ASSERTION
Entropy change in reversible adiabatic expansion of an ideal gas is zero.
REASON
The increase in entropy due to volume increase just compensate for the decrease in entropy due to fall in temperature.
A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion.
C. Assertion is correct but Reason is incorrect.
D. Assertion is incorrect but Reason is correct.
Answer
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Hint: Adiabatic process does not involve heat exchange between the system and the surroundings. Also, in a reversible process, the system remains in a state of equilibrium at every state. So, the system makes necessary arrangements to stay in equilibrium by controlling the necessary factors.
Complete step by step answer:
Entropy is a thermodynamic quantity which represents the degree of disorder or randomness in a system. It is represented by the letter S. It is a state function as when the state of the system changes, the entropy of the system also undergoes a change. The change in entropy is represented by $\Delta S$
(Final-Initial) and is defined by the mathematical relation
$\Delta S$ =\[{\text{ }}\dfrac{{{\text{q(rev)}}}}{{\text{T}}}\]
Where q(rev) is the heat absorbed by the system in a reversible manner at the temperature T. If q is expressed in Joules(\[{\text{J}}\]) and T in Kelvin (\[{\text{K}}\]),then the units of entropy becomes \[{\text{J}}{{\text{K}}^{{\text{ - 1}}}}\].
The physical meaning of entropy lies in its ability to provide a measure for degree of disorder or randomness of the system
As we know that in an adiabatic process, there is no heat exchange and which the ∆Q will be zero.
When we substitute this value of q in the above equation, it becomes;
$\Delta S$=\[\dfrac{{\text{0}}}{{\text{T}}}\]
$\Delta S$=0
Therefore, the entropy change in a reversible adiabatic process will be zero.
The reason for this can also be stated as that for a reversible process, the entropy change in a system is the negative of the entropy produced in the surroundings whereas it adds up with that of the surroundings in an irreversible process.
Also we must note that during adiabatic expansion the volume increases but in order to maintain the equilibrium, it is maintained by the fall in temperature.
So, the correct answer is Option A.
Additional Information:
Reversible and irreversible processes are defined according to the path or manner in which the process is carried out. Reversible processes are carried out in such a way that the system remains in a state of equilibrium at every state of the process. Whereas in an irreversible process, as the process is carried out rapidly and cannot be reverted without the help of an external agency, the system will not remain in an equilibrium.
Note: In a reversible process, the various changes taking place in the direct process can be exactly reversed and all such processes are ideal and cannot be realised in actual practice. But all the natural processes are irreversible. They cannot be reversed without the help of an external agency.
Complete step by step answer:
Entropy is a thermodynamic quantity which represents the degree of disorder or randomness in a system. It is represented by the letter S. It is a state function as when the state of the system changes, the entropy of the system also undergoes a change. The change in entropy is represented by $\Delta S$
(Final-Initial) and is defined by the mathematical relation
$\Delta S$ =\[{\text{ }}\dfrac{{{\text{q(rev)}}}}{{\text{T}}}\]
Where q(rev) is the heat absorbed by the system in a reversible manner at the temperature T. If q is expressed in Joules(\[{\text{J}}\]) and T in Kelvin (\[{\text{K}}\]),then the units of entropy becomes \[{\text{J}}{{\text{K}}^{{\text{ - 1}}}}\].
The physical meaning of entropy lies in its ability to provide a measure for degree of disorder or randomness of the system
As we know that in an adiabatic process, there is no heat exchange and which the ∆Q will be zero.
When we substitute this value of q in the above equation, it becomes;
$\Delta S$=\[\dfrac{{\text{0}}}{{\text{T}}}\]
$\Delta S$=0
Therefore, the entropy change in a reversible adiabatic process will be zero.
The reason for this can also be stated as that for a reversible process, the entropy change in a system is the negative of the entropy produced in the surroundings whereas it adds up with that of the surroundings in an irreversible process.
Also we must note that during adiabatic expansion the volume increases but in order to maintain the equilibrium, it is maintained by the fall in temperature.
So, the correct answer is Option A.
Additional Information:
Reversible and irreversible processes are defined according to the path or manner in which the process is carried out. Reversible processes are carried out in such a way that the system remains in a state of equilibrium at every state of the process. Whereas in an irreversible process, as the process is carried out rapidly and cannot be reverted without the help of an external agency, the system will not remain in an equilibrium.
Note: In a reversible process, the various changes taking place in the direct process can be exactly reversed and all such processes are ideal and cannot be realised in actual practice. But all the natural processes are irreversible. They cannot be reversed without the help of an external agency.
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