An ideal gas, and a block of copper, have equal volume of 0.5m^3

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Homework Help Overview

The discussion revolves around a problem involving an ideal gas and a block of copper, both having a volume of 0.5 m³ at 300 K and atmospheric pressure. The scenario describes a reversible and isothermal increase in pressure to 5 atm, prompting participants to explore the work done by both substances and the role of compressibility in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of the work done formula W = P dV and the ideal gas equation PV = nRT. Questions arise regarding how to link compressibility with work done, particularly for copper, and the implications of pressure changes on the materials involved.

Discussion Status

Some participants have made progress in calculating the work done for the ideal gas and are attempting to apply similar reasoning to the copper. There is an ongoing exploration of the concept of compressibility and its relevance to the problem. Participants are also questioning the relationship between atmospheric pressure and the pressure exerted by the copper.

Contextual Notes

Participants are navigating the complexities of the problem, including the definitions and implications of compressibility, as well as the potential discrepancies in calculated work done compared to standard answers. The discussion reflects a lack of consensus on certain assumptions, particularly regarding pressure definitions.

blade_090
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Homework Statement



An ideal gas, and a block of copper, have equal volume of 0.5m^3 at 300K and atmospheric pressure. The pressure on both is increased reversibly and isothermally to 5atm.
Find the work done of each if the compressibility of copper is 0.7X10^-6 atm-1

the atmospheric pressure the question said is 1atm?
I understand that i should used the work done formula W = P dV
bt i duno how to link the compressibility with work done
 
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Welcome to PF, blade_090! :smile:

Yes, (standard) atmospheric pressure is 1 atm.

And yes, work done is dW = P dV.

Do you have a formula for an ideal gas that links pressure to volume (and temperature)?
And what about a formula for compressibility that links pressure to volume (isothermally)?
 


so far what i know about ideal gas equation is
PV = nRT
so i have to find the volume change for the ideal gas
use it on W = P dV and i manage to find work done for ideal gas :)
however i still dun have any clue about work done on copper
and for compressibility:
K = (-1/v)(dV/dP)
im still confused about compressibility of copper...
what is the use of compressibility??
 
Last edited:


You can rewrite it as:
dW = PdV = -K Vo P dP.
Now integrate?

(We can treat V as a constant volume Vo, since it's nearly constant.)
 


ok i got ur point there...
is that any possible that copper has it own pressure beside the atmospheric pressure??
i found the work done from copper..its 0.443J
bt it is different from the standard answer it gave (0.434J)
 
Last edited:


blade_090 said:
ok i got ur point there...
is that any possible that copper has it own pressure beside the atmospheric pressure??
i found the work done from copper..its 0.443J
bt it is different from the standard answer it gave (0.434J)

I'm not sure what you mean with "its own pressure"?

The copper exerts an opposite pressure to the atmospheric pressure.
When the atmospheric pressure is increased, the copper is compressed and as a result gives a greater opposite pressure such that the balance is maintained.

Your results are pretty close...
What did you use for atmospheric pressure?
 

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