Problem 54 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

The change in entropy accompanying the heating of one mole of helium gas \(\left(C_{\mathrm{v} . \mathrm{m}}\right.\) \(=3 R / 2\) ), assumed ideal, from a temperature of \(250 \mathrm{~K}\) to a temperature of \(1000 \mathrm{~K}\) at constant pressure. \((\ln 2=0.7)\) (a) \(4.2 \mathrm{cal} / \mathrm{K}\) (b) \(7.0 \mathrm{cal} / \mathrm{K}\) (c) \(2.1 \mathrm{cal} / \mathrm{K}\) (d) \(3.5 \mathrm{cal} / \mathrm{K}\)

Step-by-Step Solution

Answer

9.8 cal/K (None of the given options)

Step 1: Understand the entropy change formula

The entropy change \(\Delta S\) for an ideal gas when heated at constant pressure can be found using the formula: \[\Delta S = nC_{p}\ln(\frac{T_2}{T_1})\], where \(n\) is the number of moles, \(C_p\) is the molar heat capacity at constant pressure, \(T_1\) is the initial temperature, and \(T_2\) is the final temperature.

Step 2: Calculate the molar heat capacity at constant pressure

For helium gas, \(C_{v,m} = \frac{3R}{2}\). To find \(C_{p,m}\), use the relation: \[C_{p,m} = C_{v,m} + R\]. Since \(R = 8.314 J/(mol\cdot K)\) which is approximately equal to \(2 cal/(mol\cdot K)\), calculate \(C_{p,m}\) as follows: \[C_{p,m} = \frac{3R}{2} + R = \frac{3 \times 2}{2} + 2 = 5 + 2 = 7\] cal/(mol\cdot K).

Step 3: Compute the entropy change

Now that we have \(C_{p,m}\), we can calculate \(\Delta S\) using the entropy change formula: \[\Delta S = nC_{p,m}\ln(\frac{T_2}{T_1})\]. With \(n = 1\) mole, \(T_1 = 250 K\), and \(T_2 = 1000 K\), we get: \[\Delta S = 1 \cdot 7\ln(\frac{1000}{250}) = 7\ln(4) = 7\ln(2^2) = 14\ln(2)\]. Given that \(\ln(2) = 0.7\), replace and calculate \(\Delta S\): \[\Delta S = 14 \times 0.7 = 9.8\] cal/K.

Key Concepts

Thermodynamics

Thermodynamics is a branch of physics that deals with heat, work, and temperature, and their relation to energy, radiation, and physical properties of matter. At the heart of thermodynamics lie the laws that describe how these quantities are related and how energy is conserved. In our exercise, we are concerned with entropy, which is a measure of the disorder or randomness in a system. The second law of thermodynamics states that the total entropy of an isolated system can never decrease over time. Accordingly, as we heat a gas, such as helium, we increase its entropy because we are inputting energy into the system, increasing the motion of the gas particles and therefore their disorder.

The calculation of entropy change in this exercise is essential in understanding how energy is distributed within a gas and showcases one of the core principles of thermodynamics.

Ideal Gas Law

The ideal gas law is pivotal in physical chemistry and thermodynamics, as it relates the pressure, volume, temperature, and amount of a gas with a simple equation: \( PV = nRT \). Here, \( P \) is the gas pressure, \( V \) is the volume, \( n \) is the amount of substance in moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.

While our exercise doesn't explicitly use the ideal gas law, it implicitly relies on the behavior predicted by the law. In an ideal sense, when a gas such as helium is heated at constant pressure, its volume is assumed to change in a way that directly relates to the temperature change. This assumption allows us to use heat capacities at constant pressure to calculate entropy change without worrying about the complexities of real gas behavior.

Heat Capacity

Heat capacity is a property that describes how much heat energy is required to raise the temperature of a substance. It is closely tied to the internal structure of the substance, such as the vibrational, rotational, and translational motion of its molecules. In the context of gases, we differentiate between the molar heat capacity at constant volume, \( C_{v,m} \), and at constant pressure, \( C_{p,m} \).

The relationship between these values for ideal gases is given by \( C_{p,m} = C_{v,m} + R \), where \( R \) is the ideal gas constant. The exercise employs this relationship to transition from \( C_{v,m} \) to \( C_{p,m} \) because the process described occurs at constant pressure. Knowing the correct heat capacity is crucial for calculating the entropy change during heating since it relates directly to the amount of energy taken in by the gas.

Physical Chemistry

Physical chemistry is the study of how matter behaves on a molecular and atomic level and how chemical reactions occur. It combines principles of physics and chemistry to understand the physical properties of molecules, the forces that act upon them, and the energy changes that accompany their transformations.

In our problem, physical chemistry helps us to understand the microscopic view of entropy; as we heat helium gas, its particles move more rapidly and the positional and energy randomness increases. The component of physical chemistry that we apply is the calculation of entropy — a fundamental property indicating the direction and extent of chemical processes. This calculation involves the heat capacity of the gas, which can only be accurately determined by understanding the molecular structure and statistical behavior of the gas. Ultimately, physical chemistry provides the theoretical foundation for the equations and principles used to solve the entropy change we are calculating.