Question

A gas undergoes two processes. In the first, the volume remains constant at 0.200 m\(^3\) and the pressure increases from 2.00 \(\times\) 10\(^5\) Pa to 5.00 \(\times\) 10\(^5\) Pa. The second process is a compression to a volume of 0.120 m\(^3\) at a constant pressure of 5.00 \(\times\) 10\(^5\) Pa. (a) In a pV-diagram, show both processes. (b) Find the total work done by the gas during both processes.

Short answer

The total work done by the gas is -4.00 x 10^4 J. The processes are represented on a pV-diagram as a vertical line (constant volume) and a horizontal line (constant pressure).

Step-by-step solution

Understanding the First Process

In the first process, the volume of the gas stays constant at 0.200 m\(^3\), while the pressure increases from 2.00 \times 10^5 \text{ Pa} to 5.00 \times 10^5 \text{ Pa}. This is an isochoric process, meaning no work is done by the gas (work is zero) as work done is given by \( W = P\Delta V \) and \( \Delta V = 0 \).

Understanding the Second Process

In the second process, the gas is compressed from a volume of 0.200 m\(^3\) to 0.120 m\(^3\) at a constant pressure of 5.00 \times 10^5 \text{ Pa}. This is an isobaric process, and work done during this process can be calculated using \( W = P \Delta V \).

Calculating Work Done in the Second Process

For the second process, we calculate the change in volume \( \Delta V = V_f - V_i = 0.120 \text{ m}^3 - 0.200 \text{ m}^3 = -0.080 \text{ m}^3 \). Note that \( \Delta V \) is negative because the volume is decreasing. The work done is \( W = P \Delta V = 5.00 \times 10^5 \text{ Pa} \times (-0.080 \text{ m}^3) = -4.00 \times 10^4 \text{ J} \).

Summing Up the Total Work Done by the Gas

The total work done by the gas is the sum of the work from both processes. Since the first process involved no work, the total work done is simply the work from the second process: \( W_{total} = 0 + (-4.00 \times 10^4 \text{ J}) = -4.00 \times 10^4 \text{ J} \).

Representing on a pV-Diagram

For the pV-diagram, the first process is a vertical line upwards (since volume is constant and pressure increases), starting from (0.200 m\(^3\), 2.00 \times 10^5 Pa) to (0.200 m\(^3\), 5.00 \times 10^5 Pa). The second process is a horizontal line leftwards (since pressure is constant and volume decreases) from (0.200 m\(^3\), 5.00 \times 10^5 Pa) to (0.120 m\(^3\), 5.00 \times 10^5 Pa).

Key concepts

Isochoric Process

An isochoric process is a fascinating concept in thermodynamics where the volume of a gas remains constant. During this process, no matter how much the pressure changes, the volume stays the same.
For this reason, we also call it a constant-volume process. One of the critical points to understand about an isochoric process is that no work is done by the gas.
This is because work is calculated based on the formula:

• \( W = P \Delta V \)
• Where \( W \) is work, \( P \) is pressure, and \( \Delta V \) is the change in volume.

Since the volume does not change \(( \Delta V = 0 )\), the work done is zero. This scenario is common in physics problems involving rigid containers. Despite no work being done, energy can still be transferred as heat, leading to changes in other properties like pressure or temperature. In a practical scenario, it would be like trying to squeeze a gas inside an unyielding metal can. The can maintains its shape, but the gas molecules inside feel more or less cramped, altering the pressure.

Isobaric Process

An isobaric process occurs when a gas changes its volume while maintaining a constant pressure. This scenario is quite common and differs from the isochoric process, where the volume stays the same.
When the volume of a gas expands or compresses under constant pressure, work is done. The work done in this process can be calculated using the formula:

• \( W = P \Delta V \)
• Where \( W \) is work, \( P \) is the constant pressure, and \( \Delta V \) is the change in volume.

During an isobaric process, the change in volume \( \Delta V \) could be positive or negative:

• If the gas expands, \( \Delta V > 0 \), resulting in positive work done by the gas.
• If it is compressed, \( \Delta V < 0 \), leading to negative work, indicating that work is done on the gas.

An example of an isobaric process can be seen in everyday situations, such as an inflated balloon heated under sunlight, where the balloon expands while the pressure inside remains fairly constant.

pV-Diagram

The pV-diagram is a powerful visual tool in thermodynamics. It provides a graphical representation of how the pressure and volume of a gas interact during various processes. Each line or curve on a pV-diagram represents a distinct thermodynamic process.
Such diagrams are instrumental for analyzing the work done by or on a gas and understanding different process types.
In this diagram:

• A vertical line represents an isochoric process because the volume doesn't change, but the pressure does.
• A horizontal line indicates an isobaric process, where the pressure stays constant while the volume changes.

For example, in the exercise, the first process is depicted as a vertical line because the volume is constant, while the pressure increases. This line moves vertically upward.
In contrast, the second process appears as a horizontal line, moving leftward, showing a decrease in volume at constant pressure.
Understanding pV-diagrams is crucial for anyone studying thermodynamics, as they allow quick insights into the nature of processes a gas undergoes.