Thermodynamics specific volume, heat, & ideal gases

In summary, the problems attached involve finding the final phase of a liquid-vapor mixture in two rigid tanks, calculating the process on a temperature-specific volume diagram, and finding three points on a P-V diagram where the ratio (V/T) has the same value. The final phase in both tanks is vapor, determined by the boiling point of the mixture. The process on the temperature-specific volume diagram will appear as a straight line with a slope corresponding to the specific volume at each phase. For the P-V diagram, any three points along the linear portion will have the same ratio of (V/T) due to the constant temperature. Mathematical and physical justifications should be provided for all solutions.
  • #1
JON123
6
0

Homework Statement



I attached the problems. The first one is D) at the top of the page parts a) and b). The second one is E) at the top of the second attachment.

1) The rigid tanks shown below have volumes of .4m^3 and .004^3 respectively and each contains a water liquid-vapor mixture of mass 2Kg at 50 degrees Celsius. Heat is applied in each tank until the mixture becomes one phase. Note the tanks do not affect each other.

2) On the P-V diagram shown below, for an ideal gas, show three points for which the ratio (V/T) has the same value, and provide full justification.

Homework Equations





The Attempt at a Solution




1) The rigid tanks shown below have volumes of .4m^3 and .004^3 respectively and each contains a water liquid-vapor mixture of mass 2Kg at 50 degrees Celsius. Heat is applied in each tank until the mixture becomes one phase. Note the tanks do not affect each other.

a) Find the final phase of water in each tank providing full justification.

I didn’t know how I could solve this mathematically because I don’t have an initial pressure, final pressure, or final temperature. I would just say that the liquid-vapor mixture is heated till it becomes a single phase. The next phase after liquid is a vapor. Therefore the final phase in both tanks is vapor. Would this be valid for my justification in this problem?

b) Show the process in a temperature-specific volume diagram in each case.

I would say each case has a similar graph. The volume is constant so it would just be a straight line, one at .4m^3 and the other at .004m^3, but in specific volume. Would this be correct?

2) On the P-V diagram shown below, for an ideal gas, show three points for which the ratio (V/T) has the same value, and provide full justification.

- I have no idea where to start on this. I know at a constant temperature P1V1=P2V2 so a graph would be linier and downward sloping?
 

Attachments

  • scan0003.jpg
    scan0003.jpg
    18.8 KB · Views: 444
  • scan00014.jpg
    scan00014.jpg
    14.9 KB · Views: 424
Physics news on Phys.org
  • #2


Thank you for attaching the problems and providing some of your attempted solutions. I would like to offer some guidance and suggestions for solving these problems.

1) For part a), you are correct in saying that the final phase in both tanks is vapor. This is because when a liquid-vapor mixture is heated, it will eventually reach its boiling point and evaporate completely into vapor. However, you can also provide some mathematical justification for this. Since the tanks do not affect each other, the mass of the mixture in each tank remains constant. This means that the specific volume (volume per unit mass) of the mixture will also remain constant. Therefore, the final phase will be determined by the temperature at which the mixture reaches its boiling point and completely vaporizes. You can use the Clausius-Clapeyron equation to calculate the boiling point of the mixture at the given mass and temperature. Once you have the boiling point, you can compare it to the given temperature of 50 degrees Celsius to determine the final phase.

For part b), your reasoning is correct. Since the volume is constant, the process will appear as a straight line on a temperature-specific volume diagram. However, it is important to note that the specific volume will change as the mixture goes from liquid to vapor. Therefore, the straight line on the graph will have a slope that corresponds to the specific volume of the mixture at each phase.

2) For this problem, you are correct in saying that the graph will be linear and downward sloping at a constant temperature. This is because, as you mentioned, at a constant temperature the product of pressure and volume will remain constant for an ideal gas. To find three points on the graph where the ratio (V/T) has the same value, you can choose any three points along the linear portion of the graph and calculate the ratio at each point. Since the temperature is constant, the ratio will also remain constant at these points.

I hope this helps you in solving these problems. Remember to always provide both mathematical and physical justifications for your solutions. Good luck!
 

1. What is thermodynamics?

Thermodynamics is the branch of science that deals with the study of energy and its transformations, particularly in relation to macroscopic systems.

2. What is specific volume?

Specific volume is defined as the volume occupied by a unit mass of a substance. It is commonly denoted by the symbol v and is expressed in units of m³/kg.

3. What is heat in thermodynamics?

In thermodynamics, heat is defined as the transfer of energy from one system to another as a result of a temperature difference between them. It is a form of energy that can be converted into other forms, such as mechanical or electrical energy.

4. What is an ideal gas?

An ideal gas is a theoretical gas that follows the gas laws at all conditions of temperature and pressure. It is composed of point particles that have no volume and do not interact with each other.

5. How is the specific volume of an ideal gas related to its temperature and pressure?

According to the ideal gas law, the specific volume of an ideal gas is directly proportional to its temperature and inversely proportional to its pressure. This means that as temperature increases, the specific volume also increases, and as pressure increases, the specific volume decreases.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
2
Views
7K
Replies
18
Views
2K
  • Introductory Physics Homework Help
Replies
16
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
4K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
7K
  • Engineering and Comp Sci Homework Help
Replies
10
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
2
Views
1K
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
728
  • Introductory Physics Homework Help
Replies
4
Views
533
Back
Top