Solveeit Logo

Question

Question: Which of the following statements is true as per the second law of thermodynamics for an isolated, o...

Which of the following statements is true as per the second law of thermodynamics for an isolated, ordered system?
A) Heat will flow into the system
B) Heat will flow out of the system
C) Work will be done by the system
D) Work will be done on the system
E) The entropy within the system will increase

Explanation

Solution

All processes are caused by energy transfers where a part of the useful work is always dissipated to thermal heat which is accompanied with entropy generation. The changes in entropy for a universe can never be negative.

Complete step by step answer:
The second law of thermodynamics states that, “the total entropy of an isolated system can never drop over time, and remains constant only if all processes are reversible.”
Isolated systems always try to evolve spontaneously towards establishing thermodynamic equilibrium.
Let the entropy change of the universe be denoted by ΔSuniv\Delta {S_{univ}} and is equal to the sums of the changes in entropy of the system and surroundings:
ΔSuniv=ΔSsys+ΔSsurr\Delta {S_{univ}} = \Delta {S_{sys}} + \Delta {S_{surr}}
ΔSuniv=qsysT+qsurrT\Rightarrow \Delta {S_{univ}} = \dfrac{{{q_{sys}}}}{T} + \dfrac{{{q_{surr}}}}{T}
For an isothermal reversible expansion, heat absorbed by the system from the surroundings is:
qrev=nRTlnV2V1{q_{rev}} = nRT\ln \dfrac{{{V_2}}}{{{V_1}}}
As the amount of heat absorbed by the system is equal to the heat lost by the surrounding, qsys=qsurr{q_{sys}} = - {q_{surr}}therefore, for a truly reversible process, the entropy change is equal to:
ΔSuniv=nRTlnV2V1T+nRTlnV2V1T=0\Delta {S_{univ}} = \dfrac{{nRT\ln \dfrac{{{V_2}}}{{{V_1}}}}}{T} + \dfrac{{ - nRT\ln \dfrac{{{V_2}}}{{{V_1}}}}}{T} = 0
For an irreversible process, the entropy change is however positive:
qsys=nRTlnV2V1>0\Rightarrow {q_{sys}} = nRT\ln \dfrac{{{V_2}}}{{{V_1}}} > 0
qsurr=nRTlnV4V3>0\Rightarrow {q_{surr}} = nRT\ln \dfrac{{{V_4}}}{{{V_3}}} > 0
ΔSuniv=qsysT+qsurrT>0\Rightarrow \Delta {S_{univ}} = \dfrac{{{q_{sys}}}}{T} + \dfrac{{{q_{surr}}}}{T} > 0
ΔSuniv=ΔSsys+ΔSsurr0\therefore \Delta {S_{univ}} = \Delta {S_{sys}} + \Delta {S_{surr}} \geqslant 0

Thus, the entropy within the system always increases in an isolated and ordered system as per second law of thermodynamics.

Option (E) is correct.

Note: The Second Law of Thermodynamics is a universal law and it is valid without any exceptions: in closed or open systems, in equilibrium or non-equilibrium, in inanimate or animate systems - that is, in all space and time scales. In reality, truly reversible processes never really occur; thus it is safe to tell that the entropy always increases for all irreversible processes.