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*Entropy:*

Entropy is the measure of disorder of the system. Disorder refers to is actually number of different microscopic states a system can be in, given that system has a particular fixed composition, pressure, volume, energy, and temperature. By microscopic states, we mean exact states of all molecules making up system.

*Change in Entropy:*

Change in entropy ΔS of the system during the process which takes system from initial state i to final state f is stated as

ΔS = S_{f} - S_{i} = ∫_{i}^{f}dQ/T

Equation implies that change in entropy depends on both energy transfer as heat during process and temperature of system. S.I unit of entropy and change in entropy is Joule per Kelvin (JK^{-1}).

**Reversible Process:**

For the reversible process which takes place in closed system, entropy always remains constant. That is, change in entropy for reversible process is zero (ΔS = 0 ). This statement can be described using Carnot cycle.

**Irreversible Process***:*

The irreversible process can be stated as one which direction can't be reversed by the infinitesimal change in some property of system. The good example of irreversible process shows the hot reservoir at temperature T_{H} in thermal contact through a conductor with a cold reservoir at temperature T_{C}. Assume arrangement is isolated from surrounding (i.e. no heat flows in or out of arrangement), heat flows from hot reservoir to cold reservoir and not in reversed direction. This procedure is the irreversible process. Assume heat Q flows from hot reservoir to cold reservoir, we can then write equation for total entropy change ΔS during the process.

The change in entropy for the cold reservoir ΔSC, using equation

ΔSC = Q/Tc

ΔSC is is positive because heat Q flows into the cold reservoir and is positive Q . Similarly the change in entropy of the hot reservoir DSH, using equation is

ΔSH = -Q/TH

ΔSH is negative because heat Q flows out of the hot reservoir (i.e. - Q ).

The total entropy is

ΔS = ΔSH + ΔSc

which gives ΔS = -Q/TH + Q/TC > 0

The total entropy ΔS is greater than zero because TH is greater than TC. We can then say that ΔS > 0 for irreversible process.

**Change in Entropy during Adiabatic and Isothermal Processes:**

Reversible adiabatic process: during adiabatic process, dQ = 0 , and this means that during reversible adiabatic process dQ_{r} =0. Then, according to change in Entropy equation:

dS = 0

Reversible isothermal process: The example of reversible isothermal process is phase change and this takes place at constant pressure during which temperature also remains constant.

S_{f} - S_{i} = ∫_{i}^{f}dQr/T = 1/T∫_{i}^{f}dQr = Q/T

Where Q is latent heat or latent heat of transformation.

*Entropy as a State Function:*

Entropy, like pressure, energy, and temperature, is the property of state of the system and is independent of how that state is reached. Above statement can be justify from equation of first law of thermodynamics

dU = dQ - dW

For a reversible isobaric process,

dQ = PdV + nCpdT

But change in entropy ΔS is already given to be dQ/T in equation

ΔS = S_{f} - S_{i} = nRlnV_{f}/V_{i} + nCplnT_{f}/T_{i}

Equation therefore means that change in entropy ΔS between initial state and final state of ideal gas depends only on properties of initial and properties of final state (i.e. ΔS doesn't depend on how gas changes between two states).

*Second Law of Thermodynamics:*

Change in entropy ΔS, for the process occurring in closed system is zero for reversible process and greater than zero for irreversible process. This is one of the statements of second law of thermodynamics. Second law of thermodynamics provides direction in which natural process will take place.

If the process takes place in the closed system, entropy of the system increases for irreversible processes and remains constant for reversible processes. Entropy of the system never decreases.

Mathematical statement of Second law of thermodynamics

ΔS ≥ 0

*Carnot Engine:*

Carnot engine is the most efficient type of heat engine and due to this it is sometime referred to as the ideal heat engine. Ideal in sense that all processes in cycle are reversible and no wasteful energy transfer take place because of friction and turbulence. During each cycle, engine (i.e. working substance) absorbs energy Q_{H} as heat from the thermal reservoir at constant temperature T_{H} and ejects energy Q_{C} as heat to second reservoir at the constant lower temperature T_{C}.

**Description of Processes:**

Process de is the isothermal process during which heat Q_{H} is transferred at temperature T_{H} to working from hot reservoir. This causes gas to go through isothermal expansion from volume Vd to Ve.

Process ef is the adiabatic expansion i.e no heat is added as working substance expands from volume Ve to V f. Temperature decreases during the process from T_{H} to T_{C}

Process fg is the isothermal process during which heat Q_{C} is transferred at temperature T_{C} from working to cold reservoir. This causes gas to go through isothermal compression from volume V f to Vg.

Processes gd is the adiabatic compression i.e. no heat is transferred as working substance compresses from volume Vg to Vd. Temperatures increases during process from T_{C} and T_{H}.

Work: From first law of thermodynamics (ΔU = Q - W), total work done during the cycle or cyclic process can be computed. For the cyclic process DU = 0, total heat transfer per cycle Q = Q_{H} - Q_{C}, and total work done is W. Then, first law of thermodynamic for Carnot cycle is

W = Q_{H} - Q_{C}

**Efficiency of the Carnot Engine:**

Efficiency of the heat engine is

E = W/QH = work output/heat (in one cycle)

The equation becomes:

E = Q_{H} - Q_{C}/Q_{H} = 1-Q_{C}/Q_{H}

But ratio of rejected heat Q_{C} to input heat Q_{H} for reversible process can be written as

Q_{C}/Q_{H} = T_{C}/T_{H}

Where temperature T_{C }and T_{H} are temperatures in Kelvin. Rearranging equation gives

Q_{H}/T_{H} = Q_{C}/T_{C}

E = 1 - T_{C}/T_{H}

This equation is efficiency of the Carnot engine operating between two heat reservoirs in which heat is taking in at temperature T_{H }and heat is ejecting at temperature T_{C}. Temperatures T_{H} and T_{C} should be in Kelvin.

**Another Statement of Second Law of Thermodynamics:**

The efficiency of an ideal heat engine is given by equation. The implication of this is that no heat engine can have efficiency greater than that of a Carnot engine. It is clear from equations that the efficiency of a Carnot engine is less that 100 %. This of course is another statement of the second law of thermodynamics. That is, the efficiency of heat engine is always less than 100 %. This statement is called Kelvin's statement.

**Entropy Change of the Reversible Process:**

The good example of reversible thermodynamic process is Carnot cycle. In the Carnot engine, there are only two reversible energy transfers as heat (that is entropy change at T_{H} and at T_{C}). Net entropy change per cycle is

ΔS = ΔS_{H} + ΔS_{C} = Q_{H}/T_{H} - Q_{C}/T_{C}

Where ΔS_{H} is positive entropy as energy Q_{H }is added to working substance (i.e. increase in entropy) and ΔS_{C} is negative entropy as energy QC is removed from working substance as heat (i.e. decrease in entropy). Then equation becomes

ΔS = ΔS_{H} + ΔS_{C} = Q_{H}/T_{H} - Q_{C}/T_{C} = 0

Therefore the entropy for a reversible process ΔS is zero (i.e. ΔS = 0).

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