How Does Internal Energy Change at Constant Temperature?

AI Thread Summary
When ten joules of heat energy are transferred to an ideal gas at constant temperature, the internal energy of the gas remains unchanged. This is because internal energy is directly proportional to absolute temperature, which does not change. The heat added to the gas must equal the work done by the gas, implying that while energy is transferred, it does not affect internal energy. Consequently, the gas experiences an increase in volume and a decrease in pressure as it does work on its surroundings. Thus, the first law of thermodynamics confirms that the internal energy remains constant despite the heat transfer.
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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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