Question

Reversible expansion of an ideal gas under isothermal and adiabatic conditions are as shown in the figure. AB $\rightarrow$ Isothermal expansion AC $\rightarrow$ Adiabatic expansion Which of the following options is not correct ?

• A. $\Delta S_{isothermal} > \Delta S_{adiabatic}$
• B. T$_A$ = TB
• C. $W_{isothermal} > W_{adiabatic}$
• D. T$_c$ > TA

Answer 1

The Correct Option is D

 To determine the correct option, let's analyze the given conditions and statements:

1. Reversible Isothermal Expansion (AB): In an isothermal expansion, the temperature remains constant, and the change in entropy, \(\Delta S\), is positive because the system absorbs heat from the surroundings. Therefore, \(\Delta S_{isothermal} > 0\).
2. Reversible Adiabatic Expansion (AC): In an adiabatic expansion, there is no heat exchange with the surroundings, so the change in entropy, \(\Delta S\), is zero because the process is reversible. Hence, \(\Delta S_{adiabatic} = 0\).
3. Therefore, \(\Delta S_{isothermal} > \Delta S_{adiabatic}\) is a correct statement.
4. In an isothermal process, the work done (\(W_{isothermal}\)) is greater than that in an adiabatic process (\(W_{adiabatic}\)) because energy is conserved by heat transfer, which is missing in adiabatic expansion. Hence, \(W_{isothermal} > W_{adiabatic}\)
5. Given that in isothermal expansion (AB), temperature remains constant, the temperature at the start of the adiabatic expansion (at point A) is the same as the end of isothermal expansion (at point B). So, \(T_A = T_B\) is accurate.
6. Adiabatic expansion results in a temperature drop because no heat is absorbed to compensate for the work done, hence \(T_C < T_A\).

The incorrect statement, therefore, is the one suggesting \(T_C > T_A\), which contradicts the behavior of adiabatic processes described here.

Thus, the correct answer is the option \(T_C > T_A\).