Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

One mole of an ideal gas is \(25^{\circ} \mathrm{C}\) is subjected to expand reversibly 10 times of its initial volume. The change in cntropy of expansion is (1) \(19.15 \mathrm{Jk}^{-1} \mathrm{~mol}^{-1}\) (2) \(16.15 \mathrm{Jk}^{-1} \mathrm{~mol}^{-1}\) (3) \(22.15 \mathrm{Jk}^{-1} \mathrm{~mol}^{-1}\) (4) \(11.25 \mathrm{Jk}^{-1} \mathrm{~mol}^{-1}\)

Short Answer

Expert verified
19.15 J/K·mol

Step by step solution

01

- Convert temperature to Kelvin

First, convert the temperature from Celsius to Kelvin. Recall that the conversion formula is: \( T(K) = T(°C) + 273.15 \). So, the temperature in Kelvin is: \( 25ºC + 273.15 = 298.15 K \)
02

- Formula for entropy change

For an ideal gas, the change in entropy (ΔS) during a reversible expansion can be calculated using the formula: \( \text{ΔS} = nR \times \text{ln}\frac{V_f}{V_i} \) where * n is the number of moles of gas, * R is the universal gas constant (8.314 J/(mol·K)), and * \( V_f \) and \( V_i \) are the final and initial volumes respectively.
03

- Identify variables

In this exercise: * n = 1 mole, * R = 8.314 J/(mol·K), * Final volume \( V_f \) = 10 times the initial volume \( V_i \).
04

- Apply the known values

Substitute the known values into the formula: \( \text{ΔS} = (1 \text{ mol}) \times (8.314 \text{ J/(mol·K)}) \times \text{ln}\frac{10V_i}{V_i} \) = \( 8.314 \times \text{ln}(10) \)
05

- Calculate natural log

Find \( \text{ln}(10) \). Using a calculator, \( \text{ln}(10) \approx 2.3026 \), so: \( \text{ΔS} = 8.314 \times 2.3026 \)
06

- Final calculation

Multiply the values to get the change in entropy: \( \text{ΔS} = 8.314 \times 2.3026 \approx 19.15 \text{ J/K·mol} \)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reversible Expansion
Reversible expansion is a key concept in understanding how gases behave when they change volume.
In thermodynamics, reversible processes are idealized processes that happen infinitely slowly, allowing the system to remain in equilibrium at all times. This means that the external pressure is always nearly equal to the internal pressure of the gas.
During reversible expansion, the gas does work on the surroundings by pushing the boundary outward.
In the context of an ideal gas expanding, the focus is on how the volume changes and how this impacts entropy, a measure of disorder in the system.
This is important for understanding how energy is efficiently transferred as work and heat during thermodynamic processes.
Natural Logarithm
The natural logarithm, denoted as ln, is a logarithm to the base e (where e is approximately equal to 2.71828).
It's a fundamental concept in calculus and is especially important in the study of exponential growth and decay processes. When dealing with entropy change in thermodynamics, the natural logarithm helps in determining the relationship between initial and final states.
This is because many thermodynamic equations naturally involve ln due to the properties of exponential functions and their inverses.
For example, when calculating the change in entropy (ΔS) during a reversible expansion, the formula involves the natural logarithm of the ratio of the final volume (V_f) to the initial volume (V_i), denoted as ln(V_f/V_i). This ratio is crucial for understanding how macroscopic changes in volume impact microscopic states of an ideal gas.
Universal Gas Constant
The universal gas constant, denoted as R, is a fundamental constant in chemistry and physics.
Its value is approximately 8.314 J/(mol·K). It appears in various gas laws that describe the behavior of ideal gases, making it a crucial value in thermodynamic calculations.
The universal gas constant bridges the macroscopic and microscopic worlds. It links the energy scale (in Joules) to the temperature scale (in Kelvin) and the amount of substance (in moles). When calculating the entropy change (ΔS) for an ideal gas undergoing reversible expansion, the formula used is ΔS = nR ln(V_f/V_i), where n is the number of moles and ln(V_f/V_i) is the natural logarithm of the volume ratio.
This equation shows how R helps quantify changes in energy distribution and disorder within the system.
Ideal Gas Law
The ideal gas law is a simple equation that describes the behavior of ideal gases. It is given by PV = nRT, where:
  • 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
  • T is the temperature of the gas in Kelvin
This equation shows the relationship between pressure, volume, temperature, and the quantity (in moles) of an ideal gas.
Despite its simplicity, the ideal gas law provides a good approximation for the behavior of many gases under a wide range of conditions.
Understanding it is essential for solving many thermodynamic problems, including those involving entropy changes. In the context of the given exercise, the ideal gas law helps confirm how volume changes when the temperature and quantity of gas are known.
This further supports the calculations involving entropy changes during reversible expansion, highlighting the practical importance of the ideal gas law in understanding real-world phenomena.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.

Sign-up for free