applications of third law of thermodynamics
The only way to use energy is to transform energy from one form to another. For example, compare the \(S^o\) values for CH3OH(l) and CH3CH2OH(l). If the system is composed of one-billion atoms that are all alike and lie within the matrix of a perfect crystal, the number of combinations of one billion identical things taken one billion at a time is = 1. Similarly, another example of the zeroth law of thermodynamics is when you have two glasses of water. A non-quantitative description of his third law that Nernst gave at the very beginning was simply that the specific heat of a material can always be made zero by cooling it down far enough. This law states that the change in internal energy for a system is equal to the difference between the heat added to the system and the work done by the system: Where U is energy, Q is heat and W is work, all typically measured in joules, Btus or calories). The third law of thermodynamics states that The entropy of a perfect crystal at absolute zero temperature is exactly equal to zero. The third law of thermodynamics establishes the zero for entropy as that of a perfect, pure crystalline solid at 0 K. With only one possible microstate, the entropy is zero. Heat was not formally recognized as a form of energy until about 1798, when Count . Thermodynamics can be defined as the study of energy, energy transformations and its relation to matter. )%2FUnit_4%253A_Equilibrium_in_Chemical_Reactions%2F13%253A_Spontaneous_Processes_and_Thermodynamic_Equilibrium%2F13.6%253A_The_Third_Law_of_Thermodynamics, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\), \[m\ce{A}+n\ce{B}x\ce{C}+y\ce{D} \label{\(\PageIndex{7}\)}\], The Third Law Lets us Calculate Absolute Entropies, http://cnx.org/contents/85abf193-2bda7ac8df6@9.110, status page at https://status.libretexts.org, Calculate entropy changes for phase transitions and chemical reactions under standard conditions. We can use the products minus reactants rule to calculate the standard entropy change (S) for a reaction using tabulated values of S for the reactants and the products. \[\begin{align*} S^o_{298} &=S^o_{298}(\ce{H2O (l)})S^o_{298}(\ce{H2O(g)})\nonumber \\[4pt] &= (70.0\: J\:mol^{1}K^{1})(188.8\: Jmol^{1}K^{1})\nonumber \\[4pt] &=118.8\:J\:mol^{1}K^{1} \end{align*}\]. So the thermal expansion coefficient of all materials must go to zero at zero kelvin. 0 Second law of thermodynamics: The state of the entropy of the entire universe, as an isolated system, will always increase over time. 3rd Law of Thermodynamics. It simply states that during an interaction, energy can change from one form to another but the total amount of energy remains constant. The third law of thermodynamics states that the entropy of a perfect crystal at a temperature of zero Kelvin (absolute zero) is equal to zero. Kids Encyclopedia Facts. The same is not true of the entropy; since entropy is a measure of the dilution of thermal energy, it follows that the less thermal energy available to spread through a system (that is, the lower the temperature), the smaller will be its entropy. Entropy increases with softer, less rigid solids, solids that contain larger atoms, and solids with complex molecular structures. Note that this is different from a freezing point, like zero degrees Celsius molecules of ice still have small internal motions associated with them, also known as heat. In other words, in any isolated system (including the universe), entropy change is always zero or positive. Use the data in Table \(\PageIndex{1}\) to calculate S for the reaction of H2(g) with liquid benzene (C6H6) to give cyclohexane (C6H12). Whether you need help solving quadratic equations, inspiration for the upcoming science fair or the latest update on a major storm, Sciencing is here to help. A great deal of attention is paid in this text to training the student in the application of the basic concepts to problems that are commonly encountered by the chemist, the biologist, the geologist, and the materials scientist. This scale is built on a particular physical basis: Absolute zero Kelvin is the temperature at which all molecular motion ceases. Legal. This constant value cannot depend on any other parameters characterizing the closed system, such as pressure or applied magnetic field. For example, \(S^o\) for the following reaction at room temperature, \[S^o=[xS^o_{298}(\ce{C})+yS^o_{298}(\ce{D})][mS^o_{298}(\ce{A})+nS^o_{298}(\ce{B})] \label{\(\PageIndex{8}\)}\], Table \(\PageIndex{1}\) lists some standard entropies at 298.15 K. You can find additional standard entropies in Tables T1 and T2. Entropy is related to the number of accessible microstates, and there is typically one unique state (called the ground state) with minimum energy. Thermodynamic cycles govern the operation of all forms of air and gas compressors, blowers, and fans. The third law was developed by chemist Walther Nernst during the years 1906 to 1912 and is therefore often referred to as Nernst's theorem or Nernst's postulate. Most heat engines fall into the category of open systems. The difference in this third law of thermodynamics is that it leads to well-defined values of entropy itself as values on the Kelvin scale. Conclusion. 15.4: Entropy and Temperature. Chemistry LibreTexts: The Third Law of Thermodynamics, Purdue University: Entropy and the 2nd and 3rd Laws of Thermodynamics. Third law of thermodynamics; . (1971). The entropy change is. This violates Eq.(8). Use the data in Table \(\PageIndex{1}\) to calculate \(S^o\) for the reaction of liquid isooctane with \(\ce{O2(g)}\) to give \(\ce{CO2(g)}\) and \(\ce{H2O(g)}\) at 298 K. Given: standard molar entropies, reactants, and products. The most common practical application of the First Law is the heat engine. \\ &=22.70\;\mathrm{J/(mol\cdot K)}\ln\left(\dfrac{388.4}{368.5}\right)+\left(\dfrac{1.722\;\mathrm{kJ/mol}}{\textrm{388.4 K}}\times1000\textrm{ J/kJ}\right) Because entropy can also be described as thermal energy, this means it would have some energy in the form of heat so, decidedly not absolute zero. At that point, the universe will have reached thermal equilibrium, with all energy in the form of thermal energy at the same nonzero temperature. This principle is the basis of the Third law of thermodynamics, which states that the entropy of a perfectly-ordered solid at 0 K is zero. The third law of thermodynamics states that the entropy of any perfectly ordered, crystalline substance at absolute zero is zero. In practice, absolute zero is an ideal temperature that is unobtainable, and a perfect single crystal is also an ideal that cannot be achieved. Standard entropies are given the label \(S^o_{298}\) for values determined for one mole of substance at a pressure of 1 bar and a temperature of 298 K. The standard entropy change (\(S^o\)) for any process may be computed from the standard entropies of its reactant and product species like the following: \[S^o=\sum S^o_{298}(\ce{products})\sum S^o_{298}(\ce{reactants}) \label{\(\PageIndex{6}\)}\], Here, \(\) represents stoichiometric coefficients in the balanced equation representing the process. [citation needed], On the other hand, the molar specific heat at constant volume of a monatomic classical ideal gas, such as helium at room temperature, is given by CV = (3/2)R with R the molar ideal gas constant. The same argument shows that it cannot be bounded below by a positive constant, even if we drop the power-law assumption. Two kinds of experimental measurements are needed: \[ S_{0 \rightarrow T} = \int _{0}^{T} \dfrac{C_p}{T} dt \label{eq20}\]. We have, By the discussion of third law above, this integral must be bounded as T0 0, which is only possible if > 0. In the limit T0 0 this expression diverges, again contradicting the third law of thermodynamics. However, ferromagnetic materials do not, in fact, have zero entropy at zero temperature, because the spins of the unpaired electrons are all aligned and this gives a ground-state spin degeneracy. It can be applied to factories that use heat to power different mechanisms. Yes the third law of thermodynamics holds for any system classical or quantum mechanical. The third law of thermodynamics states, "the entropy of a perfect crystal is zero when the temperature of the crystal is equal to absolute zero (0 K)." According to Purdue University, "the crystal . Postby Brianna Cronyn Lec3E Sat Mar 05, 2022 1:20 am. . According to the Boltzmann equation, the entropy of this system is zero. The absolute entropy of a substance at any temperature above 0 K must be determined by calculating the increments of heat \(q\) required to bring the substance from 0 K to the temperature of interest, and then summing the ratios \(q/T\). There is a condition that when a thermometer . Scientists everywhere, however, use Kelvins as their fundamental unit of absolute temperature measurement. This was true in the last example, where the system was the entire universe. The third law of thermodynamics is essentially a statement about the ability to create an absolute temperature scale, for which absolute zero is the point at which the internal energy of a solid is precisely 0. But clearly a constant heat capacity does not satisfy Eq. As the sweat absorbs more and more heat, it evaporates from your body, becoming more disordered and transferring heat to the air, which heats up the air temperature of the room. \\ 1.09\;\mathrm{J/(mol\cdot K)}&=C_{\textrm p({\alpha})}\ln\left(\dfrac{T_2}{T_1}\right)+\dfrac{\Delta H_{\textrm{fus}}}{T_{\textrm m}}+\Delta S_3+C_{\textrm p(\beta)}\ln\left(\dfrac{T_4}{T_3}\right) This law also defines absolute zero temperature. The entropy of a pure, perfect crystalline substance at 0 K is zero. We may compute the standard entropy change for a process by using standard entropy values for the reactants and products involved in the process. 16.1: Nernst's Heat Theorem. Thermal Engineering Third Law of Thermodynamics - 3rd Law The entropy of a system approaches a constant value as the temperature approaches absolute zero. The second law of thermodynamics states that a spontaneous process increases the entropy of the universe, Suniv > 0. The third law of thermodynamics says that the entropy of a perfect crystal at absolute zero is exactly equal to zero. Materials that remain paramagnetic at 0 K, by contrast, may have many nearly degenerate ground states (for example, in a spin glass), or may retain dynamic disorder (a quantum spin liquid). The third law of thermodynamics states that the entropy of any perfectly ordered, crystalline substance at absolute zero is zero. < Legal. Heat engines convert thermal energy into mechanical energy and vice versa. In contrast, graphite, the softer, less rigid allotrope of carbon, has a higher S [5.7 J/(molK)] due to more disorder in the crystal. Subtract the sum of the absolute entropies of the reactants from the sum of the absolute entropies of the products, each multiplied by their appropriate stoichiometric coefficients, to obtain \(S^o\) for the reaction. The entropy of a closed system, determined relative to this zero point, is then the absolute entropy of that system. Likewise, S is 260.7 J/(molK) for gaseous I2 and 116.1 J/(molK) for solid I2. Register to view this lesson {\displaystyle S} The third law of thermodynamics states that the entropy of a system approaches a constant value as the temperature approaches absolute zero. The third law of thermodynamics states that the entropy of a system at absolute zero is a well-defined constant. Among crystalline materials, those with the lowest entropies tend to be rigid crystals composed of small atoms linked by strong, highly directional bonds, such as diamond (\(S^o = 2.4 \,J/(molK)\)). 2. It applies to a variety of science and engineering topics such as chemical, physical, and mechanical engineering. Among crystalline materials, those with the lowest entropies tend to be rigid crystals composed of small atoms linked by strong, highly directional bonds, such as diamond [S = 2.4 J/(molK)]. . Eventually, the change in entropy for the universe overall will equal zero. It's possible to find the constant b if you fit Debye's equation to some experimental measurements of heat capacities extremely close to absolute zero (T=0 K). Similarly, the absolute entropy of a substance tends to increase with increasing molecular complexity because the number of available microstates increases with molecular complexity. To calculate S for a chemical reaction from standard molar entropies, we use the familiar products minus reactants rule, in which the absolute entropy of each reactant and product is multiplied by its stoichiometric coefficient in the balanced chemical equation. At temperatures greater than absolute zero, entropy has a positive value, which allows us to measure the absolute entropy of a substance. If we consider a container partly filled with liquid and partly gas, the entropy of the liquidgas mixture is, where Sl(T) is the entropy of the liquid and x is the gas fraction. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. The third law of thermodynamics has two important consequences: it defines the sign of the entropy of any substance at temperatures above absolute zero as positive, and it provides a fixed reference point that allows us to measure the absolute entropy of any substance at any temperature. is the Boltzmann constant, and Because the heat capacity is itself slightly temperature dependent, the most precise determinations of absolute entropies require that the functional dependence of \(C\) on \(T\) be used in the integral in Equation \ref{eq20}, i.e.,: \[ S_{0 \rightarrow T} = \int _{0}^{T} \dfrac{C_p(T)}{T} dt. It can also be used in the context of man-made energy sources, such as damns. We assume N = 3 1022 and = 1cm. The third law of thermodynamics is lesser known of all the three laws of thermodynamics, and even its applications found in our day-to-day life are fewer, though they can be seen in physical and chemical science at low temperatures. itself will also reach zero as long as the crystal has a ground state with only one configuration. \\ &+\Delta S_3+24.77\;\mathrm{J/(mol\cdot K)}\ln\left(\dfrac{368.5}{388.4}\right) Just remember that b depends on the type of substance. This definition was first proposed by Ludwig Boltzmann in 1877. Importance of third law of thermodynamics is given below: 1) It helps in calculating the thermodynamic properties. The entropy of the universe cannot increase. Even within a purely classical setting, the density of a classical ideal gas at fixed particle number becomes arbitrarily high as T goes to zero, so the interparticle spacing goes to zero. Solving for S3 gives a value of 3.24 J/(molK). Third Law of Thermodynamics - As the temperature of a system approaches absolute zero, its entropy approaches a minimum value. It is also true for smaller closed systems continuing to chill a block of ice to colder and colder temperatures will slow down its internal molecular motions more and more until they reach the least disordered state that is physically possible, which can be described using a constant value of entropy. Measurements of the heat capacity of a substance and the enthalpies of fusion or vaporization can be used to calculate the changes in entropy that accompany a physical change. To become perfectly still, molecules must also be in their most stable, ordered crystalline arrangement, which is why absolute zero is also associated with perfect crystals. As shown in Figure \(\PageIndex{2}\) above, the entropy of a substance increases with temperature, and it does so for two reasons: We can make careful calorimetric measurements to determine the temperature dependence of a substances entropy and to derive absolute entropy values under specific conditions. The entropy of a system approaches a constant value when its temperature approaches absolute zero. Download for free at http://cnx.org/contents/85abf193-2bda7ac8df6@9.110). 2) It is helpful in measuring chemical affinity. As the temperature rises, more microstates become accessible, allowing thermal energy to be more widely dispersed. Write the balanced chemical equation for the reaction and identify the appropriate quantities in Table \(\PageIndex{1}\). The Zeroth law of thermodynamics states that if two bodies are there in equilibrium with the third body in that, then they need to have a thermal equilibrium with each other. The orthorhombic () form is more stable at room temperature but undergoes a phase transition to the monoclinic () form at temperatures greater than 95.3C (368.5 K). The second, based on the fact that entropy is a state function, uses a thermodynamic cycle similar to those discussed previously. The alignment of a perfect crystal leaves no ambiguity as to the location and orientation of each part of the crystal. The idea that the entropy change for a pure substance goes to zero as the temperature goes to zero finds expression as the third law of thermodynamics. Various Applications of Thermodynamics Thermodynamics has a vast number of applications as it covers the infinite universe. The transition from S to S can be described by the thermodynamic cycle shown in part (b) in Figure \(\PageIndex{3}\), in which liquid sulfur is an intermediate. First law of thermodynamics 3. What is the Law of conservation of energy in chemistry? Our goal is to make science relevant and fun for everyone. These determinations are based on the heat capacity measurements of the substance. {\displaystyle k_{\mathrm {B} }} The reason that T = 0 cannot be reached according to the third law is explained as follows: Suppose that the temperature of a substance can be reduced in an isentropic process by changing the parameter X from X2 to X1. Because of this it is known as Nernst theorem. Third law of thermodynamics 1. \[\begin{align*} S&=k\ln \Omega \\[4pt] &= k\ln(1) \\[4pt] &=0 \label{\(\PageIndex{5}\)} \end{align*}\]. Since heat is molecular motion in the simplest sense, no motion means no heat. In simple terms, the third law states that the entropy of a perfect crystal of a pure substance approaches zero as the temperature approaches zero. The third law of thermodynamics states, regarding the properties of closed systems in thermodynamic equilibrium: .mw-parser-output .templatequote{overflow:hidden;margin:1em 0;padding:0 40px}.mw-parser-output .templatequote .templatequotecite{line-height:1.5em;text-align:left;padding-left:1.6em;margin-top:0}. 13.6: The Third Law of Thermodynamics is shared under a CC BY license and was authored, remixed, and/or curated by LibreTexts. Fourth law of thermodynamics: the dissipative component of evolution is in a direction of steepest entropy ascent. As per the third law of thermodynamics, the entropy of such a system is exactly zero. The increase in entropy with increasing temperature in Figure \(\PageIndex{2}\) is approximately proportional to the heat capacity of the substance. [9] If there were an entropy difference at absolute zero, T = 0 could be reached in a finite number of steps. While sweating also, the law of thermodynamics is applicable. In contrast, graphite, the softer, less rigid allotrope of carbon, has a higher \(S^o\) (5.7 J/(molK)) due to more disorder (microstates) in the crystal. The third law of thermodynamics, also known as the Nernst law, can be defined as, on reaching the absolute zero temperature (0 K), any physical process stops; when any system reaches absolute zero temperature, the entropy reaches a minimum constant value. If Suniv < 0, the process is nonspontaneous, and if Suniv = 0, the system is at equilibrium. Only ferromagnetic, antiferromagnetic, and diamagnetic materials can satisfy this condition. In practice, chemists determine the absolute entropy of a substance by measuring the molar heat capacity (\(C_p\)) as a function of temperature and then plotting the quantity \(C_p/T\) versus \(T\). For instance, S for liquid water is 70.0 J/(molK), whereas S for water vapor is 188.8 J/(molK). This branch was basically developed out of a desire to improve the efficiency of steam engines. [1] In such a case, the entropy at absolute zero will be exactly zero. A classical formulation by Nernst (actually a consequence of the Third Law) is: It is impossible for any process, no matter how idealized, to reduce the entropy of a system to its absolute-zero value in a finite number of operations.[3]. the more likely that a quantum state can break and become useless in technical applications. Unlike enthalpy or internal energy, it is possible to obtain absolute entropy values by measuring the entropy change that occurs between the reference point of 0 K (corresponding to \(S = 0\)) and 298 K (Tables T1 and T2). \\[4pt] & \,\,\, -\left \{[1\textrm{ mol }\mathrm{C_8H_{18}}\times329.3\;\mathrm{J/(mol\cdot K)}]+\left [\dfrac{25}{2}\textrm{ mol }\mathrm{O_2}\times205.2\textrm{ J}/(\mathrm{mol\cdot K})\right ] \right \} The molecules of solids, liquids, and gases have increasingly greater freedom to move around, facilitating the spreading and sharing of thermal energy. \[Delta S=nC_{\textrm v}\ln\dfrac{T_2}{T_1}\hspace{4mm}(\textrm{constant volume}) \tag{18.21}\]. In 1923, Lewis and Randall 1 gave a statement of the third law that is particularly convenient in chemical applications: It is also true for smaller closed systems - continuing to chill a block of ice to colder and colder . If we know the melting point of S (Tm = 115.2C = 388.4 K) and St for the overall phase transition [calculated to be 1.09 J/(molK) in the exercise in Example 6], we can calculate S3 from the values given in part (b) in Figure \(\PageIndex{3}\) where Cp() = 22.70 J/molK and Cp() = 24.77 J/molK (subscripts on S refer to steps in the cycle): \(\begin{align}\Delta S_{\textrm t}&=\Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4 Thus we can use a combination of heat capacity measurements (Equation 18.20 or Equation 18.21) and experimentally measured values of enthalpies of fusion or vaporization if a phase change is involved (Equation 18.18) to calculate the entropy change corresponding to a change in the temperature of a sample. A closed system, on the other hand, can exchange only energy with its surroundings, not matter. So the heat capacity must go to zero at absolute zero, if it has the form of a power law. Subtract the sum of the absolute entropies of the reactants from the sum of the absolute entropies of the products, each multiplied by their appropriate stoichiometric coefficients, to obtain S for the reaction. Furthermore, because it defines absolute zero as a reference point, we are able to quantify the relative amount of energy of any substance at any temperature. Hence: The difference is zero; hence the initial entropy S0 can be any selected value so long as all other such calculations include that as the initial entropy. \[\begin{align*} S^o &=S^o_{298} \\[4pt] &= S^o_{298}(\ce{products})S^o_{298} (\ce{reactants}) \\[4pt] & = 2S^o_{298}(\ce{CO2}(g))+4S^o_{298}(\ce{H2O}(l))][2S^o_{298}(\ce{CH3OH}(l))+3S^o_{298}(\ce{O2}(g))]\nonumber \\[4pt] &= [(2 \times 213.8) + (470.0)][ (2 \times 126.8) + (3 \times 205.03) ]\nonumber \\[4pt] &= 161.6 \:J/molK\nonumber \end{align*} \]. Because entropy is a state function, however, S3 can be calculated from the overall entropy change (St) for the SS transition, which equals the sum of the S values for the steps in the thermodynamic cycle, using Equation 18.20 and tabulated thermodynamic parameters (the heat capacities of S and S, Hfus(), and the melting point of S.). \\ &-\left \{[1\textrm{ mol }\mathrm{C_8H_{18}}\times329.3\;\mathrm{J/(mol\cdot K)}]+\left [\dfrac{25}{2}\textrm{ mol }\mathrm{O_2}\times205.2\textrm{ J}/(\mathrm{mol\cdot K})\right ] \right \} Calculate the standard entropy change for the following reaction at 298 K: \[\ce{Ca(OH)2}(s)\ce{CaO}(s)+\ce{H2O}(l)\nonumber\]. Putting together the second and third laws of thermodynamics leads to the conclusion that eventually, as all energy in the universe changes into heat, it will reach a constant temperature. The third law of thermodynamics states that as the temperature approaches absolute zero in a system, the absolute entropy of the system approaches a constant value. Whether we are sitting in an air-conditioned room or travelling in any vehicle, the application of thermodynamics is everywhere. Which is Clapeyron and Clausius equation. Soft crystalline substances and those with larger atoms tend to have higher entropies because of increased molecular motion and disorder. This means that a system always has the same amount of energy, unless its added from the outside. This order makes qualitative sense based on the kinds and extents of motion available to atoms and molecules in the three phases. Ans: There are two major applications of the Third law of thermodynamics, which are mentioned below: 1. This system may be described by a single microstate, as its purity, perfect crystallinity and complete lack of motion (at least classically, quantum mechanics argues for constant motion) means there is but one possible location for each identical atom or molecule comprising the crystal (\(\Omega = 1\)). \(S^o\) is positive, as expected for a combustion reaction in which one large hydrocarbon molecule is converted to many molecules of gaseous products. The third law of thermodynamics states that the entropy of a system approaches a constant value as the temperature approaches zero. What this essentially means is that random processes tend to lead to more disorder than order. Values of \(C_p\) for temperatures near zero are not measured directly, but can be estimated from quantum theory. Some crystals form defects which cause a residual entropy. Example \(\PageIndex{1}\) illustrates this procedure for the combustion of the liquid hydrocarbon isooctane (\(\ce{C8H18}\); 2,2,4-trimethylpentane). Entropy is often described in words as a measure of the amount of disorder in a system. Indeed, they are power laws with =1 and =3/2 respectively. The third law of thermodynamics states that as the temperature approaches absolute zero (0 K, 273.15 C, or 459.67 F), the temperature of the system approaches a constant minimum (the entropy at 0 K is often taken to be zero). The NernstSimon statement of the third law of thermodynamics concerns thermodynamic processes at a fixed, low temperature: The entropy change associated with any condensed system undergoing a reversible isothermal process approaches zero as the temperature at which it is performed approaches 0 K. Here a condensed system refers to liquids and solids. Learn About Boyle's Law Here (14), which yields. 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