In an isobaric process, when temperature changes from T1 to T2, \[\Delta S\] is equal to:
 A. \[2.303Cp\log (\dfrac{{{T_2}}}{{{T_1}}})\]
 B.\[2.303Cp\ln (\dfrac{{{T_2}}}{{{T_1}}})\]
 C.\[Cp\ln (\dfrac{{{T_1}}}{{{T_2}}})\]
 D.\[Cp\ln (\dfrac{{{T_2}}}{{{T_1}}})\]

Hint: Isobaric process is a process in which pressure is constant and work is being calculated. Also, \[\Delta S\] is the entropy change that occurred in the process.

Step by step answer: Isobaric process is a thermodynamic process in which the pressure stays constant i.e. \[\Delta P = 0\] . The heat transferred to the system works and changes the internal energy of the system. Since, the pressure is constant, the force exerted is constant and the work done is given as \[p\Delta V\] . Also, the heat transferred into the system at a certain rate. This is called isobaric expansion. The formula for this process is given as
 \[W = \int {pdV} \]
Here, W is work done by the system, \[p\] is the pressure and \[V\] is the volume and d is the differential which means the change over the whole process. Here, positive work means work done by the system. Now, in this process \[p\] is constant and therefore it cannot be inside integration. So the updated formula is
 \[W = p\Delta V\]
We can also apply the ideal gas law in the equation as given below
 \[W = nR\Delta T\]
Now, for change in temperature it will be given as
 \[W = nR({T_2} - {T_1})\]
Now, according to the question, \[\Delta S\] means the entropy change. Entropy is the measure of a system’s thermal energy per unit temperature that is unavailable for doing useful work. It is the measure of disorder or disorder or randomness. A positive entropy means increase in disorder. Now, for entropy change in the isobaric process is derived and given as follows:
 \[dq = nCpdT\]
 \[\Rightarrow \dfrac{{dq}}{T} = nCp\dfrac{{dT}}{T}\]
 \[\Rightarrow \int\limits_{T1}^{T2} {\dfrac{{dq}}{T}} = nCp\int\limits_{T1}^{T2} {\dfrac{{dT}}{T}} = nCpLn(\dfrac{{{T_2}}}{{{T_1}}})\]

Therefore, the correct answer is option D.


Note: Students may go wrong with the sign of the change in entropy obtained. Its sign can not be neglected as it provides the validity of the processes that have happened.