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Something that is bothering me is that isothermal process and isobaric process in a piston cylinder system behave similarly. So, in this exercise they say that temperature goes from -10 °C to 15 °C, and they say nothing about being an isobaric process, and then, in the resolution, they consider the pressure is equal in both states.

Imagine this, if I transfer heat, and if it raises the temperature of the system, it means an amount of that heat were converted in work, and the other part were converted in internal energy, so how can I be so sure that the posterior state of my particles' KE wouldn't be big enough that will increase the pressure, despite the fact Volume has increased?

In an isobaric process it increases the pressure and then decreases very fast, so I can say it is isobaric, but here I don't know if it is.

Also, the limit they give is when temperature is 15°C, they don't say to calculate something when time goes to infinity, this way I would know that pressures would equalize.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. Please do not use comments to discuss partial solutions or solution strategies, comments are for improving and clarifying the question. $\endgroup$
    – ACuriousMind
    Apr 25, 2017 at 0:17

1 Answer 1


The piston is free to move, the local atmospheric pressure is acting on the piston and it is constant. This is an isobaric process.

  • $\begingroup$ it is constant because we're inputting energy in a very slow way such that we let the piston gets in equilibrium all the times. But what if I transferred energy too quickly? Maybe I would increase the internal energy till 15°C and in this instant the internal pressure would still be bigger than external pressure + Weight. $\endgroup$ Apr 25, 2017 at 10:52

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