The heat capacity of chloroform (trichloromethane, \(\mathrm{CHCl}_{3}\) ) in the range \(240 \mathrm{~K}\) to \(330 \mathrm{~K}\) is given by \(\mathrm{C}_{\mathrm{p}, \mathrm{m}} /\left(\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right)=91.47+7.5 \times 10^{-2}(T / \mathrm{K})\). In a particular experiment, \(1.00 \mathrm{~mol} \mathrm{CHCl}_{3}\) is heated from \(273 \mathrm{~K}\) to \(300 \mathrm{~K}\). Calculate the change in molar entropy of the sample.

Short Answer

Expert verified
The change in molar entropy of the sample is \(\Delta S_{m} = (91.47(\ln{300} - \ln{273})) + (7.5 \times 10^{-2} \times 27)\) J mol^{-1} K^{-1}.

Step by step solution

01

Understand the Molar Heat Capacity Formula

The heat capacity of chloroform as a function of temperature is given by the formula \(C_{p,m}=91.47 + 7.5 \times 10^{-2}(T/K)\) where \(C_{p,m}\) is the heat capacity at constant pressure in units of J K^{-1} mol^{-1}. To find the change in molar entropy, we will integrate this formula over the temperature range of interest.
02

Set Up the Integral for Entropy Change

The change in molar entropy \(\Delta S_{m}\) is calculated by integrating the molar heat capacity with respect to temperature. Since entropy is a state function, we can write \(\Delta S_{m} = \int_{273 K}^{300 K} \frac{C_{p,m}}{T} dT\). Substituting the given heat capacity formula, the integral becomes \(\Delta S_{m} = \int_{273 K}^{300 K} \frac{91.47 + 7.5 \times 10^{-2}(T)}{T} dT\).
03

Simplify the Integral

The integral can be split into two parts: \(\Delta S_{m} = \int_{273 K}^{300 K} \frac{91.47}{T} dT + \int_{273 K}^{300 K} \frac{7.5 \times 10^{-2}}{K} dT\), which simplifies to \(\Delta S_{m} = (91.47 \ln{T}) \Big|_{273 K}^{300 K} + (7.5 \times 10^{-2})T \Big|_{273 K}^{300 K}\).
04

Evaluate the Integral

Now, calculate the definite integral by evaluating the antiderivatives at the upper and lower limits: \(\Delta S_{m} = (91.47 \ln{300} - 91.47 \ln{273}) + (7.5 \times 10^{-2})(300 - 273)\).
05

Compute the Entropy Change

Perform the calculations to find the change in molar entropy: \(\Delta S_{m} = (91.47 \ln{300} - 91.47 \ln{273}) + (7.5 \times 10^{-2})(27)\). Plug in the natural logarithm values and calculate the change: \(\Delta S_{m} = (91.47(\ln{300} - \ln{273})) + (7.5 \times 10^{-2} \times 27)\) J mol^{-1} K^{-1}.

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Key Concepts

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

Heat Capacity
Heat capacity is an intrinsic property of matter that measures the amount of heat required to change the temperature of a substance by a certain amount. When heat is added to or removed from a substance, its temperature changes according to the substance's heat capacity. The higher the heat capacity, the more heat is needed to change the temperature of the substance.

Different substances have different heat capacities due to their unique molecular structures and bonding. The heat capacity can be affected by the temperature, pressure, and phase of the substance. It's therefore an essential factor in understanding thermal systems and their behavior in various conditions.

In our chloroform example, knowing its heat capacity helps us determine the amount of energy required to heat a mole of chloroform from one temperature to another. The heat capacity, which can vary with temperature, provides a bridge between thermal energy and temperature change in a substance.
Molar Heat Capacity Formula
The molar heat capacity of a substance is the heat capacity per mole of that substance. It is often represented by the symbol \(C_{p,m}\) when measured at constant pressure, and is crucial for quantifying heat transfer in chemical reactions and physical changes.

The molar heat capacity formula for chloroform given in the textbook exercise, \(C_{p,m} = 91.47 + 7.5 \times 10^{-2}(T/K)\), shows that chloroform's heat capacity is dependent on temperature (\(T\)). This formula helps you calculate how much energy, in Joules, is needed to raise the temperature of one mole of chloroform by one Kelvin.

Understanding and using the correct molar heat capacity formula is vital because it allows for precise calculations of temperature-related entropy changes, which is a cornerstone for thermodynamics studies.
Entropy as a State Function
Entropy is a measure of the disorder or randomness within a system, and is one of the central concepts of thermodynamics. One important aspect of entropy is that it's a state function. This means that entropy is determined by the current state of the system, not by the path the system took to reach that state.

Entropy as a state function is analogous to elevation in geography; no matter how you climb a hill, the elevation at the top is the same. Similarly, for any thermodynamic system, the change in entropy between two states is independent of the transformation process. This characteristic allows us to calculate the change in molarity entropy (\(\Delta S_{m}\)) by integrating the heat capacity over a temperature range, assuming a reversible path, as done in the chloroform heating example.
Thermodynamic Integration
Thermodynamic integration is a mathematical technique used to calculate changes in functions of state, such as entropy, over a specific process. When dealing with heat capacities and changes in temperatures, integration allows us to find the total entropy change for a process, even when the heat capacity varies with temperature.

In our exercise, the provided heat capacity formula for chloroform requires integration over the temperature range of interest to determine the change in entropy. The process involves setting up a definite integral that incorporates the variable molar heat capacity with respect to temperature. This approach uses calculus to embody the nature of entropy as a state function, yielding a precise value for the entropy change that results from heating the chloroform sample from one temperature to another.

Thermodynamic integration is essential in physical chemistry because it provides an exact quantitative answer to changes in state functions, which is fundamental for predicting system behavior and conducting thermodynamic analysis.

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Most popular questions from this chapter

Suppose that \(S\) is regarded as a function of \(p\) and \(\mathrm{T}\). Show that \(T \mathrm{dS}=\) \(C_{\mathrm{p}} \mathrm{d} T-\alpha T V \mathrm{~d} p .\) Hence, show that the energy transferred as heat when the pressure on an incompressible liquid or solid is increased by \(\Delta p\) is equal to \(-\alpha T V \Delta p .\) Evaluate \(q\) when the pressure acting on \(100 \mathrm{~cm}^{3}\) of mercury at \(0^{\circ} \mathrm{C}\) is increased by \(1.0\) kbar. \(\left(\alpha=1.82 \times 10^{-4} \mathrm{~K}^{-1} .\right)\)

The cycle involved in the operation of an internal combustion engine is called the Otto cycle. Air can be considered to be the working substance and can be assumed to be a perfect gas. The cycle consists of the following steps: (1) reversible adiabatic compression from \(\mathrm{A}\) to \(\mathrm{B},(2)\) reversible constantvolume pressure increase from \(\mathrm{B}\) to \(\mathrm{C}\) due to the combustion of a small amount of fuel, (3) reversible adiabatic expansion from \(\mathrm{C}\) to \(\mathrm{D}\), and \((4)\) reversible and constant-volume pressure decrease back to state A. Determine the change in entropy (of the system and of the surroundings) tor each step of the cycle and determine an expression for the efficiency of the cycle, assuming that the heat is supplied in Step \(2 .\) Evaluate the efficiency for a compression ratio of \(10: 1\). Assume that, in state \(\mathrm{A}, \mathrm{V}=4.00 \mathrm{dm}^{3}, p=1.00 \mathrm{~atm}\), and \(\mathrm{T}=300 \mathrm{~K}\), that \(\mathrm{V}_{\mathrm{A}}==10 \mathrm{~V}_{\mathrm{B}}, \mathrm{p}_{\mathrm{C}} / \mathrm{p}_{\mathrm{B}}=5\), and that \(C_{p m}=\frac{2}{2} R\).

Find an expression for the change in entropy when two blocks of the same substance and of equal mass, one at the temperature \(T_{\mathrm{h}}\) and the other at \(T_{c}\), are brought into thermal contact and allowed to reach equilibrium. Evaluate the change for two blocks of copper, each of mass \(500 \mathrm{~g}\), with \(C_{p, m}=24.4 \mathrm{~J} \mathrm{~K}^{-1}\) \(\mathrm{mol}^{-1}\), taking \(T_{\mathrm{h}}=500 \mathrm{~K}\) and \(T_{\mathrm{c}}=250 \mathrm{~K}\)

The adiabatic compressibility, \(\kappa_{S}\), is defined like \(\kappa_{T}\) (eqn 2.44) but at constant entropy. Show that for a perfect gas \(p \gamma \kappa_{\mathrm{S}}=1\) (where \(\gamma\) is the ratio of heat capacities).

Given that \(s_{m}^{\circ}=29.79 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) for bismuth at \(100 \mathrm{~K}\) and the following tabulated heat capacities data (D.G. Archer, \(J\). Chem. Eng. Data 40 , 1015 (1995)), compute the standard molar entropy of bismuth at \(200 \mathrm{~K}\). $$ \begin{array}{lccccccl} \text { TIK } & 100 & 120 & 140 & 150 & 160 & 180 & 200 \\ C_{\mathrm{Am}} /\left(\mathrm{K} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right) & 23.00 & 23.74 & 24.25 & 24.44 & 24.61 & 24.89 & 25.11 \end{array} $$ Compare the value to the value that would be obtained by taking the heat capacity to be constant at \(24.44 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) over this range.

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