Is heat absorbed, when volume is increased?

AI Thread Summary
In a thermally insulated container with a movable piston, when the volume of gas increases, no heat is absorbed or released due to the insulation. Instead, the internal energy of the gas decreases as it does work against the piston. This decrease in internal energy leads to a reduction in temperature, contrary to the initial assumption that temperature would increase. The discussion emphasizes the application of the first law of thermodynamics, highlighting the relationship between work done, internal energy, and heat transfer. Overall, the process described is adiabatic, confirming that as volume expands, internal energy decreases, resulting in a temperature drop.
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Homework Statement


Hi! Greetings! This is not really a problem set; I just made the question up to better understand PV diagrams of Carnot cycle.

Suppose you have a gas in a container with a movable piston, but is thermally insulated. If the volume of the gas increases, is heat absorbed by the system (the gas particles) or released by the system into the environment?

Homework Equations



dQ = dU + dW
dQ = dU + PdV

The Attempt at a Solution


Hmm.. What I am thinking is that if the volume expands, the gas particles are less compressed and have more place to move around, so I guess the system should give off heat to the surroundings?
 
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Oh, I got it. It's thermally insulated, so no heat flow. It's the temperature that should change, and if it is an ideal gas, the volume is proportional with temperature and thus, the temperature should increase. Sorry for asking a silly question.. Haha..
 
physicsjn said:

Homework Statement


Hi! Greetings! This is not really a problem set; I just made the question up to better understand PV diagrams of Carnot cycle.

Suppose you have a gas in a container with a movable piston, but is thermally insulated. If the volume of the gas increases, is heat absorbed by the system (the gas particles) or released by the system into the environment?

Homework Equations



dQ = dU + dW
dQ = dU + PdV

The Attempt at a Solution


Hmm.. What I am thinking is that if the volume expands, the gas particles are less compressed and have more place to move around, so I guess the system should give off heat to the surroundings?

If the container is thermally insulated ,there is no heat exchange between the container and the surroundings
 
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physicsjn said:
Oh, I got it. It's thermally insulated, so no heat flow. It's the temperature that should change, and if it is an ideal gas, the volume is proportional with temperature and thus, the temperature should increase. Sorry for asking a silly question.. Haha..
How can it increase?

Apply the first law. If PdV>0 and Q = 0 what must happen to U?

AM
 
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Andrew Mason said:
How can it increase?

Apply the first law. If PdV>0 and Q = 0 what must happen to U?

AM

I guess the internal energy increases if PdV.

But in Carnot cycle there are two types of processes right? Adiabatic and isothermal. The process described above is adiabatic, so I thought there should be some temperature change.
 
Ah wait.. When the volume increases, the particles are doing work against the system, so the internal energy will decrease. And temperature also would decrease. Am I right?
 
physicsjn...

dU = -pdV

Now dV > 0 ,so dU <0 i.e internal energy decreases.
 
Oh I see.. Thank you very much! :)
 
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