Question

Consider \(5.5 \mathrm{~L}\) of a gas at a pressure of \(3.0 \mathrm{~atm}\) in a cylinder with a movable piston. The external pressure is changed so that the volume changes to \(10.5 \mathrm{~L}\). a. Calculate the work done, and indicate the correct sign. b. Use the preceding data but consider the process to occur in two steps. At the end of the first step, the volume is \(7.0 \mathrm{~L}\). The second step results in a final volume of \(10.5\) L. Calculate the work done, and indicate the correct sign. c. Calculate the work done if after the first step the volume is \(8.0 \mathrm{~L}\) and the second step leads to a volume of \(10.5 \mathrm{~L}\). Does the work differ from that in part b? Explain.

Short answer

a. The work done during the expansion is \(-15.0 \mathrm{~L\cdot atm}\), which indicates that work was done by the gas. b. The total work done in two steps is \(-15 \mathrm{~L\cdot atm}\), the same as in part a, and it is also done by the gas. c. The total work done with intermediate volumes of \(8.0 \mathrm{~L}\) and \(10.5 \mathrm{~L}\) is \(-15 \mathrm{~L\cdot atm}\), the same as in parts a and b, showing that the total work done only depends on the initial and final volumes, provided the external pressure remains constant during the process.

Step-by-step solution

a. Calculate the work done, and indicate the correct sign.

First, we need to determine the change in volume during this process. The initial volume is 5.5 L, and the final volume is 10.5 L. ΔV = V_final - V_initial = 10.5 L - 5.5 L = 5.0 L Now, we can use the formula for work done during an isobaric process: W = -P_ext * ΔV W = -(3.0 atm) * (5.0 L) W = -15.0 L*atm Since the work done is negative, this means that work was done by the gas (it expanded).

b. Calculate the work done when the volume changes in two steps.

Here, the process occurs in two steps. We'll need to calculate the work done during each step and then sum these values. Step 1: ΔV1 = V_final1 - V_initial = 7.0 L - 5.5 L = 1.5 L W1 = -P_ext * ΔV1 W1 = -(3.0 atm) * (1.5 L) W1 = -4.5 L*atm Step 2: ΔV2 = V_final2 - V_initial2 = 10.5 L - 7.0 L = 3.5 L W2 = -P_ext * ΔV2 W2 = -(3.0 atm) * (3.5 L) W2 = -10.5 L*atm Total work done, W_total = W1 + W2 = -4.5 L*atm - 10.5 L*atm = -15 L*atm The total work done is the same as in part a, and it is also done by the gas (negative sign).

c. Calculate the work done if after the first step the volume is 8.0 L and the second step leads to a volume of 10.5 L.

Again, we need to calculate the work done during each step and sum the values. Step 1: ΔV1 = V_final1 - V_initial = 8.0 L - 5.5 L = 2.5 L W1 = -P_ext * ΔV1 W1 = -(3.0 atm) * (2.5 L) W1 = -7.5 L*atm Step 2: ΔV2 = V_final2 - V_initial2 = 10.5 L - 8.0 L = 2.5 L W2 = -P_ext * ΔV2 W2 = -(3.0 atm) * (2.5 L) W2 = -7.5 L*atm Total work done, W_total = W1 + W2 = -7.5 L*atm - 7.5 L*atm = -15 L*atm The total work done is the same as in part b (and part a), and it is also done by the gas (the negative sign indicates this). This shows us that, despite the intermediate steps, the total work done only depends on the initial and final volumes, as long as the external pressure remains constant during the process.

Key concepts

Isobaric Process

Understanding the isobaric process is essential to comprehend the behavior of gases under constant pressure conditions. An isobaric process is a thermodynamic process that occurs at a constant pressure. This constraint simplifies various calculations, as demonstrated in the exercise, where a gas inside a cylinder with a movable piston expands from 5.5 liters to 10.5 liters against a constant external pressure.

In real-world applications, heating a gas at a constant pressure can result in isobaric expansion, such as what occurs in a steam engine. In the exercise provided, whether the gas expansion from 5.5 to 10.5 liters happens in one or multiple steps, the total work done remains the same as long as the process stays isobaric, which fundamentally means that the pressure exerted by the surroundings does not change.

Pressure-Volume Work

Pressure-volume work, often referred to as P-V work, is an important concept in thermodynamics and represents the work done by or on a gas as it expands or contracts. The formula for calculating the work done by a gas during an isobaric process is given by the equation: \[W = -P_{\text{ext}} \times \triangle V\].

Where:

• \(W\) is the work done by the gas.
• \(P_{\text{ext}}\) is the external pressure.
• \( \triangle V\) is the change in volume of the gas.

The negative sign indicates that when a gas expands, it performs work on the surroundings. Conversely, if the gas is compressed, the surroundings do work on the gas, resulting in a positive value for work. In the exercise, this concept helped us calculate that the work done when the gas expanded from 5.5 liters to either 7.0 liters or 10.5 liters, is exactly the same in the end, confirming that the P-V work is dependent only on the initial and final states, and not the path taken between them.

Thermodynamics

Thermodynamics is the study of heat, work, and energy, and their transformations. Its principles govern the behavior of systems in terms of energy exchanges and the direction of processes. It consists of a set of four laws that describe the physical quantities that characterize the energetic state of a system.

In the context of the exercise, it shows us how a system—in this case, the gas in a cylinder—can undergo transformations that obey these laws. For instance, the exercise illustrates the first law of thermodynamics, which deals with the conservation of energy. No matter how the gas reaches the expanded volume of 10.5 liters—either directly or through a two-step process—the total work done remains the same because the energy change only depends on the end states. This is vital knowledge for students studying thermodynamics as it emphasizes the state function characteristics of certain thermodynamic properties, such as internal energy, reaffirming that the path does not impact the total work done or energy change in isobaric processes.