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As per second law of thermodynamics a pr...

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 ?

A

`20 "cal "K^(-1) mol^(-1)`

B

`15 "cal "K^(-1) mol^(-1)`

C

`35 "cal "K^(-1) mol^(-1)`

D

`25 "cal "K^(-1) mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
`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
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