Is heat absorbed, when volume is increased?

  • #1

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



[itex]dQ = dU + dW[/itex]
[itex]dQ = dU + PdV[/itex]

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?
 
Last edited:

Answers and Replies

  • #2
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..
 
  • #3
1,540
135

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



[itex]dQ = dU + dW[/itex]
[itex]dQ = dU + PdV[/itex]

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
 
  • #4
Andrew Mason
Science Advisor
Homework Helper
7,671
393
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
 
  • #5
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.
 
  • #6
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?
 
  • #7
1,540
135
physicsjn...

dU = -pdV

Now dV > 0 ,so dU <0 i.e internal energy decreases.
 
  • #8
Oh I see.. Thank you very much! :)
 

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