0
$\begingroup$

Let us consider a cylinder fitted with a piston containing an ideal gas inside it.And we give some amount of heat energy to the system. Assuming ideal condition , if the piston's velocity remains effectively zero throughout then a part of the heat given is utilized to increase the internal energy and the rest is utilized by the gas in doing work.But the gas is doing work on the piston still it's speed remains effectively zero i.e. it's kinetic energy is not changing . Then in which form does the rest part of energy gets converted into?

$\endgroup$
2
  • $\begingroup$ What is the "rest part of the energy"? If the piston doesn't move, there is no work done by the gas, and the temperature simply increases along with pressure. $\endgroup$ – Drew Sep 19 '18 at 2:17
  • $\begingroup$ @Drew i want to say that in ideal condition when any gas does work on the piston the piston moves but......if it gains some velocity then the piston can never come to rest again at the end it will start oscillating ....so in ideal condition we assume that the velocity which is gained by the piston is very small i.e. effectively zero ......that's why if the energy of the piston is not changing as it's speed remaining constant then what about the work done by the gas. $\endgroup$ – It's probable Sep 19 '18 at 2:23
1
$\begingroup$

Under the described conditions, the work, performed by the gas, won't be zero only if the external pressure is not zero, in which case, the work would be performed against that pressure.

This work could be converted to a mechanical energy (lifting a weight), to an internal energy of a fluid, compressed by the piston, or some other form of energy.

$\endgroup$
8
  • $\begingroup$ are irreversible processes not quasi static? $\endgroup$ – It's probable Sep 19 '18 at 11:39
  • $\begingroup$ @RifatSafin All reversible processes are quasi static, but not all quasi static processes are reversible. So, irreversible processes could be quasi static. $\endgroup$ – V.F. Sep 19 '18 at 11:46
  • $\begingroup$ how can we change the entropy of an isolated system? $\endgroup$ – It's probable Sep 20 '18 at 18:05
  • $\begingroup$ @RifatSafin We don't have to do anything to an isolated system (that's why it is isolated) to change its entropy - if the system is not in the thermodynamic equilibrium (or not undergoing a reversible process), its entropy will be increasing by itself until it reaches the equilibrium, at which point its entropy will be maximized. $\endgroup$ – V.F. Sep 20 '18 at 18:25
  • $\begingroup$ if the system is in equilibrium can we change its entropy externally? $\endgroup$ – It's probable Sep 20 '18 at 18:46
0
$\begingroup$

If the cylinder is oriented vertically and the piston has mass, then you are increasing the potential energy of the piston as well as doing work to push back the outside atmosphere.

$\endgroup$
2
  • $\begingroup$ .....sry i forgot to mention that it is oriented horizontally....and if there is no atmosphere present will the piston continue to move on ? Considering the cylinder to be very large and frictionless $\endgroup$ – It's probable Sep 19 '18 at 2:58
  • $\begingroup$ If there is vacuum outside and the cylinder is horizontal, what is to prevent the gas from expanding to accelerate the piston even before you start heating the gas. $\endgroup$ – Chet Miller Sep 19 '18 at 3:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.