Law of Thermodynamics

a Laws of Thermodynamics and Their ApplicationFirst law of thermodynamics: Energy can neither be created nor be destroyed, it can only be transferred from one form to another.Second law of thermodynamics: The entropy of any isolated system always increases.Third law of thermodynamics: The entropy of a system approaches a constant value as the temperature approaches absolute zero.Zeroth law of thermodynamics: If two thermodynamic systems are in thermal equilibrium with a third system separately are in thermal equilibrium with each other.→ Entropy is the measure of the number of possible arrangements the atoms in a system can have.→ Enthalpy is the measurement of energy in a thermodynamic system. First Law of Thermodynamics The law of thermodynamics maybe stated in two different ways:(i) Mechanical equivalent of heat (Joule's Iaw): This form describes the equivalence between mechanical work and heat energy. It states that, "Whenever, work is transformed into heat or heat into work, the quantity of work is mechanically equivalent to the quantity of heat."i.e., W∝Q$\mathrm{W}\propto \mathrm{Q}$or, W=JQ$\mathrm{W}=\mathrm{J}\mathrm{Q}$Where, Q=$Q=$ Heat energy, W=$W=$ Mechanical work, J=$\mathrm{J}=$ Constant known as Joule's mechanical equivalent of heat. Its value is J=4.2~J/Cal$\mathrm{J}=4.2\mathrm{~}\mathrm{J}/\mathrm{C}\mathrm{a}\mathrm{l}$ or 4.2×107$4.2\times {10}^7$ ergs/cal. In S.I. systems both W$\mathrm{W}$ and Q$\mathrm{Q}$ measured in Joule, then J=1$\mathrm{J}=1$.∴ W=Q$\therefore W=Q$Note: This form is true only if the whole of work done is used in producing heat or vice versa.(ii) Law of conservation of energy (or differential form of first law of thermodynamics) It states that, "In all transformations, the energy supplied must be balanced by the external work done plus the increase in internal energy."This statement is based on the conservation of energy i.e., "energy can neither be created nor destroyed but only converted from one form to another." If dQ$\mathrm{d}\mathrm{Q}$ be the heat supplied to the system, dW $\mathrm{d}\mathrm{W}$ is external work done and dU$dU$ be the internal energy, then according principle of conservation of energy: dQ = dW + dUNote: Sign convention for first law of thermodynamics dQ : +ve if heat is given to the systemdQ :-ve if heat taken out of the systemdW: +ve if work is done by the system (gas is compressed)dW : -ve if work is done on the system (gas expands)dU$dU$ : +ve if internal energy is increased dU: -ve if internal energy is decreasedPhysical Significance of first law of thermodynamic The first law of thermodynamic establishes an exact relationship between heat and work. According to this, a definite quantity of heat will produce a definite amount of work and vice versa. It against (or denies) that work or energy can be created without anything. So, this law implies that it is impossible to construct a thermal machine which may operate without any expenditure of fuel and may thus create energy without any things. If a machine do this could run itself and in generally perpetual motion machine of the first kind. The first law rules out such machines. Therefore, it may also stated as, "perpetual motion machine of the first kind is impossible." Thus, first law of thermodynamics includes three ideas which are stated as below:(i) Heat is a forn of energy in transit.(ii) Conservation of energy takes place in thermodynamic system.(iii) Energy system in equilibrium, possesses an internal energy which is a function of state i.e., depends only upon nitial and final states of the system.Limitations of First Law of Thermodynamics1. It is unable to explain what amount of supplied heat is used to inccrease the internal energy and for the external work done.2. It could not explain the direction of flow of heat.3. It cannot explain whether heat can flow from a body of lower temperature to a body at higher temperature or not. (working of refrigerator). Second Law Thermodynamics# State second law of thermodynamics. Show that two statement of second law are equivalent. Second law of thermodynamics can be state in two forms:(i) Kelvin-Planck's (K - S) statement: "It is impossible to construct a device which operating in a cycle, has sole effect of extracting heat from a reservoir and performing an equivalent amount of work." Thís law implies that, "there is no perfect heat engine." (ii) Clausius statement: "It is impossible to construct self acting machine working in a cyclic process, without aid of external work to transfer heat from a body at a low temperature to a body at a higher temperature."(It is impossible to construct a device operating in a cycle that can transfer heat from a colder body to warmer without consuming any work.)i.e., "Heat cannot flow of itself from a colder body to a hotter body." This law implies that "there is no perfect refrigerator" So this statement is formulated by studying performance of refrigerator. These both laws apply to the reversible cyclic processes only. At first sight, these statements appear to be quite unconnected, but we shall see that they are in all respects equivalent. This can be shown as below:

Suppose a refrigerator transfer an amount of heat Q2${\mathrm{Q}}_2$ from a colder to a warmer body without any external work done on it from outside agency. Thus, this violates the Clausius statements of the second law. Let us now suppose, heat engine working between the same source and sink, absorbing heat Q1${\mathrm{Q}}_1$ from hot body and rejecting Q2${\mathrm{Q}}_2$ amount of heat to the cold body by performing: W=Q1-Q2$\mathrm{W}={\mathrm{Q}}_1-{\mathrm{Q}}_2$ work done from it. Thus, this engine does not violates any law. Now the refrigerator and heat engine are coupled together to work as a self acting device as shown in above figure. From figure it is seen that the cold body is giving out Q2${\mathrm{Q}}_2$ heat and receiving back the same amount of heat. Therefore, the net result is no change in its heat contents. On the other hand, the hot body receiving Q2 ${\mathrm{Q}}_2$ heat and giving out Q1${\mathrm{Q}}_1$ amount of heat. This in each cycle, it loses Q1-Q2${\mathrm{Q}}_1-{\mathrm{Q}}_2$ amount of heat and is completely converted into work without rejecting any heat to the cold body. Thus, this violates the Kelvin Planck's statement of second law.Similarly, we can show that violation of K$K$ - P statement leads to violation of Clausius statement of second law. Hence, the both the statement are equivalent statement of second law of thermodynamics. Heat Engine

It is the device which converts heat energy into mechanical energy. In a heat engine, working substance absorbs heat Q1${\mathrm{Q}}_1$ from a high temperature reservoir (i.e, source) at T1${\mathrm{T}}_1$ and performs some mechanical work) W$\mathrm{W}$ (i.e., external work using a part of energy of Q1${\mathrm{Q}}_1$ ). Then given out the rest of heat energy to the sink$\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{k}$ (i.e., Q2${\mathrm{Q}}_2$ ) at temperature T2${\mathrm{T}}_2$ . Now, work done (W)=Q1-Q2$(\mathrm{W})={\mathrm{Q}}_1-{\mathrm{Q}}_2$∴ $\therefore$ Efficiency of the engine: (𝜂)= W

Q1=Q1-Q2

Q1$(𝜂)=\frac{\mathrm{W}}{{\mathrm{Q}}_1}=\frac{{\mathrm{Q}}_1-{\mathrm{Q}}_2}{{\mathrm{Q}}_1}$or, 𝜂=1-Q2

Q1$𝜂=1-\frac{{\mathrm{Q}}_2}{{\mathrm{Q}}_1}$. ... (i)If Q2=0:$Q_2=0:$ then, 𝜂=1$𝜂=1$ or 100%$100\%$ (for ideal engine) , which is not practically possible. This equation (i) implies that no actual heat engine has 100%$100\%$ efficiency.RefrigeratorA refrigerator is a heat engine operated in the reversed order. It absorbs heat Q2$Q_2$ from the low temperature reservoir (i.e,, sink) at T2$T_2$ when some external work W$W$ was done on it and heat Q1$Q_1$ is rejected to$t{o}_{}$ high temperature reservoir at T1$T_1$.Now coefficient of performance 𝛽$𝛽$ for refrigerator is defined as,𝛽=Q2

W=Q2

Q1-Q2=T2

T1-T2=1-𝜂

𝜂$$\begin{array}{c}𝛽=\frac{Q_2}{W}\\ =\frac{Q_2}{Q_1-Q_2}\\ =\frac{T_2}{T_1-T_2}\\ =\frac{1-𝜂}{𝜂}\end{array}$$Where, for Carnot' cycle of engine: Q2

Q1=T2

T1$\frac{{\mathrm{Q}}_2}{{\mathrm{Q}}_1}=\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}$ and 𝜂$𝜂$ isthe efficiency of Carnot's cycle. i.e., a

𝜂	=(1-Q2 Q1)
	=(1-T2 T1)

$$\text{ i.e., }\begin{array}{rl}𝜂& =(1-\frac{Q_2}{Q_1})\\ & =(1-\frac{T_2}{T_1})\end{array}$$Thus, coefficient of performance of refrigerator depends upon temperature of source and sink only, but not nature's of working substance, Its value ranges from 2 to 6 for actual refrigerator.Carnot's EngineA heat engine is a device to convert heat energy into mechanical work. In actual practice, the efficiency of heat engine is very small and ranges between (5-25)%$(5-25)\%$. This shows that at the most 25%$25\%$ of the total heat energy supplied is converted into mechanical work. So, Sadi Carnot devised an ideal theoretical engine free from all the imperfectness of actual engines and never realized in actual practice is called Carnot's engine. It is an imaginary engine and is taken $$ as a standard against which the performance of actual engine is adjusted It consists of following parts:(i) A cylinder with perfectly non conducting walls, but perfectly conducting base containing perfect gas as the working substance and fitted with a perfectly insulating and frictionless piston upon which weights can be placed.(ii) A hot body of infinitely large heat capacity to serve as source of heat at constant temperature T1K${\mathrm{T}}_1\mathrm{K}$.(iv) A cold body of infinitely large heat capacity to serve as sink at constant temperature T2K${\mathrm{T}}_2\mathrm{K}$.(v) A perfectly insulating plat-form serving as stand for the cylinder. Here, the cylinder may be placed on any of three bodies.

Carnot's Theorem The two postulates of Carnot Theorem are:1. No engine working between two given temperatures can be more efficient than a perfectly reversible engine working between same limits of temperatures (i.e., between the same source and sink$\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{k}$ ).2. All reversible engines working between the same limits of temperature have the same efficiency whatever be the working substance. Note:These two statement explains that reversible (i.e., Carnot) engine more efficient than irreversible engine which working between same limits of temperature. The efficiency of reversible engines are equal and which depends on limits of temperature but not nature and properties working substance.Proof (i):

Consider two engine one is reversible R$R$ and another is irreversible I working between the temperatures T1${\mathrm{T}}_1$ and T2(T1>|${\mathrm{T}}_2({\mathrm{T}}_1>$ T2${\mathrm{T}}_2$ ) as shown in figure above. Here, the quantities of working substance in the two engines is so adjusted that work done by I and R$R$ per cycle is same. Here, irreversible engine absorbs Q1$Q_1$ amount of heat from source at temperature T1$T_1$ and performs the output work W$W$ and reject (Q1-W)$({\mathrm{Q}}_1-\mathrm{W})$ amount of heat to sink at temperature T2${\mathrm{T}}_2$. Then, the efficiency of irreversible engine is:(i)𝜂I=W

Q1$${𝜂}_I=\frac{\mathrm{W}}{{\mathrm{Q}}_1}$$Similarly engine R$R$ absorbs Q′1$Q_1^{\prime }$ Amount of heat from source at temperature T1${\mathrm{T}}_1$ and performs the output work W$\mathrm{W}$ and then rejects (Q'1-W)$(Q_1^{\text{'}}-W)$ amount of heat to sink. Then the efficiency of reversible engine is,(ii)𝜂R=W

Q′1 ⋯$${𝜂}_R=\frac{W}{Q_1^{\prime }}\cdots$$ Let us assume irreversible engine is more efficient than the reversible engine. Then,𝜂1>𝜂R$${𝜂}_1>{𝜂}_R$$or, W

Q1>W

Q1'$\frac{W}{Q_1}>\frac{W}{Q_1\text{'}}$or, Q′1>Q1$Q_1^{\prime }>Q_1$ (iii)i.e., Q′1-Q1:${\mathrm{Q}}_1^{\prime }-{\mathrm{Q}}_1:$ is a +ve quantity. Now, we couple these two engine I$I$ and R$R$ by a belt in such a way that the engine I$I$ is working as a heat engine (i.e., forward) and R$R$ is working as a refrigerator (i.e. backward) as shown in fig. b. The source now lose Q1$Q_1$ heat to I$I$ and gains Q'1$Q{\text{'}}_1$ from R$R$. ∴$\therefore$ Heat gained by the source =Q′1-Q1 …$=Q_1^{\prime }-Q_1\dots$ (iv) Then, the sink gains (Q1-W)$(Q_1-W)$ heat from I$I$ and loses (Q′1-W)$(Q_1^{\prime }-W)$ to R$R$.∴ $\therefore$ Heat lost by sink =(Q′1-W)-(Q1-W)$sink=(Q_1^{\prime }-W)-(Q_1-W)$ =Q′1-Q1=$=Q_1^{\prime }-Q_1=$ +ve (∵$(\because$ from equation (iii))So, external work done on the system = 0Thus the coupled engine perform a self acting machine without aid of work by external agency transfer heat continuously from a body at low temperature to a body at a higher temperature. This conclusion contradicts the second law of thermodynamics and hence our assumption is incorrect. So, finally we conclude that no engine can be more efficient than a perfectly reversible engine working between the same two temperatures. Proof (ii):To prove the second part of the theorem we replace the irreversible engine by another reversible engine R′${\mathrm{R}}^{\prime }$ proceeding as in the first part of the theorem and on assuming the engine R′${\mathrm{R}}^{\prime }$ is more efficient than the engine R$\mathrm{R}$, we will reach the stage that will contradicts the second law of thermodynamics. Therefore our assumptions 𝜂'R>𝜂R$𝜂{\text{'}}_{\mathrm{R}}>{𝜂}_{\mathrm{R}}$ must be wrong. Again considering the engine R$\mathrm{R}$ is more efficient than the engine R′${\mathrm{R}}^{\prime }$, we would obtain the same situation that will also violate the second of law of thermodynamics. That means our assumptions 𝜂R>𝜂R'${𝜂}_{\mathrm{R}}>{𝜂}_{\mathrm{R}}\text{'}$ must be wrong.Hence, we can conclude that 𝜂R${𝜂}_{\mathrm{R}}$ must be equal to the 𝜂'R$𝜂{\text{'}}_{\mathrm{R}}$ ' i.e., 𝜂R=𝜂'R${𝜂}_{\mathrm{R}}=𝜂{\text{'}}_{\mathrm{R}}$. Thus, this shows that all reversible engine working between same temperature limits are equally efficient.Absolute scale of Temperature According to the Carnot theorem, the efficiency of a reversible engine is independent of the working substance and depends only on the two temperature between which it is working. Lord Kelvin had made this reference and defined the temperature scale which dose not depend upon the properties of any particular substance. This is the kelvin's absolute thermodynamic scale of temperature. Let us consider for reversible cycle in which Q1$Q_1$ amount of heat is absorbed at temperature Q1$Q_1$ and Q2$Q_2$ amount of heat is released at temperature Q2$Q_2$. Now, efficiency of the engine 𝜂=1-Q2

Q1=f(𝜃l,𝜃2)|$𝜂=1-\frac{{\mathrm{Q}}_2}{{\mathrm{Q}}_1}=\mathrm{f}({𝜃}_{\mathrm{l}},{𝜃}_2)$Q2

Q1=1-f(𝜃1,𝜃2)$$\frac{Q_2}{Q_1}=1-f({𝜃}_1,{𝜃}_2)$$Q1

Q2=1

1-f(𝜃1,𝜃2) = F(𝜃1,𝜃2)……… (i) $$\frac{Q_1}{Q_2}=\frac{1}{1-f({𝜃}_1,{𝜃}_2)}=F({𝜃}_1,{𝜃}_2)\dots \dots \dots \text{ (i) }$$where F(𝜃1,𝜃2)$F({𝜃}_1,{𝜃}_2)$ is another function of temperatureAgain, if Q2${\mathrm{Q}}_2$ be the amount of heat absorbed at temperature (𝜃2&Q2)$({𝜃}_2\&{\mathrm{Q}}_2)$ be the amount of heat absorbed at temperature 𝜃3${𝜃}_3$.Q2

Q3=F(𝜃2,𝜃3)………| (ii) $$\frac{Q_2}{Q_3}=F({𝜃}_2,{𝜃}_3)\dots \dots \dots \text{ (ii) }$$If Q1$Q_1$ be the amount of heat absorbed at temperature (𝜃1|$({𝜃}_1$ and |Q3)$Q_3)$ be the amount of heat absorbed at temperature 𝜃3${𝜃}_3$. a

Q1 Q3=F(𝜃1,(𝜃3)………|| (iii)

Multiplying (i) and (ii) we get

Q1 Q2×Q2 Q3=F(𝜃1,(𝜃2)×F(𝜃2,(𝜃3)||||

Q1 Q3=F(𝜃1,(𝜃2)×F(𝜃2,(𝜃3)||||

$$\begin{array}{c}\\ \begin{array}{l}\frac{Q_1}{Q_3}=F({𝜃}_1,({𝜃}_3)\dots \dots \dots \text{ (iii) }\\ Multiplying(i)and(ii)weget\\ \frac{Q_1}{Q_2}\times \frac{Q_2}{Q_3}=F({𝜃}_1,({𝜃}_2)\times F({𝜃}_2,({𝜃}_3)\\ \frac{Q_1}{Q_3}=F({𝜃}_1,({𝜃}_2)\times F({𝜃}_2,({𝜃}_3)\end{array}\end{array}$$∴F(Q1,Q3)=F(𝜃1,(𝜃2)×F(𝜃2,𝜃3)………||||$\therefore F(Q_1,Q_3)=F({𝜃}_1,({𝜃}_2)\times F({𝜃}_2,{𝜃}_3)\dots \dots \dots$ (iv)This equation (iv) is the functional equation. It does not contain 𝜃2${𝜃}_2$ on left side. So to remove 𝜃2${𝜃}_2$ from right; let us consider another function as:F(𝜃l,𝜃2)=𝜙(𝜃1)

𝜙(𝜃2) and F(𝜃2,𝜃3)=𝜙(𝜃2)

𝜙(𝜃3)$$\begin{array}{c}\mathrm{F}({𝜃}_{\mathrm{l}},{𝜃}_2)=\frac{𝜙({𝜃}_1)}{𝜙({𝜃}_2)}\text{ and }\\ \mathrm{F}({𝜃}_2,{𝜃}_3)=\frac{𝜙({𝜃}_2)}{𝜙({𝜃}_3)}\end{array}$$Where, 𝜙$𝜙$ is another unknown function of temperature.∴ F(𝜃1,𝜃3)=𝜙(𝜃1)

𝜙(𝜃2)×𝜙(𝜃2)

𝜙(𝜃3)=𝜙(𝜃1)

𝜙(𝜃3)$$\therefore \mathrm{F}({𝜃}_1,{𝜃}_3)=\frac{𝜙({𝜃}_1)}{𝜙({𝜃}_2)}\times \frac{𝜙({𝜃}_2)}{𝜙({𝜃}_3)}=\frac{𝜙({𝜃}_1)}{𝜙({𝜃}_3)}$$From equation (i),Q1

Q2=F(𝜃1,𝜃2)=𝜙(𝜃1)

𝜙(𝜃2)$$\frac{{\mathrm{Q}}_1}{{\mathrm{Q}}_2}=\mathrm{F}({𝜃}_1,{𝜃}_2)=\frac{𝜙({𝜃}_1)}{𝜙({𝜃}_2)}$$Since, 𝜃1>𝜃2${𝜃}_1>{𝜃}_2$ and Q1>Q2$Q_1>Q_2$; so the function 𝜙(𝜃1)>𝜙(𝜃2)$𝜙({𝜃}_1)>𝜙({𝜃}_2)$ Thus, 𝜙(𝜃)$𝜙(𝜃)$ is a linear function of 𝜃$𝜃$ and can be used to measure temperature. Let us suppose: 𝜙(𝜃)=𝜏$𝜙(𝜃)=𝜏$ on new scale, thenQ1

Q2=𝜏1

𝜏2 ..........................(v)|$$\frac{Q_1}{Q_2}=\frac{{𝜏}_1}{{𝜏}_2}..........................(v)$$This equation (v) defines the Kelvin's absolute thermodynamic scale of temperature. Thus the ratio any two temperatures on this scale equal to the ratio of heat absorbed and rejected by a Carnot reversible engine work in between these two temperatures. Now, from equation (v), we can writeQ1-Q2

Q1=𝜏1-𝜏2

𝜏1$$\frac{Q_1-Q_2}{Q_1}=\frac{{𝜏}_1-{𝜏}_2}{{𝜏}_1}$$or, W

Q1=𝜏1-𝜏2

𝜏1$\frac{W}{Q_1}=\frac{{𝜏}_1-{𝜏}_2}{{𝜏}_1}$Thus temperature on absolute seale can be measured in terms of work. So this scale is also called work scale of temperature.Q. Show that negative temperature is not possible in absolute scale. There cannot be any temperature less than zero or -ve temperature is not possible on absolute scale. The -ve temperature means that the heat engine absorbs heat from the sink and gives back to source which is against the second law of thermodynamic. To prove it:Let 𝜏1${𝜏}_1$ be the temperature of source on absolute scale and 𝜏2${𝜏}_2$ be the temperature of sink$\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{k}$ on absolute scale.Let us assume, 𝜏2=-K${𝜏}_2=-\mathrm{K}$, then efficiency of Carnot engine is𝜂=1-Q2

Q1$$𝜂=1-\frac{Q_2}{Q_1}$$Since, Q1

Q2=𝜏1

𝜏2$\frac{{\mathrm{Q}}_1}{{\mathrm{Q}}_2}=\frac{{𝜏}_1}{{𝜏}_2}$ (for absolute scale).∴$\therefore$𝜂=1-𝜏2

𝜏1=1+K

𝜏1$$𝜂=1-\frac{{𝜏}_2}{{𝜏}_1}=1+\frac{K}{{𝜏}_1}$$i.e.,𝜂>1$$𝜂>1$$i.e., the efficiency of engine is greater than unity which is against the second law of thermodynamics and hence impossible. Hence, the -ve temperature is not possible in absolute scale. Q. Show that absolute scale is identical to the percect gas scale. The thermodynamical scale defined by utilizing the property of working of Carnot's reversible engine (i.e., as a function of temperature) is called absolute scale of temperature. It is independent of nature and properties of working substance but depends upon temperature between which working substance works. Lord Kelvin worked out the theory of such an absolutes scale, so it is also called the Kelvin's work on thermodynamical scale. Let us consider a reversible Carnot engine using perfect gas as a working substance (as perfect gas). Let Q1$Q_1$ is the heat absorbed at temperature T1$T_1$ from source and Q2$Q_2$ is the heat rejected at temperature T2${\mathrm{T}}_2$ to the sink$\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{k}$ by working substance. Then, the efficiency of Carnot's engine is𝜂=1-Q2

Q1=1-T2

T1 ....................(i)$$𝜂=1-\frac{Q_2}{Q_1}=1-\frac{T_2}{T_1}....................(i)$$According to theory of absolute scale, Q1

Q2=𝜏1

𝜏2 .........................(ii)$$\frac{Q_1}{Q_2}=\frac{{𝜏}_1}{{𝜏}_2}.........................(ii)$$ Now, the efficiency of engine on absolute scale is𝜂=1-Q2

Q1=1-𝜏2

𝜏1 .............(iii)$$𝜂=1-\frac{Q_2}{Q_1}=1-\frac{{𝜏}_2}{{𝜏}_1}.............(iii)$$ From equation (i) and (iii); we can writeT1

T2=𝜏1

𝜏2.............................(iv)$$\frac{T_1}{T_2}=\frac{{𝜏}_1}{{𝜏}_2}.............................(iv)$$So, the ratio of any two temperature on the thermodynamical scale is same as on the perfect gas scale.If T2=0${\mathrm{T}}_2=0$ in equation (iv), we get 𝜏2=0${𝜏}_2=0$; so this indicates that zero of absolute scale is identical with zero of perfect gas scale. Nów, let's take the temperature limits steam point and ice point.Then, on perfect gas scale:a

𝜂p=1-Ticc Tstam

𝜂p=Tsteam -Ticc Tsteam

$$\begin{array}{r}{𝜂}_p=1-\frac{T_{icc}}{T_{\text{stam }}}\\ {𝜂}_p=\frac{T_{\text{steam }}-T_{\text{icc }}}{T_{\text{steam }}}\end{array}$$ ∴ 𝜂p=100

Tsteam ..............$\therefore {𝜂}_p=\frac{100}{T_{\text{steam }}}..............$ (v) [∵Tsteam -Tice =100]$[\because T_{\text{steam }}-T_{\text{ice }}=100]$Again in absolute scale, we have:𝜂a=1-𝜏icc

𝜏steam $${𝜂}_a=1-\frac{{𝜏}_{icc}}{{𝜏}_{\text{steam }}}$$ or, 𝜂u=𝜏steam -𝜏ice

𝜏steam ${𝜂}_u=\frac{{𝜏}_{\text{steam }}-{𝜏}_{\text{ice }}}{{𝜏}_{\text{steam }}}$ ∴ 𝜂a=100

𝜏steam …$\therefore {𝜂}_a=\frac{100}{{𝜏}_{\text{steam }}}\dots$ ( |vi) [𝜏steam -𝜏ice =100]$vi)[{𝜏}_{\text{steam }}-{𝜏}_{\text{ice }}=100]$Since, the efficiency must be same in either case. So, from equations (v) and (vi)$(vi)$; we get: 𝜂p=𝜂a⇒100

Tsteam =100

𝜏steam ${𝜂}_p={𝜂}_a\Rightarrow \frac{100}{T_{\text{steam }}}=\frac{100}{{𝜏}_{\text{steam }}}$ ∴ Tsteam =𝜏steam $\therefore {\mathrm{T}}_{\text{steam }}={𝜏}_{\text{steam }}$Similarly,Tice=𝜏ice$${\mathrm{T}}_{\mathrm{i}\mathrm{c}\mathrm{e}}={𝜏}_{\mathrm{i}\mathrm{c}\mathrm{e}}$$In general, we have: T=𝜏$\mathrm{T}=𝜏$Therefore, Kelvin thermodynamical scale is identical with perfect gas scale. Q. How is Kelvin scale realized in practice?The Kelvin's absolute scale cannot be directly realized in practice. It is, however, exactly identical to an ideal gas scale. Hence the temperature measured by a constant volume (or constant pressure) gas thermometer filled with an ideal gas would exactly be the same as the temperature measured on an absolute scale. But, in practice no gas is ideal, hence we can not have an ideal gas thermometer. However the constant volume hydrogen thermometer. Any reading taken on the hydrogen thermomenter can be corrected to obtain on the ideal the corresponding reading on the ideal gas thermometer. Thus, kelvin's absolute scale which coincides with the ideal gas scale, is realized in practice. Entropy Generally, entropy means the measure of degree of disorder of molecules of the system. Literal meaning of entropy is "transformation". The concept of entropy was first introduced by Clausius in 1854 while working on the formulation and application of the second law of thermodynamics. Now, the entropy of a system may be defined as the function of thermodynamical variables (P,V,T,U)$(\mathrm{P},\mathrm{V},\mathrm{T},\mathrm{U})$ and its change between two states is the integral of dQ

T$\frac{\mathrm{d}\mathrm{Q}}{\mathrm{T}}$ between the states along any reversible path joining them. Mathematically, dS=dQ

TdQ= T.dS$$\begin{array}{c}\text{ Mathematically, }\mathrm{d}\mathrm{S}=\frac{\mathrm{d}\mathrm{Q}}{\mathrm{T}}\\ dQ=T.dS\end{array}$$Where, dQ=$\mathrm{d}\mathrm{Q}=$ The quantity of heat absorbed or rejected at temperature T$\mathrm{T}$. dS=$\mathrm{d}\mathrm{S}=$ Change in entropy. Similarly, the change in entropy in passing from one adiabatic to another isS2∫S1dS=S2-SI=A2∫A1dQ

T$$\underset{S_1}{\overset{S_2}{\int }}\mathrm{d}\mathrm{S}={\mathrm{S}}_2-{\mathrm{S}}_{\mathrm{I}}=\underset{{\mathrm{A}}_1}{\overset{{\mathrm{A}}_2}{\int }}\frac{\mathrm{d}\mathrm{Q}}{\mathrm{T}}$$For adiabatic process, dQ=0$\mathrm{d}\mathrm{Q}=0$∴ dS=dQ

T=0

T=0$\therefore \mathrm{d}\mathrm{S}=\frac{\mathrm{d}\mathrm{Q}}{\mathrm{T}}\mathrm{=}\frac{0}{\mathrm{T}}=0$or, dS=0$\mathrm{d}\mathrm{S}=0$On integrating, we get: S=$\mathrm{S}=$ Constant. Thus, entropy of a substance or system is that physical quantity which remains constant when the substance undergoes reversible adiabatic process. Unit: It is measured in or Joule/K. Physical Concept of Entropy Generally it is difficult to interpreted physically, the concept of entropy as there is nothing physical to represent it and it cannot felt like temperature or pressure. Simply the entropy means the measure of the degree of disorder of the molecules of system and change is entropy is Change in entropy = Heat added or subtracted

absolute temperature$\frac{Heataddedorsubtracted}{absolutetemperature}$i.e., dS=dQ

T$\mathrm{d}\mathrm{S}=\frac{\mathrm{d}\mathrm{Q}}{\mathrm{T}}$ From above expression, we may say that heat energy has the same dimensions as the product of entropy and absolute temperature. PE =g. mh dQ ds . T Since, the gravitational potential energy of a body is proportional to the product of its mass and height above some zero level. Hence, if we may take temperature measured from absolute zero equivalent to height, then we may take entropy as analogous to mass or inertia. In this way, we may think of entropy as thermal (mass) inertia which bears to heat motion. Expression of Entropy of System Let us consider I1,I2,I3…$I_1,{\mathrm{I}}_2,I_3\dots$ are isothermals at temperature, T1,T2,T3${\mathrm{T}}_1,{\mathrm{T}}_2,{\mathrm{T}}_3$. etc, as shown in diagram. Here, A1${\mathrm{A}}_1$ and A2${\mathrm{A}}_2$ are adiabatics. Consider, a Carnot reversible cycle ABCDLet Q1${\displaystyle Q_1}$ is Heat absorbed at remperature T1$T_1$, from A$A$ to B$B$. Q2${\displaystyle Q_2}$ is Heat rejected at remperature T2$T_2$, from C$C$ to D$D$.

Now, from definition of efficiency of Carnot's engine,𝜂=1-Q2

Q1=1-T2

T1$$𝜂=1-\frac{Q_2}{Q_1}=1-\frac{T_2}{T_1}$$Q1

T1=Q2

T2 .............................. (i)$$\frac{Q_1}{T_{{}_1}}=\frac{Q_2}{T_2}..............................(i)$$Again, considering reversible cyle DCEF. Let Q2${\mathrm{Q}}_2$ is Heat absorbed at temperature T2${\mathrm{T}}_2$ from D$\mathrm{D}$ to C$\mathrm{C}$ and Q3$Q_3$ is Heat rejected at temperature T3${\mathrm{T}}_3$ from E$\mathrm{E}$ to F$\mathrm{F}$. Then, we get, Q2

T2=Q3

T3 …$\frac{Q_2}{T_2}=\frac{Q_3}{T_3}\dots$ (ii)From (i) and (ii), we can write Q1

T1=Q2

T2=Q3

T3=…=$\frac{Q_1}{T_1}=\frac{Q_2}{T_2}=\frac{Q_3}{T_3}=\dots =$ Constant.In general, if Q$Q$ be the amount of heat absorbed or rejected at lemperature T$\mathrm{T}$ in going from one adiabatic to other, then Q

T$\frac{Q_{}}{T_{}}$= constant This constant ratio is called change in entropy represented by two adiabatics. If two adiabatics lie very elose to each other and dQ be the small amount of heat absorbed or rejected at temperature T$\mathrm{T}$, then change in entropy (dS)$(\mathrm{d}\mathrm{S})$ is given by, dS = dQ

T$\frac{dQ}{T}$ ...............(iii)Suppose S1${\mathrm{S}}_1$ and S2${\mathrm{S}}_2$ be entropies represented by two adiabatics,S2∫S1dS=S2-S1=A2∫A1dQ

T.......................(iv)$$\underset{S_1}{\overset{S_2}{\int }}\mathrm{d}\mathrm{S}={\mathrm{S}}_2-{\mathrm{S}}_1=\underset{{\mathrm{A}}_1}{\overset{{\mathrm{A}}_2}{\int }}\frac{\mathrm{d}\mathrm{Q}}{\mathrm{T}}.......................(iv)$$ This equation (iii) and (iv) are the required expression of change in entropy of substance. Entropy in a reversible process Let us consider a reversible carnot cycle ABCDA$ABCDA$ on moving from A to B,Q1$\mathrm{B},{\mathrm{Q}}_1$ amount of heat is absorbed at temperature T1${\mathrm{T}}_1$

The increase in entropy of working substance is Q1

T1$\frac{Q_1}{T_1}$ (the entropy of the source is decrease by Q1

T1$\frac{Q_1}{T_1}$ ) On moving from C$\mathrm{C}$ to D$\mathrm{D}$ (i.e. isothermal compression) Q2${\mathrm{Q}}_2$ amount of heat is released at temperature T2${\mathrm{T}}_2$. So increase in entropy in sink is Q2

T2$\frac{{\mathrm{Q}}_2}{{\mathrm{T}}_2}$ (entropy of working substance decreases by |Q2

T2)$\frac{{\mathrm{Q}}_2}{{\mathrm{T}}_2})$On moving from D$\mathrm{D}$ to A$\mathrm{A}$ the entropy remains constant.∴$\therefore$ Change in entropy =Q1

T1-Q2

T2…………$=\frac{Q_1}{T_1}-\frac{Q_2}{T_2}\dots \dots \dots \dots$ (i)Also, from carnot reversible process Q1

T1=Q2

T2$\frac{Q_1}{T_1}=\frac{Q_2}{T_2}$Change in entropy in (i) is, Q1

T1-Q2

T2= 0$\frac{Q_1}{T_1}-\frac{Q_2}{T_2}=0$ dS = 0Hence, we can conclude that the change in entropy of any reversible process is zero. i.e. there is no change in the entropy of the reversible system.Entropy in irreversible process Let us consider an irreversible process and Q1$Q_1$ be the amount of the heat absorbed at temperature T1${\mathrm{T}}_1$ and Q2${\mathrm{Q}}_2$ be the amount of the at released at temperature T2${\mathrm{T}}_2$. Then, efficiency of irreversible process is given by, 𝜂i=1-Q2

Q1${𝜂}_i=1-\frac{Q_2}{Q_1}$But for reversible process efficiency is,𝜂r=1-T2

T1$${𝜂}_r=1-\frac{T_2}{T_1}$$Also, we know, 𝜂i<𝜂r${𝜂}_i<{𝜂}_r$a

(1-Q2 Q1)<(1-T2 T1)

Q2 Q1>T2 T1

So, Q2 T2-Q1 T1>0

$$\begin{array}{l}(1-\frac{Q_2}{Q_1})<(1-\frac{T_2}{T_1})\\ \frac{Q_2}{Q_1}>\frac{T_2}{T_1}\\ \text{ So, }\frac{Q_2}{T_2}-\frac{Q_1}{T_1}>0\end{array}$$ dS > 0Change in entropy of an irreversible process is always positive. T - S diagramQ: Define T-S diagram. Use it to derive efficiency of a carnot cycle. Write its significance.Definition: If thermodynamic state of a substance (or system) can be determined by plotting the temperature along Y$\mathrm{Y}$ -axis and entropy along X-axis in graph, then the graphical diagram is called T-S$\mathrm{T}-\mathrm{S}$ diagram. It is used in checking of efficient working of actual engines,Consider a Carnot's cycle ABDA$\mathrm{A}\mathrm{B}\mathrm{D}\mathrm{A}$ as shown in T-S$\mathrm{T}-\mathrm{S}$ diagram. In T-S$\mathrm{T}-\mathrm{S}$ diagram:

AB$\mathrm{A}\mathrm{B}$ : an isothermal expansion at temperature T1${\mathrm{T}}_1$ BC$\mathrm{B}\mathrm{C}$ : adiabatic expansion at constant entropy S2${\mathrm{S}}_2$ but decrease in temperature from T1${\mathrm{T}}_1$ to T2${\mathrm{T}}_2$CD:$\mathrm{C}\mathrm{D}:$ isothermal compression at temperature T2${\mathrm{T}}_2$DA : adiabatic compression at constant entropy S1$S_1$ and temperature increase from T2${\mathrm{T}}_2$ to T1${\mathrm{T}}_1$. In isothermal expansion, the amount of heat absorbed by by area under the working substance (Q1)$({\mathrm{Q}}_1)$ from source curve AB$AB$.ie, Area ABEFA =T1(SB-SA)$={\mathrm{T}}_1({\mathrm{S}}_B-{\mathrm{S}}_A)$In isothermal compression, the heat is rejected to sink by working substance(Q2)$({\mathrm{Q}}_2)$ is given by area under the curve CD$\mathrm{C}\mathrm{D}$.i.e., Area DCEFD =$=$ Entropy(S) T2(SB-SA)${\mathrm{T}}_2({\mathrm{S}}_{\mathbf{B}}-{\mathrm{S}}_A)$ ∴$\therefore$ Workdone in a complete cycle =$=$ heat absorbed - heat rejected =$=$ Area ABEFA - Area DCEFD =$=$ Area ABCD$\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$∴$\therefore$ aW= T1(SB-SA) -T2(SB-SA) = (T1-T2)(SB-SA)$\begin{array}{l}\mathrm{W}={\mathrm{T}}_1({\mathrm{S}}_B-{\mathrm{S}}_A)-{\mathrm{T}}_2({\mathrm{S}}_{\mathbf{B}}-{\mathrm{S}}_A)\\ =({\mathrm{T}}_1-{\mathrm{T}}_2)({\mathrm{S}}_B-{\mathrm{S}}_A)\end{array}$Now, the efficiency of the engine isa

𝜂= Workout put (W) Heat input (Q1)=(T1-T2)(SB-SA) T1(SB-SA)

∴ 𝜂=1-T2 T1

$$\begin{array}{r}𝜂=\frac{\text{ Workout put }(\mathrm{W})}{\text{ Heat input }({\mathrm{Q}}_1)}=\frac{({\mathrm{T}}_1-{\mathrm{T}}_2)({\mathrm{S}}_B-{\mathrm{S}}_A)}{{\mathrm{T}}_1({\mathrm{S}}_B-{\mathrm{S}}_A)}\\ \therefore 𝜂=1-\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}\end{array}$$This expression is same as that obtained from PV diagram of Carnot cycle. Therefore, T-S$T-S$ diagram is a simple way to check the efficient working of actual engine. Significance of T-S$\mathrm{T}-\mathrm{S}$ diagram(i) The T-S$\mathrm{T}-\mathrm{S}$ diagram also known as tephigram is extensively used in meterology.(ii) The T-S$T-S$ diagram helps us to calculate the work done per cycle in easy way because the T-S$T-S$ diagram is a rectangle. This is why, it is used by auto-engineers to study the working efficiency of the engines.(iii) T-S$\mathrm{T}-\mathrm{S}$ diagram per cycle can be used to check the efficient working of actual engine, etc.Define 2 nd law of thermodynamies in terms of entropy? According to Clausius, Ist law of thermodynamic implies that "the energy of universe remains constant (i.e., law of conservation of energy)" and the 2 nd law implies that "the law of increase of entropy (i.e., entropy of universe tends to maximum)". Thus, 2 nd law in terms of entropy states as, "every physical or chemical process in nature takes place in such a way that to increase the entropy of the system. Let SA:${\mathrm{S}}_{\mathrm{A}}:$ Entropy of substance for state A$\mathrm{A}$ and SB:$S_B:$ Entropy of substance for state B$B$. Then, change in entropy of substance is:SB-SA=B∫AdQ

T .........................(i)$${\mathrm{S}}_{\mathrm{B}}-{\mathrm{S}}_{\mathrm{A}}=\underset{\mathrm{A}}{\overset{\mathrm{B}}{\int }}\frac{\mathrm{d}\mathrm{Q}}{\mathrm{T}}.........................(i)$$If two states are very close to each other, then equation (i) canbe written as,dS=dQ

T$$\mathrm{d}\mathrm{S}=\frac{\mathrm{d}\mathrm{Q}}{\mathrm{T}}$$∴dQ=TdS$\therefore \mathrm{d}\mathrm{Q}=\mathrm{T}\mathrm{d}\mathrm{S}$ Which is the mathematical form of 2 nd lawthermodynamic.Third Law of ThermodynamicsSince, entropy of a substance is a measure of degree of disorder of molecules of substance. At absolute zero, thermal motion of molecule completely disappears. So that disorder and hence entropy tends to zero and the molecules are in perfect order or well arranged. This is called 3rd law of thermodynamics. Third Law of EntropyAccording to Lewis and Randall, "Every system has a finite positive entropy, but at the absolute zero of temperature the entropy may become zero and does so become in the case of a pure crystalline substances."