First Law of Thermodynamics and ideal gas

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Homework Help Overview

The discussion revolves around the First Law of Thermodynamics in the context of an ideal gas undergoing isothermal expansion, specifically focusing on the work done by the gas and the associated heat transfer.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of isothermal expansion on internal energy and heat transfer, questioning the sign conventions used for work and heat.

Discussion Status

There is an ongoing examination of the sign conventions related to heat and work, with some participants providing feedback on the original poster's reasoning and calculations. Clarifications are being sought regarding the negative sign in the heat transfer calculation.

Contextual Notes

Participants are discussing the implications of the sign conventions for heat and work in thermodynamic processes, which may not align with standard interpretations.

brutalmadness
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Homework Statement


An ideal gas expands isothermally, performing 3.40x10^3 J of work in the process.
a) The change in internal energy of the gas
b) The heat absorbed during this expansion


Homework Equations


\DeltaU=Q-W


The Attempt at a Solution


a) Since it's isothermally expanding, there will be no change in internal temperature, so:
\DeltaU=0

b) Q=W, so:
Q=-3400 J
 
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Looks good, as long as positive heat means that the system is heating the environment and positive work means that the environment is doing work on the system. This sign convention is a little unconventional, though.
 
brutalmadness said:
b) Q=W, so:
Good.
Q=-3400 J
How did that negative sign sneak in? (W is positive, so Q is positive.)
 
Ah, ok. That was a silly mistake, lol :)

So Q=3400 J
 

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