How to Calculate Final Volume and Pressure in Isothermal Process"

Click For Summary

Homework Help Overview

The discussion revolves around calculating the final volume and pressure of an ideal monatomic gas undergoing an isothermal process, where the temperature remains constant (ΔT = 0). The original poster presents initial conditions, including initial volume (V1) and pressure (P1), along with the work done by the gas (W).

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between work done and volume change, referencing the equation W = nRT ln(V2/V1). There are questions about the algebraic manipulation required to isolate V2, with some participants expressing uncertainty about logarithmic properties.

Discussion Status

Some participants have provided guidance on the algebraic steps needed to solve for V2, while others are still exploring their understanding of the logarithmic relationships involved. The discussion reflects a mix of confidence and uncertainty regarding the mathematical processes.

Contextual Notes

Participants note challenges with algebra and calculus, indicating a focus on improving their mathematical skills in the context of the problem. There is an acknowledgment of the importance of self-guided learning in tackling the question.

mrmonkah
Messages
24
Reaction score
0
Hey all,

If i have an isothermal process of an ideal monotomic gas, IE. deltaT = 0 and i have an initial volume (V1) and an initial pressure (P1). Then I am given the work done BY the gas (W = -### as it is the environment providing the energy)

How do i calculate the final volume and the final pressure? I've been toying with different equations of state and like the look of w = nRT ln(V2/V1)... it is probably just my algebra that needs to improve in order for me to re-arrange for V2... am i on the right track?

Cheers in advance.

Dan
 
Physics news on Phys.org
P1V1=nRT

so you have W=P1V1ln(V2/V1)

you can get V2 now I believe.
 
mrmonkah said:
Hey all,

If i have an isothermal process of an ideal monotomic gas, IE. deltaT = 0 and i have an initial volume (V1) and an initial pressure (P1). Then I am given the work done BY the gas (W = -### as it is the environment providing the energy)

How do i calculate the final volume and the final pressure? I've been toying with different equations of state and like the look of w = nRT ln(V2/V1)... it is probably just my algebra that needs to improve in order for me to re-arrange for V2... am i on the right track?

Cheers in advance.

Dan
If you are given the work, you can find the volume change using:

W= \int PdV = nRT\int dV/V = nRTln(V2/V1)

as you have suggested. I am not sure why you are hesitating or unsure.

AM
 
okay cool, so the equation i have is right, I am just un-sure about re-arranging for V2 now, what with the ln(v2/v1) business... am i right in thinking that i can take ln(v2) - ln(v1) as being equal to ln(v2/v1)?

Like i say its my calculus/algebra that is poor at the moment.
 
mrmonkah said:
okay cool, so the equation i have is right, I am just un-sure about re-arranging for V2 now, what with the ln(v2/v1) business... am i right in thinking that i can take ln(v2) - ln(v1) as being equal to ln(v2/v1)?

Like i say its my calculus/algebra that is poor at the moment.
You don't need to do that. Just take the anti-log of both sides:

e^{W/nRT} = \frac{V_2}{V_1}

You are given W, and nRT so this is just a matter doing the algebra to calculate V2

AM
 
Hi Andrew,

Thank you very much for the pointer, i was slowly coming to that conclusion i think once i knuckled down and did some revision. Now i can hopefully conquer the rest of the question.

Ill post any issues i have, but hopefully ill get there under my own steam. Best way to learn right!

Cheers,

Dan
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
1K
Replies
21
Views
2K
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
14
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
943