How to Calculate Final Volume and Pressure in Isothermal Process"

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SUMMARY

The discussion focuses on calculating the final volume (V2) and final pressure (P2) of an ideal monatomic gas during an isothermal process, where the temperature remains constant (ΔT = 0). The work done by the gas (W) is related to the initial volume (V1) and initial pressure (P1) using the equation W = nRT ln(V2/V1). Participants confirm that the algebraic manipulation to isolate V2 involves taking the anti-logarithm of both sides, leading to the formula V2 = V1 * e^(W/nRT). This method provides a clear path to determine the final state of the gas.

PREREQUISITES
  • Understanding of ideal gas laws, specifically the equation P1V1 = nRT
  • Familiarity with isothermal processes and their characteristics
  • Basic calculus, particularly logarithmic and exponential functions
  • Knowledge of thermodynamic work calculations in gas systems
NEXT STEPS
  • Study the derivation of the ideal gas law and its applications
  • Learn about isothermal processes in thermodynamics
  • Explore logarithmic and exponential equations in algebra
  • Investigate the implications of work done in thermodynamic systems
USEFUL FOR

Students and professionals in physics, engineering, and chemistry who are studying thermodynamics, particularly those interested in gas behavior during isothermal processes.

mrmonkah
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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
 
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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
 
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?

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