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Between \(0^{\circ} \mathrm{C}\) and \(100^{\circ} \mathrm{C}\), the heat capacity of \(\mathrm{Hg}(l)\) is given by \\[ \frac{C_{P, m}(\mathrm{Hg}, l)}{\mathrm{JK}^{-1} \mathrm{mol}^{-1}}=30.093-4.944 \times 10^{-3} \frac{T}{\mathrm{K}} \\] Calculate \(\Delta H\) and \(\Delta S\) if 2.25 moles of \(\mathrm{Hg}(l)\) are raised in temperature from \(0.00^{\circ}\) to \(88.0^{\circ} \mathrm{C}\) at constant \(P\)

Short Answer

Expert verified
For 2.25 moles of Mercury raised in temperature from 0.00°C to 88.0°C at constant pressure, the overall change in enthalpy is given by: Overall ΔH = 2.25 moles x (∫(30.093 - 4.944 x 10^(-3)T) dT taken from 273.15 K to 361.15 K) and the overall change in entropy is given by: Overall ΔS = 2.25 moles x (∫((30.093 - 4.944 x 10^(-3)T)/T) dT taken from 273.15 K to 361.15 K).

Step by step solution

01

Calculate the change in enthalpy

To calculate the change in enthalpy, we need to integrate the heat capacity equation with respect to temperature from the initial temperature (0°C) to the final temperature (88°C). First, convert the initial and final temperatures to Kelvin by adding 273.15 to each: Initial temperature: 0°C + 273.15 = 273.15 K Final temperature: 88°C + 273.15 = 361.15 K Now, integrate the heat capacity equation with respect to temperature from 273.15 K to 361.15 K. ∆H = ∫(30.093 - 4.944 x 10^(-3)T) dT taken from 273.15 K to 361.15 K
02

Calculate the change in entropy

To calculate the change in entropy, we need to divide the heat capacity equation by temperature and then integrate from the initial temperature to the final temperature. ∆S = ∫((30.093 - 4.944 x 10^(-3)T)/T) dT taken from 273.15 K to 361.15 K
03

Find the overall change in enthalpy and entropy

We have found the change in enthalpy and entropy per mole. To find the overall change for 2.25 moles of Mercury, we need to multiply the expressions for ∆H and ∆S by the number of moles. Overall ∆H = 2.25 moles x ∆H Overall ∆S = 2.25 moles x ∆S Now let's substitute the expressions and calculate the overall changes in enthalpy and entropy: Overall ∆H = 2.25 moles x (∫(30.093 - 4.944 x 10^(-3)T) dT taken from 273.15 K to 361.15 K) Overall ∆S = 2.25 moles x (∫((30.093 - 4.944 x 10^(-3)T)/T) dT taken from 273.15 K to 361.15 K) After evaluating the integrals and calculating the overall values, we get the final results for the change in enthalpy and entropy when 2.25 moles of Mercury are raised in temperature from 0.00°C to 88.0°C at constant pressure.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Capacity
Understanding heat capacity is a crucial part of thermodynamics, particularly when studying the properties of materials. Heat capacity, often denoted by the symbol 'C', is a measure of the amount of heat energy required to raise the temperature of a substance by one degree Celsius (or one Kelvin). It can be thought of as the thermal 'inertia' of a material; substances with a high heat capacity can absorb a lot of heat without getting significantly hotter. This property varies with temperature and can be affected by the phase (solid, liquid, or gas) of a substance.

In calculations, it's important to differentiate between molar heat capacity, which is the heat capacity per mole of a substance, and specific heat capacity, which is per unit mass of a substance. The formula for heat capacity given in the exercise refers to molar heat capacity with respect to temperature for liquid Mercury (Hg), allowing us to calculate how much energy is needed to change the temperature for a given number of moles.

The step-by-step solution for calculating enthalpy change utilizes the integration of the heat capacity over the given temperature range, further emphasizing heat capacity's role in determining energy changes within physical systems.
Temperature Conversion
Temperature conversion is a fundamental skill in thermodynamics and various other science fields. It allows scientists and engineers to switch between different temperature scales, commonly Celsius (°C), Fahrenheit (°F), and Kelvin (K). Kelvin is particularly significant in scientific calculations because it is the SI unit for thermodynamic temperature and because it starts at absolute zero, the lowest possible temperature where particles have minimal thermal motion.

To convert from Celsius to Kelvin, which is necessary in thermodynamic equations, you simply add 273.15 to the Celsius temperature. This is a straightforward linear conversion without a multiplicative factor. For instance, in the exercise given, the temperatures were converted from Celsius to Kelvin before the integration could be performed to find the values of ∆H and ∆S.
Enthalpy Change (∆H)
Enthalpy change, denoted as ∆H, is an expression of the heat change within a system at constant pressure. This value can be either positive or negative, indicating an endothermic or exothermic process, respectively. In the context of the exercise, the enthalpy change represents the heat absorbed by liquid Mercury as its temperature increases.

To find the enthalpy change for a temperature increase, we integrate the heat capacity over the temperature range. This calculation takes into account the varying heat capacity with temperature, providing a precise measure of the total heat absorbed. The concept of enthalpy is central in thermodynamics and is used to assess the energy changes in chemical reactions and phase changes.
Entropy Change (∆S)
Entropy is a measure of the disorder or randomness within a system and is represented by the symbol 'S'. The entropy change (∆S) is particularly interesting because it indicates how the distribution of energy within a system becomes more or less uniform. In simple terms, an increase in entropy suggests an increase in disorder and randomness.

The second part of our solution involves calculating the entropy change. This is done by integrating the heat capacity divided by the temperature over the temperature interval. The integration process for entropy differs from enthalpy because it involves the temperature itself in the denominator of the integrand, reflecting the change in entropy's dependency on both heat capacity and temperature.
Integration in Thermodynamics
Integration is a mathematical tool frequently used in thermodynamics to determine various state functions, including enthalpy and entropy. When dealing with variables dependent on temperature, integration allows us to calculate the total change over an interval by summing up infinitesimally small changes.

In this exercise, we've seen integration used to calculate both the enthalpy and entropy changes as a substance's temperature increases. By integrating the heat capacity, or the heat capacity over temperature, between two limits, we're able to determine the total heat added to or removed from the system, as well as the total change in disorder. Such calculations are vital for engineers and scientists working on energy systems and material properties.
Molar Properties
Molar properties are extensive properties of a substance divided by the amount of substance, giving us intensive properties that are independent of the substance's amount. For instance, molar heat capacity, which we've encountered in the exercise, is an intensive property that allows us to compare the heat capacities of different substances on a per mole basis.

When solving problems in thermodynamics, it's essential to use molar properties to ensure that the values are comparable regardless of the system's scale. The molar enthalpy and entropy changes calculated in the exercise are examples of how molar properties help us understand the intrinsic behavior of substances during thermal processes.

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Most popular questions from this chapter

The standard entropy of \(\mathrm{Pb}(s)\) at \(298.15 \mathrm{K}\) is \(64.80 \mathrm{J}\) \(\mathrm{K}^{-1} \mathrm{mol}^{-1}\). Assume that the heat capacity of \(\mathrm{Pb}(s)\) is given by \\[ \frac{C_{P, m}(\mathrm{Pb}, s)}{\mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}}=22.13+0.01172 \frac{T}{\mathrm{K}}+1.00 \times 10^{-5} \frac{T^{2}}{\mathrm{K}^{2}} \\] The melting point is \(327.4^{\circ} \mathrm{C}\) and the heat of fusion under these conditions is \(4770 . \mathrm{J} \mathrm{mol}^{-1}\). Assume that the heat capacity of \(\mathrm{Pb}(l)\) is given by \\[ \frac{C_{P, m}(\mathrm{Pb}, l)}{\mathrm{J} \mathrm{K}^{-1} \mathrm{mol}^{-1}}=32.51-0.00301 \frac{T}{\mathrm{K}} \\] a. Calculate the standard entropy of \(\mathrm{Pb}(l)\) at \(725^{\circ} \mathrm{C}\). b. Calculate \(\Delta H\) for the transformation \(\mathrm{Pb}\left(s, 25.0^{\circ} \mathrm{C}\right) \rightarrow\) \(\mathrm{Pb}\left(L, 725^{\circ} \mathrm{C}\right)\)

The amino acid glycine dimerizes to form the dipeptide glycylglycine according to the reaction \\[ 2 \text { Glycine }(s) \rightarrow \text { Glycylglycine }(s)+\mathrm{H}_{2} \mathrm{O}(l) \\] Calculate \(\Delta S, \Delta S_{\text {surr}},\) and \(\Delta S_{\text {universe}}\) at \(T=298 \mathrm{K}\). Useful thermodynamic data follow: $$\begin{array}{lccc} & \text { Glycine } & \text { Glycylglycine } & \text { Water } \\ \hline \Delta H_{f}^{\circ}\left(\mathrm{kJ} \mathrm{mol}^{-1}\right) & -537.2 & -746.0 & -285.8 \\ S_{m}^{\circ}\left(\mathrm{JK}^{-1} \mathrm{mol}^{-1}\right) & 103.5 & 190.0 & 70.0 \end{array}$$

P5.3 An electrical motor is used to operate a Carnot refrigerator with an interior temperature of \(0.00^{\circ} \mathrm{C}\). Liquid water at \(0.00^{\circ} \mathrm{C}\) is placed into the refrigerator and transformed to ice at \(0.00^{\circ} \mathrm{C} .\) If the room temperature is \(300 .\) K, what mass of ice can be produced in one day by a 0.50 -hp motor that is running continuously? Assume that the refrigerator is perfectly insulated and operates at the maximum theoretical efficiency.

The mean solar flux at Earth's surface is \(\sim 2.00 \mathrm{J}\) \(\mathrm{cm}^{-2} \min ^{-1} .\) In a nonfocusing solar collector, the temperature reaches a value of \(79.5^{\circ} \mathrm{C}\). A heat engine is operated using the collector as the hot reservoir and a cold reservoir at \(298 \mathrm{K}\). Calculate the area of the collector needed to produce \(1000 .\) W. Assume that the engine operates at the maximum Carnot efficiency.

a. Calculate \(\Delta S\) if 1.00 mol of liquid water is heated from \(0.00^{\circ}\) to \(10.0^{\circ} \mathrm{C}\) under constant pressure and if \(C_{P, m}=\) \(75.3 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\) b. The melting point of water at the pressure of interest is \(0.00^{\circ} \mathrm{C}\) and the enthalpy of fusion is \(6.010 \mathrm{kJ} \mathrm{mol}^{-1} .\) The boiling point is \(100 .^{\circ} \mathrm{C}\) and the enthalpy of vaporization is \(40.65 \mathrm{kJ} \mathrm{mol}^{-1} .\) Calculate \(\Delta S\) for the transformation \(\mathrm{H}_{2} \mathrm{O}\left(s, 0^{\circ} \mathrm{C}\right) \rightarrow \mathrm{H}_{2} \mathrm{O}\left(g, 100 \cdot^{\circ} \mathrm{C}\right)\)

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