Question

Which of the following processes (all constant-temperature expansions) produces the most work? a) An ideal gas consisting of 1 mole of argon at \(20^{\circ} \mathrm{C}\) expands from \(1 \mathrm{~L}\) to \(2 \mathrm{~L}\). b) An ideal gas consisting of 1 mole of argon at \(20^{\circ} \mathrm{C}\) expands from \(2 \mathrm{~L}\) to \(4 \mathrm{~L}\). c) An ideal gas consisting of 2 moles of argon at \(10^{\circ} \mathrm{C}\) expands from \(2 \mathrm{~L}\) to \(4 \mathrm{~L}\). d) An ideal gas consisting of 1 mole of argon at \(40^{\circ} \mathrm{C}\) expands from \(1 \mathrm{~L}\) to \(2 \mathrm{~L}\). e) An ideal gas consisting of 1 mole of argon at \(40^{\circ} \mathrm{C}\) expands from \(2 \mathrm{~L}\) to \(4 \mathrm{~L}\).

Short answer

Short Answer: Process c) produces the most work, with a work value of \(3986.930\, \mathrm{J}\).

Step-by-step solution

Understand the isothermal work formula for an ideal gas.

The work done by an ideal gas during an isothermal process can be calculated using the formula: $$ W = nRT \ln \left(\frac{V_f}{V_i}\right) $$ where, \(W\) = work done \(n\) = number of moles of the gas \(R\) = universal gas constant (\(8.314\, \mathrm{J\, mol^{-1}K^{-1}} \)) \(T\) = temperature (in Kelvin) \(V_i\) = initial volume \(V_f\) = final volume

Convert the given temperatures to Kelvin.

As the given temperatures are in Celsius, we need to convert them to Kelvin using the formula: $$ T_\mathrm{K} = T_\mathrm{C} + 273.15 $$ For the processes a), b), and d) the temperatures are: - Process a) and b): \( T = 20^\circ \mathrm{C} = 293.15\, \mathrm{K} \) - Process c): \( T = 10 ^\circ \mathrm{C} = 283.15\, \mathrm{K} \) - Process d) and e): \( T = 40 ^\circ \mathrm{C} = 313.15\, \mathrm{K} \)

Calculate the work done for each process.

Using the isothermal work formula and the given information for each process, we can calculate the work done as follows: - Process a): $$ W = (1)(8.314)(293.15) \ln \left(\frac{2}{1}\right) = 2047.72\, \mathrm{J} $$ - Process b): $$ W = (1)(8.314)(293.15) \ln \left(\frac{4}{2}\right) = 2047.72\, \mathrm{J} $$ - Process c): $$ W = (2)(8.314)(283.15) \ln \left(\frac{4}{2}\right) = 3986.930\, \mathrm{J} $$ - Process d): $$ W = (1)(8.314)(313.15) \ln \left(\frac{2}{1}\right) = 2173.267\, \mathrm{J} $$ - Process e): $$ W = (1)(8.314)(313.15) \ln \left(\frac{4}{2}\right) = 2173.267\, \mathrm{J} $$

Compare the work done for each process and identify the largest.

Now that we have calculated the work done for each process, we can compare them to find out which process produces the most work. By comparing the calculated work values, we see that process c) produces the most work, with \(W = 3986.930\, \mathrm{J}\).

Key concepts

Ideal Gas Expansion

When we talk about the expansion of an ideal gas, we're discussing a change in the gas's volume while it may be subjected to various conditions such as temperature, pressure, or the number of gas molecules present (moles). In the context of our exercise, we are focusing on an isothermal expansion, which means the temperature remains constant throughout the process.

During expansion, an ideal gas can do work on its surroundings, for instance, by pushing against a piston in an engine. This is important because it's one of the ways an ideal gas can transfer energy. The more a gas expands, the more work it can potentially do, but this also depends on other factors such as temperature and the amount of gas present.

In our textbook exercise, we're particularly interested in comparing how much work is done when an ideal gas expands under different conditions but at constant temperatures. This requires understanding not just the change in volume, but also how this change interacts with other variables under isothermal conditions.

Work Done by Ideal Gas

The work done by an ideal gas during expansion (or compression) is a measure of the energy transferred by the gas to its surroundings. In the realm of thermodynamics, work is a way of transferring energy, and in the case of gases, it's often visualized as the gas pushing against a boundary, such as a piston in a cylinder.

For an isothermal process, the work done is dependent on the temperature, the number of moles of gas, and the logarithm of the volume ratio (final volume over initial volume). Since temperature is constant in an isothermal process, the amount of work done is directly related to how much the gas expands and the number of moles of gas present.

• If the volume doubles, the work done is more significant than if the volume increases by a smaller fraction.
• If there are more moles of gas, they can collectively do more work during the expansion.

A key aspect of this process is that the internal energy of the gas remains constant, hence, all energy expended in doing work comes from the heat absorbed by the gas.

Isothermal Work Formula

The isothermal work formula is a mathematical representation of the work done by a gas during an isothermal process. This formula is critical in our exercise for calculating and comparing the work done in different scenarios. The formula is:

\[ W = nRT \ln \left(\frac{V_f}{V_i}\right) \]
where,

• \(W\) is the work done by the gas (in Joules),
• \(n\) is the number of moles of gas,
• \(R\) is the universal gas constant, valued at \(8.314\, \mathrm{J\, mol^{-1}K^{-1}}\),
• \(T\) is the absolute temperature (in Kelvin),
• \(V_i\) and \(V_f\) are the initial and final volumes respectively.

When applying this formula, it's essential to convert temperatures to Kelvin for accuracy. As seen in the step-by-step solution, each part of the exercise used this formula to calculate work done during the expansion processes, correctly accounting for the number of moles and temperature in each scenario. Comparing the results allowed us to conclude which process produced the most work, guiding the students toward understanding the practical application of the isothermal work formula in thermodynamics.