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## Reversible, Irreversible and Cyclic Process, Chemistry tutorial

Introduction:It is been observed that the transfer of energy between a system and its surroundings occurs via heat and work. This is regulated by the first law of thermodynamics which states that the increase in the energy of a system should be accompanied via an equivalent decrease in the energy of the Surroundings and vice-versa. Though, it doesn't tell us anything regarding the feasibility and direction of flow of energy. According to the first law of thermodynamics, all methods in which energy is conserved are possible. For illustration, if a cup of hot tea is left on a table then according to the first law it may be cooled by transferring energy to the surroundings or be heated via absorbing energy from the surroundings. However we all are familiar from daily experience that the cup of tea will forever cool till it needs the temperature of the surroundings. Likewise, if a bottle of perfume is opened in a room, the perfume spreads all through the room. The reverse procedure in which all the perfume vapours are collected in the bottle does not take place. These are illustrations of what are known as spontaneous processes that are irreversible and carry out only in one direction. Again, according to the first law, there is a direct relationship between the heat and work. However, it doesn't state us whether heat can be fully transformed into work and if so, what is the effect on the system and the surroundings.

:Reversible, Irreversible and cyclic processesAs we are familiar that all the thermodynamic properties are state functions and are independent of the path adopted via the system. As well, the internal energy change of a system is represented by the equation ΔU = q + w. Here, ΔU is independent of the path selected however q and w certainly based on it. Therefore for the same ΔU, different values of q and w are possible by bringing about the method in different ways. It was as well illustrated that the work done by a system is maximum if a reversible path is adopted and this maximum work can be found out from the initial and final states of the system. Let us take a reversible and an irreversible procedure in which ΔU is same. No matter how we carry out the process (that is, reversible or irreversible), ΔU based only on the initial and final states of the system. Therefore,

ΔU = q

_{rev}+ w_{rev }for reversible processAnd ΔU = q

_{rev}+ w_{irrev}for irreversible processAnd therefore q

_{rev }+ w_{rev }= q_{rev }+ w_{irrev}We are familiar that the work done by a system under the reversible conditions (-w

_{rev}) is bigger than the work done by a system under irreversible conditions (-w_{irrev}).That is, -w

_{rev }> -w_{irrev }Therefore, w

_{rev}< w_{irrev }This is true whenever we compare the work done on the system under reversible and irreversible conditions. Let us suppose that the driving forces beneath the two conditions illustrated above are fairly different. As well, let us suppose that the processes are not adiabatic in such a way q

_{rev }or q_{irrev }is not equivalent to zero. By using the equations q_{rev }+ w_{rev}= q_{rev}+ w_{irrev}and w_{rev}< w_{irrev}, we can represent,q

_{rev }> q_{irrev }This signifies that in a non-adiabatic process, heat absorbed via a system from the surroundings is more under reversible conditions than under irreversible conditions. This relationship will be employed while computing the entropy changes of isolated systems.

Now, suppose ΔU, q and w are the values of a system in a cyclic process. As we know a cyclic process is one in which the system after undergoing any number of processes return to its initial state. This signifies that ΔU = 0; therefore, the work system throughout all such processes must be equivalent to the heat absorbed by the system.

That is, q = q1 + q2 + .... = - (w1 + w2 + ...) = - w

Here q and w are the heat absorbed and work done on the system in the whole cyclic processes comprising of some processes; in the individual processes, q

_{1}, q_{2}.... and so on, are the heat absorbed via the system and, w_{1}, w_{2}.... and so on, are the work done on the system.:The Carnot CycleCarnot analyzed the functioning of an engine by the given features:

a) The engine works in cycles.

b) It absorbs heat from the reservoir termed as source.

c) It does some work out of the heat absorbed.

d) It returns the unused portion of the heat to the other reservoir, termed as sink.

e) Lastly it returns to its original state.

Such an engine is termed as Carnot engine. The temperature of the source (T

_{H}) is higher than that of the sink (T_{C}). The source and the sink are supposed to be of infinite heat capacity; that is, the temperatures of the source and the sink are not influenced by small amounts of heat exchange.Carnot illustrated that the whole amount of the heat absorbed can't be converted into work in a cyclic process, no matter how ideal the heat engine is. He expressed that only a fraction of the net heat absorbed is converted into work and this fraction is termed as the efficiency of the Carnot engine. Now we are going to derive an equation helpful in computing its efficiency.

For the sake of simplicity, let us suppose that the engine comprises of a cylinder and a piston having lone mole of an ideal gas in between the two. The cylinder has perfectly insulated; walls and a perfectly conducting base; the piston is frictionless. This is just for the sake of convenience that we have taken that the engine has ideal gas in reality there can be any appropriate fluid. We make use of the given expressions (for one mole of gas).

w

_{isothermal}= RT ln (V_{initial}/V_{final})w

_{adiabatic}= C_{V}‾(T_{final }- T_{initial})Here, 'w' is the work done on the system and C

_{V}‾ is the molar heat capacity of the gas.The plot of the pressure-volume graph is as shown below:

Fig: Carnot Cycle graph

The Carnot cycle operation can be explained by the given four steps:

1) Isothermal Expansion

2) Adiabatic expansion

3) Isothermal compression

4) Adiabatic compression

Step I: Isothermal Expansion:

Primarily the gas consists of pressure p

_{1}and volume V_{1}. The cylinder is positioned on a heat source maintained at the temperature T_{H}. The gas is isothermally and reversibly expanded to a volume V_{2}and pressure p_{2}. Let the work done on the gas be w_{1 }and the heat absorbed from the source be q_{H}. In the isothermal process, ΔU = 0.Therefore, by using equation ΔU = U

_{f}- U_{i}= q + w and W = - nRT ln(V_{2}/V_{1}) = nRT ln(V_{1}/V_{2})w

_{1}= q_{H}= RTH ln(V_{1}/V_{2})Or q

_{H}= RT_{H}ln(V_{2}/V_{1})Step II: Adiabatic Expansion:

The cylinder is now put on a thermally insulated stand and the gas is adiabatically and reversibly expanded till it attains a pressure p

_{3}, volume V_{3}and temperature T_{C}. Throughout this period, no heat is absorbed by the system. The work done on the gas, w_{2}, as the gas gets cooled from T_{H}to T_{C}is represented by using ΔU = W = nC_{V}‾ (T_{2}- T_{1}) = nC_{V}‾ΔT as,w

_{2}= C_{V}‾ (T_{C}- T_{H})Step III: Isothermal Compression:

The cylinder is now put on a sink at temperature T

_{C}and the gas is isothermally and reversibly compressed to the volume V_{4}at pressure p_{4}. Throughout the process the work done on the gas is w_{3}and the heat evolved to the sink is q_{c}(or q_{c}is the heat absorbed from the sink).By using equation ΔU = U

_{f}- U_{i }= q + w and W = - nRT ln(V_{2}/V_{1}) = nRT ln(V_{1}/V_{2})w

_{3}= - q_{c}= RT_{C}ln (V_{3}/V_{4})Or q

_{C }= RT_{C}ln (V_{4}/V_{3})Step IV: Adiabatic Compression:

In the last step, the cylinder is again put on an insulating stand and the gas is adiabatically and reversibly compressed till it reaches its initial state of volume V

_{1}and temperature T_{H}. Therefore the work done on the gas, w_{4}is given by the equation ΔU = W = nC_{V}‾ (T_{2}- T_{1}) = nC_{V}‾ΔT as,w

_{4}= C_{V}(T_{H }- T_{C})The total work done on the system is represented by:

w = w

_{1}+ w_{2 }+ w_{3}+ w_{4}Or w = RT

_{H }ln V_{1}/V_{2}+ C_{V}‾ (T_{C}- T_{H}) + RT_{C }ln V_{3}/V_{4}+ C_{V}‾ (T_{H}- T_{C})That is, w = RT

_{H}ln V_{1}/V_{2 }+ RT_{C}ln V_{3}/V_{4}The equation T

_{2}/T_{1}= ln (V_{1}/V_{2})^{γ - 1 }can be applied to relate the initial and final values of volume and temperature of the two adiabatic processes illustrated in steps II and IV.Applying the above equation to step II we obtain,

T

_{C}/T_{H}= (V_{2}/V_{3})^{γ - 1}Likewise applying the equationT

_{2}/T_{1}= ln (V_{1}/V_{2})^{γ - 1}to step IV, we obtain:T

_{C}/T_{H}= (V_{4}/V_{1})^{γ - 1}or T_{C}/T_{H}= (V_{1}/V_{4})^{γ - 1}Therefore, (V

_{2}/V_{3})^{γ - 1}= (V_{1}/V_{4})^{γ - 1}That is, V

_{2}/V_{3}= V_{1}/V_{4}Or V

_{3}/V_{4}= V_{2/}V_{1}By using the above in equation w = RT

_{H}ln V_{1}/V_{2}+ RT_{C }ln V_{3}/V_{4}, we obtain:w = RT

_{H}ln V_{1}/V_{2}+ RT_{C}ln V_{2}/V_{1}Therefore, total work done on the system = w = R (T

_{H}- T_{C}) ln V_{1}/V_{2}Or total work done by the system, = w' = - w = R (T

_{H}- T_{C}) ln V_{2}/V_{1 }As work done by the system = - (work done on the system)

We are familiar that the heat exchange between the gas and the source or sink occurs only in the isothermal processes (steps 1 and 3); in adiabatic processes (steps 2 and 4), there is no heat exchange. Again, q

_{H}is the heat absorbed from the source in step 1 and q_{c}is the heat absorbed from the sink in the step 3. The net heat absorbed by the system:q = q

_{H}+ q_{C}As predicted, q turns out to be equivalent to - w or w' as for the overall cyclic process, ΔU = 0

Therefore, q= q

_{H}+ q_{C }= w' = - w = R (T_{H}- T_{C}) ln V_{2}/V_{1 }It might though be noted that out of the heat q

_{H}, (= RT_{H}ln V_{2}/V_{l}as per equation absorbed from the source, only some of it is converted to helpful work and the rest is lost to the sink. Let us now compute w'/q_{H}, that is, the ratio between the total work done by the system throughout one cycle and the heat absorbed in the first step. This quantity is known as the efficiency 'η' of a Carnot engine.Efficiency (η) = Total work done by the system/Heat Absorbed from the source at higher temperature

Efficiency (η) = w'/q

_{H}By using the equations q

_{H}= RT_{H}ln (V_{2}/V_{1}) and q= q_{H}+ qC = w' = - w = R (T_{H }- T_{C}) ln V_{2}/V_{1}(η) = (q

_{H }+ q_{C})/q_{H }= [R (T_{H}- T_{C}) ln V_{2}/V_{1}]/[RT_{H}ln V_{2}/V_{1}](η) = (q

_{H }+ q_{C})/q_{H }= (T_{H}- T_{C})/T_{H}As T

_{C}and T_{H }are always positive and T_{C}/T_{H }is less than one, equation above can be rearranged as:(η) = 1 T

_{C}/T_{H}= 1 + (q_{C}/q_{H}) = < 1As q

_{C}is negative and q_{H}is positive, (q_{C}/q_{H}) is a negative quantity; 1 + (q_{C}/q_{H}) is as well less than one.This signifies that the efficiency is for all time less than one, that is, all the heat absorbed at a higher temperature is not converted into work. This is as well clear that efficiency will be more if the ratio T

_{C}/T_{H}is small. Therefore, for efficient working of the engine, it must absorb heat at as high a temperature as possible and refuse it at as low a temperature as possible. It must as well be noted that the efficiency is independent of the nature of the fluid. This is termed as Carnot theorem which can as well be represented as:In all the cyclic engines working between the similar temperatures of the source and the sink, the efficiency similar.

It should be pointed out that in the Carnot cycle; all the processes have been taken out reversibly. Therefore, maximum and minimum amount of work are comprised in expansion and compression, correspondingly; this means that there can't be any engine more efficient than Carnot engine. In real engines, there is irreversibility due to unexpected expansion and compression and as well due to the friction of the piston.

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