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imagine a closed vessel filled with gas with a piston and stones on top of piston. The system will be at constant pressure and is at equilibrium. That means the weight of stones are perfectly balanced by the pressure exerted by the gas.If the system is to do work by lifting the stones then the internal pressure should increase and be greater than the weight of the stone but this means the system is no longer an isobaric system

so how actually is the work done by isobaric systems.

I am a 12 th class student so please keep the answer simple.

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  • $\begingroup$ What you described is not an isobaric expansion. In fact, in the process you described, the gas pressure actually decrease as you remove stones. $\endgroup$
    – Chet Miller
    Nov 8, 2020 at 18:11
  • $\begingroup$ @Chet Miller Oops I thought he was heating the gas. Missed the stone removal $\endgroup$
    – Bob D
    Nov 8, 2020 at 19:32
  • $\begingroup$ @Bob D I’m not so sure what he meant. $\endgroup$
    – Chet Miller
    Nov 8, 2020 at 19:37
  • $\begingroup$ @Chet Miller I think you may be right. I revised the answer to try to cover it either way. Since he is interested in an isobaric process, at minimum the external pressure must be constant. $\endgroup$
    – Bob D
    Nov 8, 2020 at 19:45

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That means the weight of stones are perfectly balanced by the pressure exerted by the gas.

Be careful. It is the pressure exerted by the stones plus atmospheric pressure that is balanced by the pressure of the gas. When you apply gas laws you use absolute pressure, not gauge pressure.

If the system is to do work by lifting the stones then the internal pressure should increase and be greater than the weight of the stone but this means the system is no longer an isobaric system

It is the external pressure that determines the work done by the gas. That pressure is initially constant (pressure of stones plus pressure of atmosphere) until you start removing stones in which case the external pressure drops. Now the process is not isobaric.

If you remove one small stone at a time the pressure of the gas will drop very slowly as well and always approximately equal the external pressure. But since the pressure is dropping, the process is not isobaric.

If you want the process to be isobaric, don't remove any stones. Instead, heat the gas so that it expands and does work raising the piston against the constant external pressure of the stones and atmosphere. Now the process will be isobaric insofar as the external pressure is concerned which is sufficient to calculate the work. If you want the process to be isobaric with respect to gas equilibrium pressure, then heat the gas very slowly so that the pressure of the gas is always only slightly greater than the external pressure. Then you have a reversible isobaric process.

so how actually is the work done by isobaric systems.

It is done by heating the gas very slowly while keeping the external pressure constant.

Hope this helps.

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