Question

Entropy change in reversible adiabatic process is:
A. Infinite
B. Zero
C. Equal to $ {C_v}\Delta T $
D. Equal to $ nR\ln \left( {\dfrac{{{V_2}}}{{{V_1}}}} \right) $

Answer 1

Hint :Entropy is a thermodynamic term used to measure the extent of disorder or randomness of the system. Reversible Adiabatic processes are characterized by an increase in entropy, or degree of disorder but in irreversible adiabatic processes, there is no change in entropy if they are reversible. Adiabatic processes cannot decrease the entropy.

Complete Step By Step Answer:
We know that the entropy of an object is a measure of the number of possible arrangements of the atoms in a system. In simple words, entropy is a measure of uncertainty or randomness in a thermodynamic process.
Consider a cylinder fitted with a frictionless and Weight less piston which contains a gas and is in contact with a large heat reservoir. During isothermal and reversible. Expansion of the gas from volume $ {V_1} $ and $ {V_2} $ , let there be absorption of heat, $ q $ at temperature $ T $ .
 $ \therefore $ Change in entropy of the system $ \left( {\Delta {S_{sys}}} \right) = \dfrac{{{q_{rey}}}}{T} $
Since an equivalent amount of heat will be lost by the reservoir, change in entropy of reservoir will be $ \Delta {S_{surr}} = \dfrac{{ - {q_{rev}}}}{T} $
Total change in entropy $ \left( {\Delta {S_T}} \right) = \Delta {S_{sys}} + \Delta {S_{surr}} $ .
 $ \therefore \Delta {S_T} = \dfrac{{{q_{rev}}}}{T} - \dfrac{{{q_{rev}}}}{T} $
Thus $ \Delta {S_T} = 0 $
For reversible process, $ \Delta {S_T} = 0,{\text{ }}\Delta {{\text{S}}_{sys}} = 0,{\text{ }}\Delta {{\text{S}}_{surr}} = 0 $
For adiabatic reversible processes, the change in heat remains constant, in an adiabatic process zero heat is exchanged with the surroundings. If the Adiabatic process is reversible then the process should be slow enough that the system remains in equilibrium throughout the whole process.
 $ \Delta q = 0 $
 $ \therefore q = 0 $
Then
  $ \Delta {S_{sys}} = + \dfrac{{{q_{rev}}}}{T} = 0 $
 $ \Delta {S_{surr}} = + \dfrac{{{q_{rev}}}}{T} = 0 $
 $ \therefore \Delta {S_T} = 0 $
So, entropy change in reversible adiabatic process is zero.
Hence, the correct option is (B).

Note :
Entropy is a state function or state variable because it depends upon the initial and final state of the system. Entropy is an extensive property. An extensive property is the property which depends on size or mass, entropy = q/T and q in itself is dependent on the mass, therefore, it is an extensive property. Mass is also an extensive property.