PV Diagrams & Work: Videos & Practice Problems

PV diagrams plot pressure (P) against volume (V) to visualize thermodynamic processes. For constant pressure, work done (W) can be calculated using the equation $w=p\mathrm{\Delta v}$, where $\mathrm{\Delta v}$ is the change in volume. The area under the curve on a PV diagram represents work done, with positive work for expansion and negative for compression, emphasizing the importance of direction in these processes.

Work and PV Diagrams

The ideal gas law and the first law of thermodynamics are foundational concepts in understanding how gases behave under various conditions. The ideal gas law relates pressure (P), volume (V), and temperature (T) of an ideal gas, while the first law of thermodynamics connects heat (Q), work (W), and internal energy (U) in a system undergoing a thermodynamic process. A key tool for visualizing these processes is the PV diagram, which plots pressure on the y-axis and volume on the x-axis, allowing for a graphical representation of how a gas transitions between states.

In a PV diagram, thermodynamic processes are represented as curves or lines. For example, when a gas expands at constant pressure, the process can be illustrated as a horizontal line. If a gas expands from a volume of 2 m³ to 5 m³ at a constant pressure of 100 kPa, the work done by the gas can be calculated using the formula:

$$ W = P \cdot \Delta V $$

Here, \( \Delta V \) is the change in volume, which in this case is \( 5 - 2 = 3 \) m³. Thus, the work done is:

$$ W = 100 \, \text{kPa} \cdot 3 \, \text{m}^3 = 300 \, \text{J} $$

This work can also be interpreted as the area under the process curve on the PV diagram, which in this case forms a rectangle with a base of 3 m³ and a height of 100 kPa, confirming that the area (and thus the work done) is 300 J.

In contrast, when a gas is compressed, the situation changes. For instance, if a gas compresses from 5 m³ to 2 m³ while the pressure increases from 100 kPa to 220 kPa, the process is represented by a line that slopes upwards on the PV diagram. Since the pressure is not constant, the work done cannot be calculated using the simple \( P \cdot \Delta V \) formula. Instead, the area under the curve must be calculated by breaking it down into simpler geometric shapes, such as a triangle and a rectangle.

The area of the triangle can be calculated as:

$$ A_1 = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 3 \cdot 120 = 180 \, \text{J} $$

However, since the gas is being compressed, the work done is negative, resulting in:

$$ W = -180 \, \text{J} $$

The area of the rectangle can be calculated as:

$$ A_2 = \text{base} \cdot \text{height} = -3 \cdot 100 = -300 \, \text{J} $$

Adding these two areas gives the total work done during the compression:

$$ W = -180 + (-300) = -480 \, \text{J} $$

This negative value indicates that work is done on the gas during compression. A crucial takeaway is that the direction of the process on the PV diagram significantly affects the sign of the work done: processes moving from left to right yield positive work, while those moving from right to left yield negative work. Understanding these principles is essential for analyzing thermodynamic systems effectively.

Finding Value of V on Axis

In this thermodynamic problem, we are tasked with determining the volume represented by the first tick mark on the volume axis, given a specific work done during a process. The work done is stated as \( W = 2 \times 10^5 \, \text{J} \). Since the pressure is not constant throughout the process, we cannot use the simple formula \( W = P \Delta V \). Instead, we need to relate the work done to the area under the curve of the pressure-volume (P-V) graph.

The area under the curve for a trapezoidal shape can be calculated using the formula:

A = \frac{1}{2} (b_1 + b_2) h

In this context, \( b_1 \) and \( b_2 \) represent the pressures at the two ends of the process, while \( h \) represents the change in volume. Here, we identify \( b_1 = 1 \times 10^5 \, \text{Pa} \) and \( b_2 = 3 \times 10^5 \, \text{Pa} \). The height of the trapezoid corresponds to the difference in volume, which is expressed as \( h = V_2 - V_1 = 2V_1 \), where \( V_1 \) is the volume at the first tick mark.

Substituting these values into the area formula gives us:

W = \frac{1}{2} (1 \times 10^5 + 3 \times 10^5) (2V_1)

Setting this equal to the work done, we have:

2 \times 10^5 = \frac{1}{2} (4 \times 10^5) (2V_1)

By simplifying, we find:

2 \times 10^5 = 4 \times 10^5 V_1

Dividing both sides by \( 2 \times 10^5 \) leads to:

1 = 2V_1

Thus, solving for \( V_1 \) yields:

V_1 = 0.5 \, \text{m}^3

This indicates that the first tick mark on the volume axis corresponds to \( 0.5 \, \text{m}^3 \). The subsequent tick marks would then represent \( 1 \, \text{m}^3 \) and \( 1.5 \, \text{m}^3 \) respectively. Verifying this by substituting back into the trapezoid area equation confirms the accuracy of our calculations, reaffirming that the work done matches the given value.

Here’s what students ask on this topic:

What is a PV diagram and how is it used in thermodynamics?

A PV diagram is a graphical representation that plots pressure (P) on the y-axis against volume (V) on the x-axis. It is used in thermodynamics to visualize the changes in a system, particularly gases, as they undergo various processes. These diagrams help in understanding how pressure and volume change during processes like expansion and compression. By analyzing the area under the curve in a PV diagram, one can calculate the work done by or on the system. For instance, the work done during a constant pressure process can be calculated using the equation $W=P\mathrm{\Delta V}$, where $\mathrm{\Delta V}$ is the change in volume.

How do you calculate work done in a PV diagram for a constant pressure process?

To calculate the work done in a PV diagram for a constant pressure process, you can use the equation $W=P\mathrm{\Delta V}$, where $P$ is the constant pressure and $\mathrm{\Delta V}$ is the change in volume. For example, if a gas expands from 2 m³ to 5 m³ at a constant pressure of 100 Pa, the work done is calculated as $W=100(5-2)=300 J$. The area under the curve in the PV diagram, which is a rectangle in this case, also represents the work done.

What does the area under the curve in a PV diagram represent?

The area under the curve in a PV diagram represents the work done by or on the system during a thermodynamic process. For a constant pressure process, this area is a rectangle, and the work done can be calculated using the equation $W=P\mathrm{\Delta V}$. For processes where pressure changes, the area might be more complex, such as a combination of rectangles and triangles. The sign of the work depends on the direction of the process: positive for expansion (left to right) and negative for compression (right to left).

How do you determine the direction of a process on a PV diagram?

The direction of a process on a PV diagram is indicated by arrows that show the transition from the initial to the final state. This is crucial because the direction affects the sign of the work done. If the process moves from left to right, it indicates expansion, and the work done is positive. Conversely, if the process moves from right to left, it indicates compression, and the work done is negative. Always ensure to mark the direction with arrows to avoid errors in calculations.

Can you use the equation W = PΔV for processes with changing pressure?

No, the equation $W=P\mathrm{\Delta V}$ is only valid for processes with constant pressure. For processes where the pressure changes, you need to calculate the work done by finding the area under the curve in the PV diagram. This might involve breaking the area into simpler shapes like rectangles and triangles and summing their areas. For example, if the pressure increases linearly, the area under the curve will be a combination of a rectangle and a triangle.