Question

The total entropy change for a system and its surroundings increases if the process is:
A. Reversible
B. Irreversible
C. Exothermic
D. Endothermic

Answer 1

Hint: This question is based on the statement of the Second Law of Thermodynamics. Recalling the statement, more specifically, the version of the statement involving total entropy change of the system and its surroundings will give us the solution to this question.

Complete Step by Step Solution:

The efficiency (\[\eta \]) of a Carnot engine (which is a reversible cycle of isothermal and adiabatic expansions and compressions) is given by:
\[\eta = \dfrac{{{q_2} - {q_1}}}{{{q_2}}} = \dfrac{{{T_2} - {T_1}}}{{{T_2}}}\] … (1)
Where \[{q_1}\]= heat emitted to the surroundings during isothermal compression (negative sign according to to sign convention)
             \[{q_2}\] = heat absorbed from the surroundings during an isothermal expansion (positive sign according to to sign convention)
             \[{T_1}\] = isothermal compression occurs at this temperature
             \[{T_2}\] = isothermal expansion occurs at this temperature
Equation (1) can be generalised in the following way:
\[\dfrac{{{q_2} - {q_1}}}{{{q_2}}} = \dfrac{{{T_2} - {T_1}}}{{{T_2}}}\]
\[ \Rightarrow 1 - \left( {\dfrac{{{q_1}}}{{{q_2}}}} \right) = 1 - \left( {\dfrac{{{T_1}}}{{{T_2}}}} \right)\]
\[ \Rightarrow \dfrac{{{q_1}}}{{{T_1}}} = \dfrac{{{q_2}}}{{{T_2}}}\] … (2)
Applying the sign convention to \[{q_1}\] and\[{q_2}\] in equation (2) we get:
\[\dfrac{{ - {q_1}}}{{{T_1}}} = \dfrac{{{q_2}}}{{{T_2}}}\]
\[ \Rightarrow \dfrac{{{q_2}}}{{{T_2}}} + \dfrac{{{q_1}}}{{{T_1}}} = 0\] … (3)
Thus, when isothermal and adiabatic processes are carried out reversibly in a Carnot cycle, the sum of the \[\dfrac{q}{T}\] terms would be zero.

Let’s consider a reversible cyclic process where a system is taken from state A to state B and back to state A again. Any reversible, cyclic process can be made of an infinite number of extremely small Carnot cycles. Knowing that for a Carnot cycle the sum of the \[\dfrac{q}{T}\] terms would be zero, it follows that for the reversible cycle ABA, the summation becomes
\[\sum {\dfrac{q}{T}} = 0\] … (4)
When the changes are infinitesimal, equation (4) can be expressed as:
\[\sum {\dfrac{{dq}}{T} = 0} \]
Since the cycle is performed in two steps, from A to B and from B to A, it follows that:
\[\sum {\dfrac{{dq}}{T} = \int\limits_A^B {\dfrac{{dq}}{T} + \int\limits_B^A {\dfrac{{dq}}{T}} } = 0} \]
\[ \Rightarrow \int\limits_A^B {\dfrac{{dq}}{T} = - \int\limits_B^A {\dfrac{{dq}}{T}} } \] … (5)

From equation (5), it follows that \[\int\limits_A^B {\dfrac{{dq}}{T}} \] is a definite quantity that depends only on the states A and B. Thus, \[\int\limits_A^B {\dfrac{{dq}}{T}} \]is a state function. This state function is called entropy and it is denoted by S. Change in entropy over a particular process is given as \[\Delta S = \int\limits_A^B {\dfrac{{dq}}{T}} \] . At a constant temperature, for a finite change,
\[\Delta S = \dfrac{q}{T}\] … (6)

If we consider and isothermal expansion of an ideal gas occurring reversibly, the gas performs some external work (\[w = - P\Delta V\] ) and an equivalent amount of heat is absorbed reversibly by the system from its surroundings (\[{q_{rev}}\]). Thus, the entropy of the system increases by \[\dfrac{{{q_{rev}}}}{T}\] . Since \[{q_{rev}}\]amount of heat is lost by the surroundings, its entropy decreases by \[\dfrac{{{q_{rev}}}}{T}\]as well. Thus, the net change in entropy of the system and surroundings is given by \[\dfrac{{{q_{rev}}}}{T} - \dfrac{{{q_{rev}}}}{T} = 0\] .
Therefore, we can conclude that in a thermodynamically reversible process, there is no net change in entropy of the system and surroundings. A corollary to this would be that in a thermodynamically irreversible process, the net entropy of the system and surroundings will always increase. Mathematically:
\[\left( {\Delta {S_{sys}} + \Delta {S_{sur}}} \right) = 0\] for a reversible process
\[\left( {\Delta {S_{sys}} + \Delta {S_{sur}}} \right) > 0\] for an irreversible process
This is the statement of the Second Law of Thermodynamics.
Thus, option B is correct.

Note: Some students might think option C to be the correct answer to this question. We know the relation \[\Delta G = \Delta H - T\Delta S\]. In this relation, \[\Delta S\] is the change in entropy of the system only. It is not the net entropy change of the system and its surroundings. Thus, option C is not correct.