Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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

Expert verified
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

01

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 \( \Delta V = 0 \).
02

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 \( W = P \Delta V \).
03

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 \( W = P \Delta V = 5.00 \times 10^5 \text{ Pa} \times (-0.080 \text{ m}^3) = -4.00 \times 10^4 \text{ J} \).
04

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: \( W_{total} = 0 + (-4.00 \times 10^4 \text{ J}) = -4.00 \times 10^4 \text{ J} \).
05

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).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A gas in a cylinder expands from a volume of 0.110 m\(^3\) to 0.320 m\(^3\). Heat flows into the gas just rapidly enough to keep the pressure constant at 1.65 \(\times\) 10\(^5\) Pa during the expansion. The total heat added is 1.15 \(\times\) 10\(^5\) J. (a) Find the work done by the gas. (b) Find the change in internal energy of the gas. (c) Does it matter whether the gas is ideal? Why or why not?

A cylinder with a piston contains 0.150 mol of nitrogen at 1.80 \(\times\) 10\(^5\) Pa and 300 K. The nitrogen may be treated as an ideal gas. The gas is first compressed isobarically to half its original volume. It then expands adiabatically back to its original volume, and finally it is heated isochorically to its original pressure. (a) Show the series of processes in a \(pV\)-diagram. (b) Compute the temperatures at the beginning and end of the adiabatic expansion. (c) Compute the minimum pressure.

Two moles of an ideal gas are compressed in a cylinder at a constant temperature of 65.0\(^\circ\)C until the original pressure has tripled. (a) Sketch a \(pV\)-diagram for this process. (b) Calculate the amount of work done.

Five moles of monatomic ideal gas have initial pressure 2.50 \(\times\) 10\(^3\) Pa and initial volume 2.10 m\(^3\). While undergoing an adiabatic expansion, the gas does 1480 J of work. What is the final pressure of the gas after the expansion?

Three moles of argon gas (assumed to be an ideal gas) originally at 1.50 \(\times\) 10\(^4\) Pa and a volume of 0.0280 m\(^3\) are first heated and expanded at constant pressure to a volume of 0.0435 m\(^3\), then heated at constant volume until the pressure reaches 3.50 \(\times\) 10\(^4\) Pa, then cooled and compressed at constant pressure until the volume is again 0.0280 m\(^3\), and finally cooled at constant volume until the pressure drops to its original value of 1.50 \(\times\) 10\(^4\) Pa. (a) Draw the \(pV\)-diagram for this cycle. (b) Calculate the total work done by (or on) the gas during the cycle. (c) Calculate the net heat exchanged with the surroundings. Does the gas gain or lose heat overall?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free