Question

Which of the following is correct in terms of increasing work done for the same initial and final state?
A. $Adiabatic < Isothermal < Isobaric$
B. $Isobaric < Adiabatic < Isothermal$
C. $Adiabatic < Isobaric < Isothermal$
D. None of these

Answer 1

Hint: In this problem, to find out the correct order of work done in different processes such as isobaric, isothermal and adiabatic; we have to first analyze the conditions (such as pressure, temperature, heat) of given processes. Then, Plot a P-V graph as the area of the P-V graph is proportional to the work done.

Complete answer:
We know that for the Isobaric process, $P = $ constant
For the Isothermal process, $T = $ constant
and for the Adiabatic Process, there is no heat transfer i.e., $P{V^\gamma } = $ constant

Also, we know that in any thermodynamic process work done is always equal to the area covered by $P - V$ curve with the volume axis.

Let us draw $P - V$ graph of each of the given processes below: -


We can easily observe from the above graph that work done for all the three given processes i.e., Isobaric, Isothermal, and Adiabatic process is in the order:
${W_{Adiabatic}} < {W_{Isothermal}} < {W_{Isobaric}}$

Thus, the correct increasing order for work done is -$Adiabatic < Isothermal < Isobaric$. Hence, the correct option is (A) $Adiabatic < Isothermal < Isobaric$.

Note: Since this is a theoretically based problem, it is necessary to scientifically evaluate each and every option provided and plot a graph carefully for each isobaric, isothermal, and adiabatic process which will help in selecting an option that is more relevant and delivering the solution with proper explanation.