How Does Internal Energy Change at Constant Temperature?

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Homework Help Overview

The discussion revolves around the change in internal energy of an ideal gas when heat is transferred at constant temperature. Participants are exploring the implications of the first law of thermodynamics in this context.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the relationship between heat transfer and internal energy at constant temperature, with some attempting to confirm whether internal energy remains unchanged. Others are exploring what changes occur in the system despite the constant temperature.

Discussion Status

The discussion is active, with some participants confirming the original poster's understanding that internal energy remains unchanged. Others are engaging in further clarification about the implications of this situation, particularly regarding work done by the gas.

Contextual Notes

There is an emphasis on the first law of thermodynamics, and participants are considering the roles of heat and work in the context of an ideal gas at constant temperature. The nature of the problem suggests that assumptions about the behavior of the gas and the definitions of work and heat are under examination.

mike115
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[SOLVED] Internal Energy and Heat

Homework Statement


Ten joules of heat energy are transferred to a sample of an ideal gas at constant temperature. As a result, the internal energy of the gas:
a) increases by 10 J
b) increases by less than 10 J
c) increases by more than 10 J
d) remains unchanged


Homework Equations


delta U = Q + W


The Attempt at a Solution


I thought the internal energy of the gas remains unchanged since internal energy is proportional to absolute temperature and the temperature remains the same. Can anybody confirm this please?
 
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confirmed.
 
So the total internal energy remains unchanged? Then what changes?
 
e(ho0n3 said:
So the total internal energy remains unchanged? Then what changes?

Are you asking because you don't know, or for pedagogical reasons?

If the former: heat and work both are non-zero, the system does work. If the latter: the OP didn't ask what changes he just asked for confirmation.
 
I ask because I'm confused. So because the gas remains at a constant temperature, its internal energy remains unchanged. By the first law, that means that W is non-zero and so either the gas's pressure decreased and/or its volume increased?
 
Yeah, that's correct, since the temperature is constant we must have zero change in energy and so the heat Q which flows into the gas must exactly equal the work W done by the gas.
<br /> Q=W_{\rm by}<br />

In most circumstances one uses the expression dW_{\rm by}=pdV. Let's assume--as usual--that the previous equation holds. Then, although it's not generally true, in this case (the case of ideal gas of fixed number of particles at constant temperature) both the volume must increase and the pressure must decrease... and it's not too hard to calculate exactly how much the volume must increase and pressure decrease in terms of Q and the temperature and the number of particles and the initial volume. ;)
 

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