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Chapter 18: Problem 8

State the third law of thermodynamics and explain its usefulness in calculating entropy values.

Short Answer

Expert verified
The third law of thermodynamics states that as temperature approaches absolute zero, the entropy of a system approaches a minimum constant value. This law provides a 'zero point' for entropy, allowing us to calculate absolute entropy of any substance at any temperature, thus being very useful in calculating entropy values.

Step by step solution

01

Statement of the third law of thermodynamics

The third law of thermodynamics states that as the temperature in a closed system approaches absolute zero, the entropy of that system approaches a minimal, constant value. This law establishes that absolute zero is the point at which molecules cease to move and thus cease to create disorder.
02

Explanation of the third law of thermodynamics

This law is helpful for several reasons. Firstly, it establishes a consistent baseline, or 'zero point', for entropy measurement. This allows the measurement of entropy to be absolute, rather than relative to some arbitrary reference point. Secondly, it underpins the concept that energy is not conserved at very low temperatures and thus provides important insights into the fundamental nature of matter.
03

The usefulness of this law in calculating entropy values

The third law, in setting absolute zero as a point of reference, allows for the calculation of absolute entropy of any substance at any given temperature. The entropy of a substance at absolute zero is usually close to zero (for a perfect crystalline substance it is zero). Therefore, the entropy of a substance at any other temperature would be the difference between the entropy at that temperature and the entropy at absolute zero.

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Most popular questions from this chapter

The molar heat of vaporization of ethanol is \(39.3 \mathrm{~kJ} / \mathrm{mol}\) and the boiling point of ethanol is \(78.3^{\circ} \mathrm{C}\) Calculate \(\Delta S\) for the vaporization of \(0.50 \mathrm{~mol}\) ethanol.

Which of the following are not state functions: \(S, H\) \(q, w, T ?\)

The standard enthalpy of formation and the standard entropy of gaseous benzene are \(82.93 \mathrm{~kJ} / \mathrm{mol}\) and \(269.2 \mathrm{~J} / \mathrm{K} \cdot\) mol, respectively. Calculate \(\Delta H^{\circ}, \Delta S^{\circ}\) and \(\Delta G^{\circ}\) for the process at \(25^{\circ} \mathrm{C}\). Comment on your answers. $$ \mathrm{C}_{6} \mathrm{H}_{6}(l) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(g) $$

The internal combustion engine of a \(1200-\mathrm{kg}\) car is designed to run on octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right),\) whose enthalpy of combustion is \(5510 \mathrm{~kJ} / \mathrm{mol}\). If the car is moving up a slope, calculate the maximum height (in meters) to which the car can be driven on 1.0 gallon of the fuel. Assume that the engine cylinder temperature is \(2200^{\circ} \mathrm{C}\) and the exit temperature is \(760^{\circ} \mathrm{C},\) and neglect all forms of friction. The mass of 1 gallon of fuel is \(3.1 \mathrm{~kg} .\) [Hint: The efficiency of the internal combustion engine, defined as work performed by the engine divided by the energy input, is given by \(\left(T_{2}-T_{1}\right) / T_{2},\) where \(T_{2}\) and \(T_{1}\) are the engine's operating temperature and exit temperature (in kelvins). The work done in moving the car over a vertical distance is \(m g h,\) where \(m\) is the mass of the car in \(\mathrm{kg}, g\) the acceleration due to gravity \(\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right),\) and \(h\) the height in meters. \(]\)

The equilibrium constant \(\left(K_{P}\right)\) for the reaction $$ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) $$ is 4.40 at \(2000 \mathrm{~K}\). (a) Calculate \(\Delta G^{\circ}\) for the reaction. (b) Calculate \(\Delta G\) for the reaction when the partial pressures are \(P_{\mathrm{H}_{2}}=0.25 \mathrm{~atm}, P_{\mathrm{CO}_{2}}=0.78 \mathrm{~atm}\) \(P_{\mathrm{H}_{2} \mathrm{O}}=0.66 \mathrm{~atm},\) and \(P_{\mathrm{CO}}=1.20 \mathrm{~atm}\)
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