset(T_(2))int (n C_(v)dT)/(T) = (0.2T)/(T) dT` `= 02 (T_(2) - T_(1)) = 20` cal/mol- k