Heat of friction in cylinder-piston

  • #1
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I believe I'm missing something, probably a very stupid question incoming:

I am considering process with ideal gas expanding without piston-friction in constant temperature environment. Amount of heat taken from the reservoir is of the same amount as the work done by the expansion of the gas.

ΔU = Q - W

Since there is no change in temperature:

Q - W = 0

Let's consider process where the same ideal gas expands for the same amount of ΔV but this time with piston-friction.

ΔU = Q + Q(Friction) - W

Since there is no change in temperature:

Q + Q(Friction) - W = 0

So the "sum of heats" is of same amount as the work done, so less heat had to be taken from the reservoir, as the friction added some heat itself.

But how do I explain this practically? What it means for me is that I need less heat taken from the source for the same amount of work. How is that bad? What am I missing?
 

Answers and Replies

  • #2
A.T.
Science Advisor
11,402
2,775
What am I missing?
What are the final velocities of the pistons?
 
  • #3
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Some of the work is expended in overcoming the force of friction. So, in the frictional case, W includes both the frictional work and the work done on the surroundings. Actually, since part of the force exerted by the gas is used to overcome friction, the force on the surroundings and the work done on the surroundings are less than if the piston were frictionless. The net heat absorbed from the surroundings is less and the work done on the surroundings is less.

Chet
 
  • #4
24
0
Some of the work is expended in overcoming the force of friction. So, in the frictional case, W includes both the frictional work and the work done on the surroundings. Actually, since part of the force exerted by the gas is used to overcome friction, the force on the surroundings and the work done on the surroundings are less than if the piston were frictionless. The net heat absorbed from the surroundings is less and the work done on the surroundings is less.

Chet
Thanks!
 

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