Problem 76
Question
Oxygen gas weighing \(64 \mathrm{~g}\) is expanded from 1 atm to \(0.25\) atm at \(30^{\circ} \mathrm{C}\). What is the entropy change, assuming the gas to be ideal? \((\ln 4=1.4, R=8.3 \mathrm{~J} / \mathrm{K}-\mathrm{mol})\) (a) \(23.24 \mathrm{~J} / \mathrm{K}\) (b) \(34.86 \mathrm{~J} / \mathrm{K}\) (c) \(46.48 \mathrm{~J} / \mathrm{K}\) (d) \(11.62 \mathrm{~J} / \mathrm{K}\)
Step-by-Step Solution
Verified Answer
ΔS = 2 mol * 8.3 J/mol-K * 1.4 = 23.24 J/K
1Step 1: Calculate the amount of substance of oxygen
Use the molar mass of oxygen (32 g/mol for O2) and the given weight to calculate the amount of substance (moles) using the formula: Number of moles (n) = weight / molar mass. In this case, n = 64 g / (32 g/mol).
2Step 2: Use the isothermal expansion entropy change formula
For an isothermal expansion of an ideal gas, the entropy change (ΔS) can be calculated using the formula: ΔS = nRln(V2/V1) = nRln(P1/P2), since P1V1 = P2V2 for an isothermal process. Here, P1 is the initial pressure, and P2 is the final pressure.
3Step 3: Calculate the entropy change (ΔS)
Substitute the known values (moles from Step 1, R = 8.3 J/K-mol, and the given pressures) into the entropy change formula to find ΔS. ΔS = nRln(P1/P2) = 2 mol * 8.3 J/mol-K * 1.4.
4Step 4: Perform the calculation
Calculate ΔS = 2 mol * 8.3 J/mol-K * 1.4 to get the entropy change in joules per kelvin.
Key Concepts
Isothermal ExpansionIdeal Gas LawEntropy in ThermodynamicsPhysical Chemistry Problems
Isothermal Expansion
The process of isothermal expansion occurs when a gas increases in volume while maintaining the same temperature. The term 'isothermal' implies 'equal temperature,' indicating that there is no change in internal energy since the temperature of the system remains constant. This scenario applies perfectly to the provided exercise, where oxygen gas is expanded isothermally.
In practical terms, this could happen in a piston where the gas expands slowly enough that the temperature is able to remain the same. This type of expansion is governed by the ideal gas law, which is central to calculations involving changes in volume and pressure.
The significance of isothermal expansion in thermodynamics lies in its connection to entropy change. Entropy can be seen as a measure of disorder or randomness in a system, and during an isothermal expansion, the entropy of a gas increases because the molecules have more space to occupy, hence increasing the randomness of the system.
In practical terms, this could happen in a piston where the gas expands slowly enough that the temperature is able to remain the same. This type of expansion is governed by the ideal gas law, which is central to calculations involving changes in volume and pressure.
The significance of isothermal expansion in thermodynamics lies in its connection to entropy change. Entropy can be seen as a measure of disorder or randomness in a system, and during an isothermal expansion, the entropy of a gas increases because the molecules have more space to occupy, hence increasing the randomness of the system.
Ideal Gas Law
The ideal gas law is a fundamental equation in physical chemistry and thermodynamics, represented by the formula \( PV = nRT \). Here, \( P \) stands for pressure, \( V \) for volume, \( n \) for the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvins.
This law is based on the assumption of an 'ideal' gas, where the particles don't attract or repel each other and occupy negligible space. Though real gases may deviate from ideal behavior under high pressure or low temperature, the ideal gas law generally provides an excellent approximation under standard conditions.
In the context of the exercise at hand, the ideal gas law helps to understand the relationship between pressure and volume during the isothermal expansion of oxygen gas. Despite the change in volume and pressure, the temperature and the amount of substance remain constant, ensuring the applicability of this law.
This law is based on the assumption of an 'ideal' gas, where the particles don't attract or repel each other and occupy negligible space. Though real gases may deviate from ideal behavior under high pressure or low temperature, the ideal gas law generally provides an excellent approximation under standard conditions.
In the context of the exercise at hand, the ideal gas law helps to understand the relationship between pressure and volume during the isothermal expansion of oxygen gas. Despite the change in volume and pressure, the temperature and the amount of substance remain constant, ensuring the applicability of this law.
Entropy in Thermodynamics
Entropy is a core concept in thermodynamics and it represents the degree of disorder or randomness in a system. In thermodynamics, the second law states that the total entropy of an isolated system can never decrease over time. This means that natural processes tend to move toward a state of maximum entropy or disorder.
Mathematically, for a reversible isothermal process, the change in entropy (\( \triangle S \)) can be determined using the formula \( \triangle S = nR\text{ln}(V_2/V_1) \) or, equivalently, \( \triangle S = nR\text{ln}(P_1/P_2) \), depending on whether volume or pressure is constant or varying. Here, \( R \) is the ideal gas constant and \( n \) is the number of moles.
In the exercise provided, we're focusing on entropy during the isothermal expansion of oxygen gas. The question drives us to understand how the randomness of the gas molecules increases as the gas expands from a state of higher pressure to a state of lower pressure.
Mathematically, for a reversible isothermal process, the change in entropy (\( \triangle S \)) can be determined using the formula \( \triangle S = nR\text{ln}(V_2/V_1) \) or, equivalently, \( \triangle S = nR\text{ln}(P_1/P_2) \), depending on whether volume or pressure is constant or varying. Here, \( R \) is the ideal gas constant and \( n \) is the number of moles.
In the exercise provided, we're focusing on entropy during the isothermal expansion of oxygen gas. The question drives us to understand how the randomness of the gas molecules increases as the gas expands from a state of higher pressure to a state of lower pressure.
Physical Chemistry Problems
Physical chemistry problems often involve understanding the properties of matter on both the macroscopic and atomic levels. These problems require a knowledge of theories and laws, such as the ideal gas law, to solve complex scenarios encountered in the study of gases, solutions, thermodynamics, and kinetics.
When tackling physical chemistry problems, especially those related to thermodynamics like in our exercise, it’s important to carefully understand the system in question and the processes involved. This usually involves identifying key properties such as pressure, volume, temperature, and mass or moles, and how these properties are interrelated through physical laws.
Success in solving these problems comes with practice and a stepwise approach—breaking down the larger problem into smaller, more manageable steps, such as calculating the number of moles, understanding the process type (like isothermal expansion), and then applying the relevant equations to find the final answer.
When tackling physical chemistry problems, especially those related to thermodynamics like in our exercise, it’s important to carefully understand the system in question and the processes involved. This usually involves identifying key properties such as pressure, volume, temperature, and mass or moles, and how these properties are interrelated through physical laws.
Success in solving these problems comes with practice and a stepwise approach—breaking down the larger problem into smaller, more manageable steps, such as calculating the number of moles, understanding the process type (like isothermal expansion), and then applying the relevant equations to find the final answer.
Other exercises in this chapter
Problem 75
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