Question

Computing entropies. The heat capacity of an ideal diatomic gas is \(C_{V}=(5 / 2) N k\) for \(N\) molecules. Assume \(T=300 \mathrm{~K}\) where needed. (a) One mole of \(\mathrm{O}_{2}\) gas fills a room of \(500 \mathrm{~m}^{3}\). What is the entropy change \(\Delta S\) for squeezing the gas into \(1 \mathrm{~cm}^{3}\) in the corner of the room? (b) An adult takes in about 3000 kcal per day from food ( 1 food \(\mathrm{Cal}=1 \mathrm{kcal}\) ). What is \(\Delta S\) for this process? (c) One mole of \(\mathrm{O}_{2}\) gas is in a room of \(500 \mathrm{~m}^{3}\). What is the entropy change \(\Delta S\) for heating the room from \(T=270 \mathrm{~K}\) to \(330 \mathrm{~K}\) ? (d) The free energy of a conformational motion of a loop in a protein is \(\Delta G=2 \mathrm{kcal} \mathrm{mol}^{-1}\). The enthalpy change is \(\Delta H=0.5 \mathrm{kcal} \mathrm{mol}^{-1}\). Compute \(\Delta S\).

Short answer

(a) -158.8 J/K (b) 4184 J/K (c) 4.3 J/K (d) -20.92 J/K

Step-by-step solution

Calculating entropy change for squeezing the gas

Use the formula for entropy change for a gas undergoing an isothermal reversible compression or expansion: \[ \Delta S = nR \ln \left( \frac{V_2}{V_1} \right) \]Given: \( n = 1 \) mole (mole of \( O_2 \)) \( R = 8.314 \ J \ mol^{-1} \ K^{-1} \) \( V_1 = 500 \ m^3 \) \( V_2 = 1 \ cm^3 = 1 \times10^{-6} \ m^3 \). Plug in these values to get \[ \Delta S = (1 \ mol)(8.314 \ J \ mol^{-1} \ K^{-1}) \ln \left( \frac{1 \times 10^{-6} \ m^3}{500 \ m^3} \right) \].

Simplifying the logarithmic term

Calculate the ratio of volumes: \[ \frac{V_2}{V_1} = \frac{1 \times 10^{-6}}{500} = 2 \times 10^{-9} \]. Now calculate \[ \ln (2 \times 10^{-9}) \approx -19.1 \]..

Plug in the values into entropy formula

Now, \[ \Delta S = 8.314 \times (-19.1) \approx -158.8 \ J \ K^{-1} \]. Entropy change for squeezing the gas is approximately -158.8 J/K.

Calculating entropy change for food intake

Calculate the entropy change using \[ \Delta S = \frac{Q}{T} \], where \( Q \) is the energy intake and \( T = 300 \ K \). Given: \( Q = 3000 \ kcal \times 4184 \ J/kcal = 1.2552 \times 10^7 \ J \). Now, \[ \Delta S = \frac{1.2552 \times 10^7 \ J}{300 \ K} \].

Calculating the value of entropy change

Perform the division: \[ \Delta S = \frac{1.2552 \times 10^7}{300} = 4184 \ J \ K^{-1} \]. Entropy change for the food intake is approximately 4184 J/K.

Calculating entropy change for heating the gas

Use the formula for entropy change for heating a gas at constant volume: \[ \Delta S = nC_V \ln \left( \frac{T_2}{T_1} \right) \]. Given: \( n = 1 \) mole, \( C_V = \frac{5}{2} R \), and the temperatures \( T_1 = 270 \ K \), \( T_2 = 330 \ K \). Calculate \( C_V = \frac{5}{2} \times 8.314 = 20.785 \ J \ mol^{-1} \ K^{-1} \).

Plug in the temperature values

Now, \[ \Delta S = (1 \ mol) \times 20.785 \ J \ mol^{-1} \ K^{-1} \times \ln \left( \frac{330}{270} \right) \]..

Calculate the value of entropy change

First, calculate the logarithmic term \[ \ln \left( \frac{330}{270} \right) \approx 0.206 \]. Now, \[ \Delta S = 20.785 \times 0.206 = 4.3 \ J \ K^{-1} \]. Entropy change for heating the gas is approximately 4.284 J/K.

Calculating entropy change for protein conformational motion

Use Gibbs free energy relation: \[ \Delta G = \Delta H - T\Delta S \]. Given: \( \Delta G = 2 \ kcal/mol \), \( \Delta H = 0.5 \ kcal/mol \). Rearrange to isolate \( \Delta S \): \[ \Delta S = \frac{\Delta H - \Delta G}{T} = \frac{0.5 - 2}{T} = \frac{-1.5}{T} \ kcal/mol \]..

Convert and simplify

Convert \( kcal \) to \( J \) and use \( T = 300 \ K \) \[ \Delta S = \frac{-1.5 \times 4184}{300} \approx -20.92 \ J \ K^{-1} \]. The entropy change for the conformational motion is approximately -20.92 J/K.

Key concepts

Ideal Gas Law

Understanding the Ideal Gas Law is crucial in solving entropy change problems with gases. The Ideal Gas Law is given by the equation:
\[ PV = nRT \]
Here,

• P is the pressure of the gas,
• V is the volume of the gas,
• n is the number of moles of the gas,
• R is the universal gas constant (8.314 J/mol·K), and
• T is the absolute temperature in Kelvin.

For example, in the exercise where 1 mole of O2 fills a room of 500 m^3, if we assume constant temperature, the volume reduction plays a major role in the entropy change calculation. The law helps establish the relationship between these gas variables under ideal conditions, simplifying many thermodynamic computations.
In the exercise, this law helped determine the changes in volume for entropy calculations by providing a foundational understanding of how gases behave, which is vital in statistical mechanics and thermodynamics contexts.

Thermodynamics

Thermodynamics deals with heat, work, and various forms of energy transfer. This field of physics explains how changes in energy affect the state of matter. Key concepts include:

• Entropy (S): A measure of the disorder or randomness in a system. It quantifies the number of microscopic configurations that correspond to a thermodynamic system's macroscopic state. Higher entropy means higher disorder.
• Enthalpy (H): The total heat content of a system, important for understanding heat exchange at constant pressure.
• Free Energy (G): Represented by the Gibbs free energy equation \[ G = H – TS \]. It determines the system's spontaneity — spontaneous processes have negative ΔG.

**Entropy Changes**:
In the provided problem, entropy change is calculated for different scenarios.
1. **Isothermal Compression**: In part (a), squeezing O2 gas into a much smaller volume results in a negative entropy change, representing decreased disorder.
2. **Heat Transfer**: Part (c) investigates heating at constant volume, using molar heat capacity (Cv) and the temperature change ratio.
3. **Food Intake**: In part (b), ΔS is determined using the energy intake (Q) and temperature (T).
Understanding these thermodynamic quantities and their calculations allows us to navigate through problems dealing with energy changes in various processes.

Statistical Mechanics

Statistical Mechanics bridges the macroscopic and microscopic worlds by statistically studying the behavior of large numbers of particles. It offers a deeper understanding of thermodynamic principles. Main concepts include:

• Microstates and Macrostates: A microstate is a specific detailed configuration of a system, while a macrostate is defined by macroscopic quantities (e.g., temperature, pressure).
• Probability Distributions: Describes how particles are distributed among available energy states, influencing entropy calculations.
• Boltzmann's Entropy Formula: Given by \[ S = k \,\ln \, W \], where S is the entropy, k is the Boltzmann constant, and W is the number of microstates corresponding to the macrostate.

**Entropy Calculation Connects to Statistical Mechanics**:
In part (a) of the exercise, as the gas is squeezed into a smaller volume, the number of available microstates drops, leading to lower entropy. This ties back to Boltzmann's formula, which interprets entropy as a measure of system disorder probabilistically. Statistical mechanics thus provides a microscopic basis for thermodynamic properties, deepening our understanding of why certain processes (like compression or heating) result in specific entropy changes.