Question

1 mole of an ideal gas at \(300 \mathrm{~K}\) is expanded isother-mally and reversible from an initial volume of 1 litre to 10 litre. The work in this process is \(\left(\mathrm{R}=2 \mathrm{cal} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\right)\) : (a) \(163.7 \mathrm{cal}\) (b) zero (c) \(-1381.8 \mathrm{cal}\) (d) 9 litreatm

Short answer

The work done is approximately -1381.8 cal.

Step-by-step solution

Understand the Problem

We are given a scenario where 1 mole of an ideal gas undergoes an isothermal and reversible expansion from an initial volume of 1 litre to a final volume of 10 litres at a constant temperature of 300 K. We need to find the work done in this process.

Recall the Formula for Isothermal Expansion Work

The work done by an ideal gas expanding isothermally and reversibly is given by the formula:\[ W = -nRT \ln \left(\frac{V_f}{V_i}\right) \]where \(n\) is the number of moles, \(R\) is the ideal gas constant, \(T\) is the temperature in Kelvin, \(V_f\) is the final volume, and \(V_i\) is the initial volume.

Insert the Known Values into the Formula

We have:- \(n = 1\) mole- \(R = 2 \text{ cal} \text{ mol}^{-1} \text{ K}^{-1}\)- \(T = 300 \text{ K}\)- \(V_f = 10 \text{ l}\)- \(V_i = 1 \text{ l}\)Plug these values into the formula:\[ W = -1 \times 2 \times 300 \times \ln \left(\frac{10}{1}\right) \]

Calculate the Natural Logarithm

Calculate the natural logarithm \(\ln \left(\frac{10}{1}\right)\):\[ \ln (10) \approx 2.302 \]

Compute the Work Done

Substitute \(\ln (10) \approx 2.302\) back into the work formula:\[ W = -1 \times 2 \times 300 \times 2.302 \]\[ W = -1381.2 \text{ cal}\]

Choose the Nearest Answer

The calculated work is \(-1381.2 \text{ cal}\). The closest answer from the options provided is option (c) \(-1381.8 \text{ cal}\).

Key concepts

Ideal Gas

An ideal gas is a hypothetical gas composed of many randomly moving point particles that interact only through elastic collisions. It is an important concept in thermodynamics because it simplifies the mathematical modelling of gases. While real gases can behave like ideal gases at high temperatures and low pressures, ideal gas calculations provide useful approximations for many practical scenarios.

• Ideal gases follow the Ideal Gas Law: \[\text{PV} = nRT\]
• \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is temperature in Kelvin.
• For calculations, the ideal gas constant \(R\) can vary depending on the units used, such as \(2 \text{ cal mol}^{-1} \text{ K}^{-1}\) in the given exercise.

Understanding the concept of ideal gases helps in predicting the outcomes of thermodynamic processes, such as the one described in the given problem.

Reversible Process

A reversible process is a thermodynamic process that can be reversed by infinitesimal changes in some property of the system, with no change in entropy. These processes are idealizations since most natural processes are irreversible due to factors like friction and heat loss.

• In a reversible process, the system remains in thermodynamic equilibrium at all times.
• This means changes happen infinitely slowly, allowing the system to adjust to new equilibrium states without generating excess entropy.
• Reversible processes maximize the work done by or on a system, making them crucial for calculating theoretical limits.

The isothermal expansion of an ideal gas outlined in the problem is considered reversible, which allows us to use specific equations to calculate the work involved precisely.

Thermodynamics

Thermodynamics is the science of energy transfer, particularly regarding heat and work. It explores how these energies can be converted and how they affect matter. It is guided by a set of laws that describe these processes in systematic ways.

• First Law of Thermodynamics: Energy cannot be created or destroyed, only transferred or changed from one form to another.
• Second Law of Thermodynamics: Total entropy of an isolated system can never decrease over time.
• Thermodynamic processes can be isothermal (constant temperature), adiabatic (no heat transfer), isobaric (constant pressure), or isochoric (constant volume).

In the exercise, we're dealing with an isothermal process, meaning the temperature stays constant. Understanding these foundational principles of thermodynamics is key to analyzing energy changes in a system, such as calculating the work done during the expansion.

Work Calculation

Work calculation in the context of thermodynamics, especially during gas expansion, involves determining how energy is transferred as work by or on the system as it changes state. For an isothermal and reversible expansion of an ideal gas, the formula used is:
\[W = -nRT \ln\left(\frac{V_f}{V_i}\right)\]

• \(W\) is the work done.
• \(n\) is moles of gas.
• \(R\) is the gas constant, \(T\) is the temperature.
• \(V_i\) and \(V_f\) are the initial and final volumes.

By substituting known values into this formula, such as \(n = 1\), \(R = 2 \text{ cal mol}^{-1} \text{ K}^{-1}\), \(T = 300 \text{ K}\), \(V_i = 1\), and \(V_f = 10\), we find the work done equals \(-1381.8 \text{ cal}\). Knowing the steps for plugging in these numbers is crucial for students to accurately perform these calculations.