Question

Consider 2.00 moles of an ideal gas that are taken from state \(A\) \(\left(P_{A}=2.00 \mathrm{atm}, V_{A}=10.0 \mathrm{L}\right)\) to state \(B\left(P_{B}=1.00 \mathrm{atm}, V_{B}=\right.\) 30.0 \(\mathrm{L}\) ) by two different pathways: These pathways are summarized on the following graph of \(P\) versus \(V :\) Calculate the work (in units of \(\mathrm{J} )\) associated with the two path- ways. Is work a state function? Explain.

Short answer

The work done in Pathway 1, which is an isothermal expansion, is approximately -676.92 J. In Pathway 2, which consists of an isothermal compression followed by an isothermal expansion, the work done is approximately -338.46 J. Since the work done is different for the two pathways despite having the same initial and final states, work is not a state function.

Step-by-step solution

Pathway 1: Isothermal Expansion

In the first pathway, the gas undergoes an isothermal expansion from state A to state B. Therefore, we can use the formula mentioned above: $$ W_1 = -P_{A} V_{A} \ln \frac{V_{B}}{V_{A}} $$ Plugging in the given values into the formula: $$ W_1 = -\left(2.00\,\text{atm}\right)\left(10.0\,\text{L}\right) \ln \frac{30.0\,\text{L}}{10.0\,\text{L}} $$ To convert the work from L.atm to J, we can use the conversion factor: 1 L.atm ≈ 101.3 J. Therefore, we have: $$ W_1 = -\left(2.00\,\text{atm}\right)\left(10.0\,\text{L}\right) \ln 3 \times 101.3\,\frac{\text{J}}{\text{L.atm}} $$ After calculation, we get: $$ W_1 \approx -676.92\,\text{J} $$

Pathway 2: Isothermal Compression and Expansion

In the second pathway, the gas first undergoes an isothermal compression at constant volume, followed by an isothermal expansion at constant pressure. Since the volume does not change during the first part of the process, no work is done. Therefore, we only need to calculate the work done during the second part, where the gas expands at constant pressure: $$ W_2 = -P_{B} V_{B} \ln \frac{V_{B}}{V_{A}} $$ Plugging in the given values into the formula: $$ W_2 = -\left(1.00\,\text{atm}\right)\left(30.0\,\text{L}\right) \ln \frac{30.0\,\text{L}}{10.\,\text{L}} $$ Multiplying by the conversion factor from L.atm to J: $$ W_2 = -\left(1.00\,\text{atm}\right)\left(30.0\,\text{L}\right) \ln 3 \times 101.3\,\frac{\text{J}}{\text{L.atm}} $$ After calculation, we get: $$ W_2 \approx -338.46\,\text{J} $$

Is work a state function?

A state function is a property that depends only on the initial and final states of a system, and not on the path taken to reach those states. Since the work done is different for the two pathways, even though the initial and final states are the same, work is not a state function.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a central concept in thermodynamics and describes the relationship between pressure, volume, and temperature for an ideal gas. The equation is expressed as \[ PV = nRT \] where:

• \( P \) is the pressure of the gas
• \( V \) is the volume
• \( n \) is the number of moles of gas
• \( R \) is the universal gas constant (approximately 8.314 J/mol·K)
• \( T \) is the temperature in Kelvin

This equation provides a simple way to understand how gases behave when they are heated, cooled, compressed, or expanded. It's especially useful in simplifying calculations in both chemistry and physics when dealing with gases in various conditions. Understanding the Ideal Gas Law helps in analyzing processes like isothermal changes described in the given exercise.

Isothermal Processes

In an isothermal process, the temperature of the system remains constant while some other changes occur, such as changes in pressure and volume. During these processes, the internal energy of the system doesn't change, because the temperature remains the same.For an ideal gas undergoing an isothermal process, the work done can be calculated using the formula: \[ W = -nRT \ln \left( \frac{V_f}{V_i} \right) \] where:

• \( V_f \) is the final volume
• \( V_i \) is the initial volume

Isothermal processes are especially significant in understanding how energy transfer occurs when gases exchange heat with their surroundings. The process described in the exercise exemplifies this, as the gas expands or compresses while maintaining constant temperature, making isothermal processes a foundational concept in thermodynamics.

Work in Thermodynamics

Work is a critical concept in thermodynamics, representing the energy transferred by a system to its surroundings due to volume changes. In the context of gases, work is done during expansion and compression processes. When a gas expands, it does work on its surroundings and when it compresses, work is done on the gas.Work done during an expansion or compression can be evaluated as: \[ W = -\int P \, dV \] In simpler terms for isothermal processes, where pressure changes but temperature remains constant, the work formula becomes: \[ W = -P V \ln \left( \frac{V_f}{V_i} \right) \]This equation shows that work depends on the path taken, as it involves integrating the pressure over volume changes. This path dependency explains why, in the exercise, different works are calculated for different pathways even between the same states A and B, showing that work is inherently a process-dependent quantity.

State Functions

State functions are properties that depend only on the initial and final states of a system, not on how the system transitioned between these states. Common state functions include enthalpy, internal energy, and entropy. A great way to think about state functions is to compare them to the difference in elevation between two points. No matter what path is taken to move from one point to another, the difference in elevation is the same. In contrast, the actual path influences process-dependent quantities, like work and heat. In the context of the exercise, it's demonstrated that work is not a state function because the amount of work done differs depending on the pathway followed from state A to state B. While the initial and final states in terms of pressure and volume are the same, the different pathways yield different work values, underscoring work's path-dependent nature.