Calculating work done by a gas

Click For Summary
SUMMARY

The discussion focuses on calculating the work done by a gas using the provided initial and final pressures (Pi = 101325 Pa, Pf = 357000 Pa), initial and final volumes (Vi = 0.35 m³, Vf = 0.281 m³), and temperature (T = 848.615 K). The user attempts to derive the constant 'a' using the equation P = aV + Pnot, resulting in a value of 909875.4448 Pa/m³. However, the user expresses uncertainty regarding the calculation of work, indicating a need for clarification on the correct methodology.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically the work-energy theorem.
  • Familiarity with the ideal gas law and its applications.
  • Knowledge of pressure-volume work calculations in thermodynamics.
  • Proficiency in using LaTeX for formatting mathematical equations.
NEXT STEPS
  • Review the derivation of work done by a gas in non-constant pressure scenarios.
  • Study the application of the ideal gas law in calculating work done during expansion or compression.
  • Learn about the integration of pressure with respect to volume to find work done in thermodynamic processes.
  • Explore LaTeX formatting techniques for presenting complex equations clearly.
USEFUL FOR

This discussion is beneficial for students and professionals in thermodynamics, mechanical engineers, and anyone involved in calculating work done by gases in various processes.

hraghav
Messages
48
Reaction score
5
Homework Statement
An isolated container of a monatomic ideal gas is compressed in such a way that the pressure is linearly proportional to the volume. The pressure can be parametrized as P=αV+Pnot. The gas is initially at atmospheric pressure and room temperature 300K. The initial volume of the container is 0.35m^3. After compression, the gas is allowed to return to room temperature again at a constant volume. What is the temperature inside the container after compression if the final volume is 0.281m^3 and pressure is 357000Pa?
What is the work done by the gas in the container during compression?
What is the pressure of the gas inside the container when the gas has reached room temperature?
What is the change in entropy of the gas as it returns to room temperature?
Relevant Equations
PV = nRT
P = aV + Pnot
W = integral from Vi to Vf of P(V)dv
I have found the answers for T = 848.615K, P = 126137.7705 Pa and change in S = -184.27008 J/K. But my answer for work is not correct and I am not sure where I am making an error. Could someone please help me out with how to calculate work? My steps for work is :

We are given:
Pi = 101325 Pa
Pf = 357000 Pa
Ti = 300 K
Vi = 0.35 m^3
Vf = 0.281 m^3

Using the given equation P = aV + Pnot we can find 'a' which is (357000 - 101325)/(0.281) = 909875.4448 Pa/m^3

For work done we have:
1719620469449.png
 
Last edited:
Physics news on Phys.org
See the link below for the LaTeX guide. I doubt anyone will want to try to decipher your work in the format you're using.
 
Mister T said:
See the link below for the LaTeX guide. I doubt anyone will want to try to decipher your work in the format you're using.
I did update the question thanks
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 4 ·
Replies
4
Views
1K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
7K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 15 ·
Replies
15
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 10 ·
Replies
10
Views
2K