Question

Which of the following processes (all constanttemperature 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

Using the formula provided for each process, we find the following values: - \(W_A\) = -1714.7 J - \(W_B\) = -1714.7 J - \(W_C\) = -3270.8 J - \(W_D\) = -1831.3 J - \(W_E\) = -1831.3 J Out of these processes, Process C produces the most work with a value of -3270.8 J.

Step-by-step solution

Process A: Work Calculation

Given: \(n = 1\,\text{mol}\), \(T = 20^{\circ}\text{C} (= 293.15\,\text{K})\), \(V_i = 1\,\text{L}\), and \(V_f = 2\,\text{L}\). Calculate work using the formula: $$ W_A = -1 \times 8.314\,\text{J/(mol} \cdot \text{K})\times 293.15\,\text{K}\times \ln\left(\dfrac{2\,\text{L}}{1\,\text{L}}\right) $$

Process B: Work Calculation

Given: \(n = 1\,\text{mol}\), \(T = 20^{\circ}\text{C} (= 293.15\,\text{K})\), \(V_i = 2\,\text{L}\), and \(V_f = 4\,\text{L}\). Calculate work using the formula: $$ W_B = -1 \times 8.314\,\text{J/(mol} \cdot \text{K})\times 293.15\,\text{K}\times \ln\left(\dfrac{4\,\text{L}}{2\,\text{L}}\right) $$

Process C: Work Calculation

Given: \(n = 2\,\text{mol}\), \(T = 10^{\circ}\text{C} (= 283.15\,\text{K})\), \(V_i = 2\,\text{L}\), and \(V_f = 4\,\text{L}\). Calculate work using the formula: $$ W_C = -2 \times 8.314\,\text{J/(mol} \cdot \text{K})\times 283.15\,\text{K}\times \ln\left(\dfrac{4\,\text{L}}{2\,\text{L}}\right) $$

Process D: Work Calculation

Given: \(n = 1\,\text{mol}\), \(T = 40^{\circ}\text{C} (= 313.15\,\text{K})\), \(V_i = 1\,\text{L}\), and \(V_f = 2\,\text{L}\). Calculate work using the formula: $$ W_D = -1 \times 8.314\,\text{J/(mol} \cdot \text{K})\times 313.15\,\text{K}\times \ln\left(\dfrac{2\,\text{L}}{1\,\text{L}}\right) $$

Process E: Work Calculation

Given: \(n = 1\,\text{mol}\), \(T = 40^{\circ}\text{C} (= 313.15\,\text{K})\), \(V_i = 2\,\text{L}\), and \(V_f = 4\,\text{L}\). Calculate work using the formula: $$ W_E = -1 \times 8.314\,\text{J/(mol} \cdot \text{K})\times 313.15\,\text{K}\times \ln\left(\dfrac{4\,\text{L}}{2\,\text{L}}\right) $$ After calculating the work for each process, compare the values and find the process that produces the most work.

Key concepts

The Ideal Gas Law: Making Sense of Work in Thermodynamics

The ideal gas law is a cornerstone of thermodynamics and provides a mathematical relationship between pressure (P), volume (V), temperature (T), and the number of moles (n) of an ideal gas. Described by the formula
\( PV = nRT \)
where R is the ideal gas constant (8.314 J/(mol·K)), this equation allows us to predict how a gas will behave under different conditions. When it comes to calculating the work done during the expansion or compression of a gas, this law becomes especially useful.

During an isothermal expansion—where the temperature (T) remains constant—the work (W) done by the gas on its surroundings can be calculated by integrating the pressure over the change in volume. Since, from the ideal gas law, \( P = \frac{nRT}{V} \), we can express work done in the form of an equation involving the natural logarithm (ln) of the volume ratio. Speaking of which, let's include the formula for work here:
\( W = -nRT \ln\left(\frac{V_f}{V_i}\right) \)
where \(V_i\) and \(V_f\) are the initial and final volumes, respectively. Notice the negative sign, which indicates that work is done by the system on the environment. The elegant way this law ties together the variables at play makes it a powerful tool to calculate work done in process A through E from the original exercise.

Isothermal Expansion: Understanding Work and Gas Behavior

Isothermal expansion, as the name suggests, is an expansion that happens at a constant temperature. This concept is critical in understanding processes A through E in the given exercise, as it provides the framework to determine which expansion does the most work. What is unique about isothermal processes is that the internal energy of an ideal gas does not change because the temperature remains constant. Thus, the work done by the gas is equal to the heat absorbed, based on the first law of thermodynamics.

In the specific context of the ideal gas law, isothermal expansion means that while the gas expands, its pressure decreases in such a way that the product of pressure and volume remains constant. This behavior is quantified by the natural logarithm involved in the work calculation mentioned earlier. This logarithmic relationship between volumes signifies that during an isothermal expansion, even if the change in volume is the same, the amount of work done can differ depending on the initial and final volumes—as shown by comparing processes B and E from our problem.

Molar Specific Heat at Constant Volume

Understanding molar specific heat is essential for grasping how gases store and transfer energy. Molar specific heat at constant volume (\(C_V\)) is the amount of heat required to raise the temperature of one mole of a gas by one degree Celsius (or one Kelvin), with the condition that the volume remains constant. For an ideal gas, the molar specific heat is a constant, meaning it does not change with temperature. This is not directly involved in the work calculations of our original problem but plays an important role in understanding the overall energy changes in the gas.

For an ideal gas undergoing an isothermal process, however, it is interesting to note that the change in internal energy (\(\Delta U\)) is zero because the temperature is constant. Thus, molar specific heat does not directly enter into our calculations here, but it lurks in the background, influencing the energy dynamics of the system. To connect this with our exercise, it's this energy perspective that would differ if the problem involved processes at constant pressure or other conditions where the specific heats at constant volume or constant pressure become relevant for calculating the heat added or lost.