Problem 77 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

Two moles of an ideal monoatomic gas is heated from \(27^{\circ} \mathrm{C}\) to \(627^{\circ} \mathrm{C}\), reversibly and isochorically. The entropy of gas (a) increases by \(2 R \ln 3\) (b) increases by \(3 R \ln 3\) (c) decreases by \(2 R \ln 3\) (d) decreases by \(3 R \ln 3\)

Step-by-Step Solution

Answer

The entropy of the gas increases by \(3 R \ln 3\).

Step 1: Convert temperatures to Kelvin

First, convert the given temperatures from Celsius to Kelvin using the formula: Kelvin = Celsius + 273.15. For the initial temperature: T1 = 27°C + 273.15 = 300.15 K. For the final temperature: T2 = 627°C + 273.15 = 900.15 K.

Step 2: Calculate the entropy change

Use the formula for entropy change for an isochoric process of an ideal gas, which is: \(\Delta S = n C_v \ln(\frac{T2}{T1})\), where \(n\) is the number of moles, \(C_v\) is the molar heat capacity at constant volume for a monoatomic ideal gas (\(C_v = \frac{3}{2}R\)), and \(T1\) and \(T2\) are the initial and final temperatures in Kelvin, respectively. Then, calculate \(\Delta S\): \(\Delta S = 2 \cdot \frac{3}{2}R \ln(\frac{900.15}{300.15})\).

Step 3: Simplify and Find the Final Answer

Simplify the expression for \(\Delta S\) by canceling out similar terms in temperature ratio and proceed with the calculation: \(\Delta S = 3R \ln(\frac{900.15}{300.15}) = 3R\ln(3)\) since \(300.15 K\) cancels out when you divide \(900.15 K\) by \(300.15 K\). Thus, the entropy increases by \(3R\ln(3)\).

Key Concepts

Isochoric Process

An isochoric process, also known as an isovolumetric or constant-volume process, is a thermodynamic process in which the volume of the closed system remains constant. Since there is no change in volume, the system does no work on the environment, as work in thermodynamics is defined by the process of volume change under pressure. For an ideal gas, any heat added or removed from the system during an isochoric process will solely affect the internal energy, leading to a change in temperature.

Understanding an isochoric process is critical for calculating entropy changes in such a system, which is particularly straightforward since it eliminates the work term from the first law of thermodynamics equation, simplifying many thermodynamic calculations.

Ideal Gas Law

The ideal gas law is a fundamental equation in physics and chemistry that relates the pressure, volume, temperature, and amount of an ideal gas. It can be expressed as PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the universal gas constant, and T is the temperature in Kelvin.

This law assumes that the gas particles are point particles that interact only through elastic collisions and that they have no intermolecular forces acting between them. While real gases do not always behave like ideal gases, the ideal gas law provides a good approximation for many conditions, especially at high temperatures and low pressures.

Entropy of Monoatomic Gas

Entropy, in the context of thermodynamics, is a measure of the randomness or disorder in a system. For a monoatomic ideal gas, which consists of single-atom molecules, the entropy change can be calculated using the formula \(\Delta S = n C_v \ln(\frac{T2}{T1})\), where \(n\) is the number of moles, \(C_v\) is the molar heat capacity at constant volume, and \(T1\) and \(T2\) are the initial and final temperatures, respectively, in Kelvin.

As temperature increases, the kinetic energy of the gas particles increases, leading to a greater degree of randomness in their distribution and speeds, which in turn increases the entropy of the gas. Recognizing the relationship between temperature and entropy is crucial for understanding the thermodynamic behavior of ideal gases.

Heat Capacity at Constant Volume

The heat capacity at constant volume, denoted as \(C_v\), is a measure of how much heat energy is required to raise the temperature of a substance by a certain amount while maintaining a constant volume. For an ideal monoatomic gas, the molar heat capacity at constant volume is given by \(C_v = \frac{3}{2}R\), where \(R\) is the ideal gas constant.

The given \(C_v\) value is derived from the equipartition theorem, which states that energy is distributed equally among the degrees of freedom of the gas particles. Since monoatomic gases only have translational degrees of freedom, each degree contributes \(\frac{1}{2}R\) to the molar heat capacity. Thus, for three translational degrees of freedom, we obtain the aforementioned value for \(C_v\), vital for calculating the entropy change during isochoric processes.