Adiabatic process in a monatomic ideal gas

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Homework Help Overview

The discussion revolves around a quasistatic adiabatic process involving a monatomic ideal gas, specifically focusing on the work done during a compression to half its initial volume. The original poster presents the relationship PV^(5/3) = constant and seeks to derive an expression for the work done on the gas.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the definition of work done on the gas and the implications of the adiabatic process. There are attempts to relate pressure and volume changes, and questions arise regarding the next steps in the calculation.

Discussion Status

The discussion is ongoing, with some participants providing hints and guidance on how to approach the problem, such as finding the final pressure and temperature. However, there is no explicit consensus on the method to proceed, and multiple interpretations of the problem are being explored.

Contextual Notes

Participants are navigating the constraints of the adiabatic process, including the condition that dQ = 0 and the need to express pressure as a function of volume for integration. There is also a mention of the relationship between internal energy and work done, which is central to the discussion.

Jenkz
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Homework Statement


In a quasistaic adiabatic process in a monatomic ideal gas PV^5/3 = constant [DO
NOT PROVE]. A monatomic ideal gas initially has a pressure of P0 and a volume of
V0. It undergoes a quasistatic adiabatic compression to half its initial volume. Show
that the work done on the gas is

W = 3/2 P0V0 ( 2^(2/3) - 1)

Homework Equations



dU= dQ + dW
dW= -p dV
V1= V0/2

The Attempt at a Solution


dU= dW as adiabatic procees means dQ=0

dU= 3/2 NKbT = -p dV

And I don't know what to do next.

3/2 NKbT= P0Vo[1-V1/V0]
 
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You have given the definition of the work done on the gas:

W=\int P dV

HINT: Now if you were given P as a function of V you could calculate the integral explicitly.
 
I'm sorry, I don't quite understand what I should do.
 
Use PV^{5/3}=\text{constant} to find the final P. Then find the temperature from the Pf and Vf. Use that to determine the change in internal energy, which as you have stated must equal the work done by the gas.

AM
 
Ok, thank you!
 

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