Question

Two moles of an ideal monatomic gas go through the cycle \(abc\). For the complete cycle, 800 J of heat flows out of the gas. Process \(ab\) is at constant pressure, and process \(bc\) is at constant volume. States \(a\) and \(b\) have temperatures \(T_a\) = 200 K and \(T_b\) = 300 K. (a) Sketch the \(pV\)-diagram for the cycle. (b) What is the work \(W\) for the process \(ca\)?

Short answer

Work for process ca is -800 J, assuming the net energy balance aligns with the heat removed.

Step-by-step solution

Understand the Cycle

The gas undergoes a thermodynamic cycle from state \(a\) to \(b\) at constant pressure, then to \(c\) at constant volume, and finally back to \(a\). Since the process \(ab\) is isobaric and \(bc\) is isochoric, the process \(ca\) must be at constant temperature because it forms a closed cycle.

Draw the pV Diagram

Identify that in a \(pV\) diagram, constant pressure is depicted by a horizontal line, and constant volume by a vertical line. The cycle will look like a rectangle, moving from \(a\) to \(b\) horizontally, then vertically from \(b\) to \(c\), and finally back to \(a\) diagonally, due to the isothermal process.

Calculate Work in Process CA

The work done during an isothermal process can be calculated using the formula: \[ W = nRT \ln\left(\frac{V_a}{V_c}\right) \]Given that the process returns to the initial state \(a\), \(T_a = 200 \text{ K}\). Since exact volumes or initial pressure are not given, consider using the fact that overall heat lost during the cycle is known and work done in a full cycle can be related to heat flow.

Relate Heat and Work

For the entire cycle, using the first law of thermodynamics: \( Q = \Delta U + W \). Since \(\Delta U = 0\) for a complete cycle, \( Q = W \). Given \( 800 \text{ J} \) heat flows out, \(W_{total} = -800 \text{ J}\). This is the net work done on the gas for the cycle, with \(W_{ca}\) being part of this.

Solve for W_ca

The total work \(W_{total}\) is given by the work done in individual processes. Since only \(W_{ab}\) and \(W_{ca}\) involve changing volume, for \(n = 2\), since \(ab\) is horizontal the work done is \[ W_{ab} = P(V_b - V_a) \]During \(ca\), we assume isothermal conditions. Without specific volume constraints, assume all other energy changes are internal so: \[ W_{ca} = -W_{ab} \] Hence, considering energy symmetry in an ideal situation, solve for the desired unknown using the cycle properties. Since direction determines sign, \(W_{ca} = -800 \text{ J}\).

Key concepts

Ideal Gas Processes

An ideal gas process involves a gas that perfectly follows the laws of thermodynamics without any deviation from the ideal gas law, which is articulated as \(PV = nRT\). Here, \(P\) is pressure, \(V\) is volume, \(n\) is the amount of substance in moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin.
In the context of the exercise, the gas undergoes several types of processes, including constant pressure (isobaric) and constant volume (isochoric). Each of these processes results in different changes to the internal energy and work done by the system.
These processes are part of a cycle, which is pivotal in thermodynamics as it allows us to understand how heat and work are exchanged in a closed system. In any complete cycle, the change in internal energy (\(\Delta U\)) is zero, leading to a clear relationship between heat exchanged (\(Q\)) and work done (\(W\)). This relationship is critical for solving problems related to cycles and comprehending the efficiency of thermodynamic systems.

Isothermal Process

An isothermal process is a type of thermodynamic process where the temperature remains constant throughout. It is quite significant in cycles, particularly because it helps understand processes where energy exchange occurs without changing temperature.
In an isothermal process involving an ideal gas, the internal energy stays constant since it is a function of temperature, which does not change. Therefore, the work done (\(W\)) by or on the gas is equal to the heat exchanged (\(Q\)) with the surroundings:
\[ Q = W \]
The formula to calculate work done during an isothermal expansion or compression is:\
\[ W = nRT \ln\left(\frac{V_f}{V_i}\right) \]
where \(V_f\) and \(V_i\) are the final and initial volumes, respectively. This formula derives from the integral of pressure with respect to volume under constant temperature conditions. This understanding is crucial when analyzing processes like \(ca\) in the exercise, where such conditions apply.

pV Diagram

The pV (pressure-volume) diagram is an essential tool in thermodynamics used to visualize processes and cycles that an ideal gas undergoes. It is a graphical representation of how the pressure of the gas relates to its volume during different processes.
In this exercise, the cycle involves moving from one state to another under constant pressure and volume, forming a series of straight lines and thus creating a rectangle or loop on the diagram.
- **Isobaric Process (ab):** Represented by a horizontal line, showing constant pressure with changing volume.
- **Isochoric Process (bc):** Depicted as a vertical line, illustrating constant volume while pressure changes.
The final part, process \(ca\), typically draws a curve (often an isothermal line), indicating simultaneous changes in pressure and volume at a constant temperature.
Using such diagrams makes it easier to calculate the work done during each process, as the area under the curve (or within the loop) invariably relates to the work performed by or on the system. Understanding the pV diagram equips you with deeper insights into various thermodynamic processes and how systems evolve from one state to another.