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Consider a container filled with ideal gas and is covered by a frictionless and lightweight piston that exerts pressure on the gas.

I understand that during an isobaric process, the container is in contact with a pressure reservoir, ensuring that the gas is in mechanical equilibrium with its surroundings for the whole process. This process happens quasistatically.

I am uncertain how this process works step-by-step. Suppose a small amount of heat is added to the container, it causes the gas to have greater pressure than its surroundings (in quasistatic process, the pressure of the gas is infinitesimally greater than its surroundings so that $\Delta P$ is negligible which makes pressure constant). In order to respond to this pressure imbalance, the gas expands until mechanical equilibrium is reached. According to the ideal gas law $P\Delta V=nR\Delta T$, volume increase causes the temperature to increase. Did I understand it correctly?

Also there is one thing that bothers me: If gas expands, then it performs work on the surroundings right? In that case, its internal energy should decrease which in turn decreases temperature, according to this formula $\Delta U=\frac{f}{2}Nk_B\Delta T$, no? I am not sure why temperature increases during an isobaric process.

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    $\begingroup$ I think you mean that 'a temperature increase causes the volume to increase', not 'volume increase causes the temperature to increase'. $\endgroup$
    – Dedados
    Commented Apr 8, 2023 at 9:32
  • $\begingroup$ See: physics.stackexchange.com/questions/137595/… $\endgroup$
    – Dedados
    Commented Apr 8, 2023 at 9:36

2 Answers 2

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I understand that during an isobaric process, the container is in contact with a pressure reservoir, ensuring that the gas is in mechanical equilibrium with its surroundings for the whole process.

Maintaining a constant external pressure does not ensure that the gas is in mechanical equilibrium with its surroundings for the whole process. Carrying out the process quasi statically and without friction ensures it.

According to the ideal gas law $P\Delta V=nR\Delta T$, volume increase causes the temperature to increase. Did I understand it correctly?

You've got it backwards. The temperature increase causes the volume to increase, with the ratio of temperature to volume being constant for a constant pressure process.

Also there is one thing that bothers me: If gas expands, then it performs work on the surroundings right?

Correct.

In that case, its internal energy should decrease which in turn decreases temperature, according to this formula $\Delta U=\frac{f}{2}Nk_B\Delta T$, no? I am not sure why temperature increases during an isobaric process.

The internal energy does not decrease, it increases. That's because, for a constant pressure process, part of the energy transfer to the gas by heat performs work while part increases its temperature (and thus internal energy). From the first law $$Q=\Delta U+W$$

I am uncertain how this process works step-by-step.

The key to the process is keeping the external pressure constant. For an example, see the Figure below where the piston/cylinder is vertically oriented and a weight is placed on top of the piston.

Ignoring the mass of the piston, the total external pressure is as shown. It is fixed. The piston/cylinder is placed in a thermal bath or reservoir of constant temperature infinitesimally greater than the gas temperature. This results in an infinitesimal transfer of heat to the gas increasing the gas temperature and pressure and doing an infinitesimal amount of work raising the weight. The gas then comes into thermal equilibrium and mechanical equilibrium with surroundings. The process is repeated with additional higher temperature thermal reservoirs continually taking in heat, increasing temperature, and performing work, always coming into equilibrium with the same constant external pressure.

Hope this helps.

enter image description here

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  • $\begingroup$ Thank you but I still have hard time understanding how it is possible for pressure to remain constant during quasistatic expansion. If the process is not quasistatic, then the (large) temperature increase will cause both the increase in volume and the increase in pressure, right? $\endgroup$
    – Ray Siplao
    Commented Apr 8, 2023 at 20:10
  • $\begingroup$ @RaySiplao I will add a diagram and explanation when I get back to my computer $\endgroup$
    – Bob D
    Commented Apr 8, 2023 at 20:20
  • $\begingroup$ @RaySiplao See update to my answer. $\endgroup$
    – Bob D
    Commented Apr 9, 2023 at 19:22
  • $\begingroup$ Thank you very much for the addendum! BTW what kind of software did you use to create this diagram? $\endgroup$
    – Ray Siplao
    Commented Apr 9, 2023 at 20:34
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    $\begingroup$ @RaySiplao PowerPoint. Export it as a JPEG file which you can drag and drop onto your answer $\endgroup$
    – Bob D
    Commented Apr 9, 2023 at 20:37
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When a gas expands against external pressure, as in the isobaric example you are describing, it does work. In isobaric expansion the energy for this work must come from heating, as a result of which the temperature of the gas increases. This explains the $\Delta T$ in your analysis.

Since we are powering the expansion by heating, the energy for the expansion does not come from internal energy. In fact, the internal energy of the gas increases because its temperature increases. If we do the expansion adiabatically (but then it is not isobaric), then the temperature of the expanding gas will decrease because internal energy is the only possible source for the work of the expansion.

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