Question

A volume of \(6.00 \mathrm{~L}\) of a monatomic ideal gas, originally at \(400 . \mathrm{K}\) and a pressure of \(3.00 \mathrm{~atm}\) (called state 1 ), undergoes the following processes, all done reversibly: \(1 \rightarrow 2\) isothermal expansion to \(V_{2}=4 V_{1}\) \(2 \rightarrow 3\) isobaric compression \(3 \rightarrow 1\) adiabatic compression to its original state Find the entropy change for each process.

Short answer

Question: Calculate the entropy change for each process that a monatomic ideal gas undergoes during the following three process cycle: 1) Isothermal expansion from V1 to V2 = 4 * V1, 2) Isobaric compression, and 3) Adiabatic compression back to the original state. The initial conditions of the gas are P1 = 3.00 atm, V1 = 6.00 L, and T1 = 400 K. Answer: The entropy changes for each process are given by: 1) Isothermal expansion: ΔS₁→₂ = nR ln(4) 2) Isobaric compression: ΔS₂→₃ = n(5/2)R ln(T₂/T₁) 3) Adiabatic compression: ΔS₃→₁ = 0

Step-by-step solution

Isothermal expansion

To find the entropy change in an isothermal process, we can use the formula: \(\Delta S = nR\ln \left(\frac{V_2}{V_1}\right)\) We first need to find the number of moles, n, using the initial conditions of the gas: \(P_1 = 3.00\ \mathrm{atm},\ V_1 = 6.00\ \mathrm{L},\ T_1 = 400\ \mathrm{K}\) Using the ideal gas law \(PV=nRT\), we have: \(n = \frac{P_1V_1}{RT_1} = \frac{3.00\ \mathrm{atm} \cdot 6.00\ \mathrm{L}}{0.0821\ \mathrm{L\ atm/mol\ K} \cdot 400\ \mathrm{K}}\) Now, we can find the entropy change in the isothermal expansion: \(\Delta S_{1\rightarrow2} = nR\ln \left(\frac{V_2}{V_1}\right) = nR\ln(4)\)

Isobaric compression

For an isobaric process, the formula for the entropy change is: \(\Delta S = nC_P\ln \left(\frac{T_2}{T_1}\right)\) Since the process is isobaric, the pressure at state 2 is the same as state 1. We can use the ideal gas law to determine the temperature at state 2: \(P_1V_1 = nRT_1 = nRT_2\) Rearranging for T2 gives: \(T_2 = \frac{P_1V_1}{nR} = \frac{3.00\ \mathrm{atm} \cdot 6.00\ \mathrm{L}}{0.0821\ \mathrm{L\ atm/mol\ K}}\) Because the gas is monatomic, we have: \(C_P = \frac{5}{2}R\) Now, we can find the entropy change in the isobaric compression: \(\Delta S_{2\rightarrow3} = nC_P\ln \left(\frac{T_2}{T_1}\right) = n\left(\frac{5}{2}R\right)\ln\left(\frac{T_2}{T_1}\right)\)

Adiabatic compression

In an adiabatic process, the entropy change is \(0\) since there is no heat transfer: \(\Delta S_{3\rightarrow1} = 0\) Finally, we calculate the entropy change for each process using the equations derived in the previous steps: \(\Delta S_{1\rightarrow2} = nR\ln(4)\) \(\Delta S_{2\rightarrow3} = n\left(\frac{5}{2}R\right)\ln\left(\frac{T_2}{T_1}\right)\) \(\Delta S_{3\rightarrow1} = 0\) Using these entropy changes, we are able to describe the entropy changes in the three processes which the monatomic ideal gas undergoes.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in thermodynamics that relates the pressure, volume, temperature, and number of moles of an ideal gas. The law is expressed as \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature. This equation helps us understand how gases behave under different conditions and is essential for calculating various properties of gases.
For the problem at hand, the Ideal Gas Law allows us to determine the number of moles of the gas using the initial conditions: pressure of 3.00 atm, volume of 6.00 L, and temperature of 400 K. By substituting these values into the equation, we can solve for \( n \). This provides a foundation to calculate other thermodynamic properties such as entropy change in various processes.

Entropy Change

Entropy is a measure of the disorder or randomness in a system. In thermodynamics, an increase in entropy signifies a move toward greater randomness. When analyzing gas processes, understanding entropy changes helps us determine how energy is distributed within the system.
During an isothermal expansion, entropy increases because the gas expands to occupy a larger volume without a change in temperature, leading to more disorder. The change in entropy \( \Delta S \) during this process can be calculated using the formula \( \Delta S = nR\ln \left( \frac{V_2}{V_1} \right) \), where \( V_2 \) and \( V_1 \) are the final and initial volumes, respectively.
For isobaric processes, the entropy change involves a change in temperature at constant pressure. It's calculated as \( \Delta S = nC_P\ln \left( \frac{T_2}{T_1} \right) \), indicating how the system's energy distribution changes with temperature.
In the case of adiabatic processes, \( \Delta S = 0 \). There's no heat exchange with the surroundings, so entropy remains constant, reflecting no change in internal disorder.

Isothermal Process

An isothermal process occurs when a system changes its state at a constant temperature. For an ideal gas, this means any work done by or on the gas is exactly compensated by heat exchange with the surroundings.
For the isothermal expansion from state 1 to state 2 in the exercise, the gas is expanding while maintaining the same temperature, which impacts entropy. To calculate entropy change in this specific isothermal expansion, you can use the formula \( \Delta S = nR\ln \left( \frac{V_2}{V_1} \right) \). This increase in entropy represents the increased randomness as the gas spreads to occupy a larger volume.
Isothermal processes are often idealized as slow enough to allow heat exchange, ensuring constant temperature - important for a reversible process in thermodynamics.

Isobaric Process

An isobaric process involves a change in a gas's state where the pressure remains constant. This type of process allows us to observe how temperature changes can lead to changes in volume or vice versa.
In the exercise's isobaric compression from state 2 to state 3, the gas is compressed at constant pressure, impacting both temperature and volume. The entropy change during an isobaric process is calculated using \( \Delta S = nC_P\ln \left( \frac{T_2}{T_1} \right) \), where \( C_P \) is the specific heat at constant pressure. This calculation helps quantify how energy distribution changes with a temperature change during a constant-pressure process.
Understanding isobaric processes is crucial for practical applications such as heat engines and refrigerators, where pressure often remains constant while doing work.

Adiabatic Process

An adiabatic process is characterized by no heat transfer between the system and its surroundings. In other words, all energy changes are due to work done on or by the system. This often results in significant temperature changes.
During the adiabatic compression from state 3 back to state 1 in the exercise, the gas is compressed without any heat exchange, leading to a return to its original state. Since no heat is transferred, the entropy change for this process is zero: \( \Delta S = 0 \).
Adiabatic processes are important in applications where rapid pressure changes occur, such as in engine cylinders. They illustrate how work can change internal energy and temperature without altering entropy. Understanding adiabatic processes is essential for calculating efficiency and understanding cycles of thermodynamic engines.