
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.\[Cp\ln (\dfrac{{{T_2}}}{{{T_1}}})\]
Answer
483.6k+ views
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.
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.
Recently Updated Pages
Master Class 12 Social Science: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Class 10 Question and Answer - Your Ultimate Solutions Guide

Master Class 10 Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Trending doubts
The gas that burns in oxygen with a green flame is class 12 chemistry CBSE

The probability that a leap year will have only 52 class 12 maths CBSE

Describe the poetic devices used in the poem Aunt Jennifers class 12 english CBSE

And such too is the grandeur of the dooms We have imagined class 12 english CBSE

What does the god that failed refer to class 12 english CBSE

Which country did Danny Casey play for class 12 english CBSE
