Question

The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of \(10 \mathrm{dm}^{3}\) to a volume of \(100 \mathrm{dm}^{3}\) at \(27^{\circ} \mathrm{C}\) is: (a) \(35.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\) (b) \(38.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\) (c) \(45.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\) (d) \(23.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\)

Short answer

The entropy change is approximately \(38.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\).

Step-by-step solution

Understand the Question

We need to find the change in entropy for the isothermal reversible expansion of 2 moles of an ideal gas. The experiment begins at a volume of 10 dm³ and ends at 100 dm³, with the process occurring at 27°C, which is equivalent to 300 K.

Formula for Entropy Change

For an isothermal reversible expansion of an ideal gas, the change in entropy \( \Delta S \) is given by the formula: \( \Delta S = nR\ln\left(\frac{V_f}{V_i}\right) \) where \(n\) is the number of moles, \(R\) is the ideal gas constant \(8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\), and \(V_f\) and \(V_i\) are the final and initial volumes respectively.

Substitute Values

Substitute the given values into the formula: \( n = 2 \) moles, \( V_f = 100 \mathrm{~dm}^3 \), \( V_i = 10 \mathrm{~dm}^3 \), \( R = 8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \). Therefore, \( \Delta S = 2 \times 8.314 \times \ln\left(\frac{100}{10}\right) \).

Calculate the Natural Logarithm

Calculate the natural logarithm part: \( \ln\left(\frac{100}{10}\right) = \ln(10) \approx 2.302 \).

Calculate the Entropy Change

Using the values calculated, \( \Delta S = 2 \times 8.314 \times 2.302 \). This simplifies to \( 38.28 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \).

Choose the Closest Answer

The closest answer to the calculated entropy change \( 38.28 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \) is option (b) which is \( 38.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \).

Key concepts

Isothermal Reversible Expansion

Isothermal reversible expansion is a fundamental concept in thermodynamics, particularly when dealing with ideal gases. In an isothermal process, the temperature remains constant. Since the internal energy of an ideal gas only depends on its temperature, keeping it constant means there's no change in the internal energy during this process. This signifies that the energy added to the system through expansion work is offset entirely by heat absorbed from the surroundings.
During this process, if carried out reversibly, the system is in near-perfect equilibrium at all times. This gives allowance for the maximum amount of work, ensuring energy conversion is as efficient as possible without loss. It's crucial to note that real-world processes can never be fully reversible, but the assumption of reversibility allows us to model and understand complex systems with greater simplicity.

• Temperature Maintenance: In isothermal processes, external sources provide or absorb heat to maintain a constant temperature.
• Efficiency and Work: Reversible processes are considered ideal for their efficiency, extracting or applying the maximum amount of work possible.
• Equilibrium Consideration: The process is so slow that the system remains in equilibrium, an essential characteristic for reversibility.

Ideal Gas Law

The ideal gas law serves as a foundational principle for understanding the behavior of ideal gases. The law is elegantly simple and described by the equation \( PV = nRT \), where \(P\) is the pressure, \(V\) is the volume, \(n\) denotes the number of moles of the gas, \(R\) represents the ideal gas constant, and \(T\) is the absolute temperature in Kelvin.
One of the remarkable aspects of the ideal gas law is its ability to relate these four primary physical properties of gases. It combines Boyle's, Charles's, and Avogadro's laws into a single equation, making it a versatile tool for problem-solving in chemistry and physics.
In the context of isothermal reversible expansion, we capitalize only on the initial and final volumes, as the pressure and temperature return to singular focus due to their constant values in isothermal conditions.

• A unifying equation that ties together pressure, volume, moles, and temperature for ideal gases.
• Is most accurate under low pressure and high temperature where real gases behave more ideally.
• For isothermal processes, temperature is constant, simplifying calculations.

Natural Logarithm Calculation

Natural logarithms are indispensable in thermodynamics, particularly in calculating processes involving exponential changes and entropy. It uses the base \(e\) (approximately 2.718), often symbolized as \(\ln\).
In our exercise, the change in entropy is calculated using the natural logarithm of the ratio of the final volume to the initial volume. This mathematical operation allows us to capture and quantify how these volumes change in a manner that's directly correlatable with entropy.
Given the base \(e\), natural logs appear frequently in calculations modeling growth and decay processes, embodying the principle of continual multiplication or division. In the problem's context, we computed \( \ln(10) \approx 2.302 \), serving as a crucial step between understanding physical expansions and their resulting entropy.

• The natural logarithm is the inverse of the exponential function \(e^x\).
• Commonly used in calculative models for growth, decay, and compounding processes.
• For entropy change: \( \Delta S = nR\ln(V_f/V_i) \), it's vital to compute the natural log accurately for solutions.