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I approach these expansion problems like so: The gas and the surroundings(piston+outside) are at the same pressure at first. We heat the gas. The pressure rises inside the syringe a bit. The gas expands so the pressure remains constant. Then I use P(the constant pressure of the gas) *dV. What I want to confirm is my reasoning on using this equation. It was derived assuming P(internal) = constant. But it does change momentarily. Is the reason we ignore it in the "a bit" nature?

Also for compression, the force exerted on the gas by surroundings (piston+outside) is taken as the force the gas exerts on the piston. Is this Newton's third law?

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  • $\begingroup$ What do you mean by "Is the reason we ignore it in the "a bit" nature?" $\endgroup$ – KV18 Apr 13 at 6:43
  • $\begingroup$ As in the as soon as pressure rises because of the temp rise the volume expands to "counter it". So pressure never really changes much. So it's okay to ignore it for calculations $\endgroup$ – Sal_99 Apr 13 at 7:13
  • $\begingroup$ @KV18^^^^^>>>>> $\endgroup$ – Sal_99 Apr 13 at 7:19
  • $\begingroup$ @KV18, I forgot to tag you. So I put this up sign to refer you up. $\endgroup$ – Sal_99 Apr 13 at 7:31
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    $\begingroup$ Nah. It is alright. It does display in my feed anyway. $\endgroup$ – KV18 Apr 13 at 7:33
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As in the as soon as pressure rises because of the temp rise the volume expands to "counter it". So pressure never really changes much. So it's okay to ignore it for calculations

If the piston moves by a little bit, then the pressure is literally the same. Considering $\Delta V \to 0$ which would mean $\Delta V \approx dV$.

Also for compression, the force exerted on the gas by surroundings (piston+outside) is taken as the force the gas exerts on the piston. Is this Newton's third law?

Yes, this is true.

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  • $\begingroup$ the problem I was doing had the piston moving somewhat more than what I guess would be appropriate to call a little. Can't the relatively constant pressure argument be used for that? $\endgroup$ – Sal_99 Apr 13 at 7:58
  • $\begingroup$ That is why we integrate. We neglect infinitesimal changes in the pressure although we consider the change for the volume to be $dV$, and we integrate along the volume to find the work done. It is more of a mathematical trick. Sure, when we integrate, it becomes clear that the change in the pressure of the gas is a lot. $\endgroup$ – KV18 Apr 13 at 15:16
  • $\begingroup$ apologies for the late reply. In such situations, my teacher told me to use W=PdV since P is more or less constant. Is this also since P doesn't change much ? $\endgroup$ – Sal_99 Apr 18 at 7:07
  • $\begingroup$ Yes, for that infinitesimal change $dV$, the pressure does not change much. $\endgroup$ – KV18 Apr 18 at 8:31
  • $\begingroup$ we did questions where we just multiplied P ( taken constant) and the whole change in volume. The argument here is that since pressure never changed much through the process, we get to use this? No integration or differentiation was used. $\endgroup$ – Sal_99 Apr 20 at 16:45
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If the heat addition occurs very slowly such that the pressure and temperature gradients in the gas approach zero, the process can be considered quasi-static and the pressure of the gas will always be very close to the external pressure. If the process is also frictionless, then we can say it is a reversible isobaric expansion. At every point along the process the ideal gas equation applies

$$\frac{V}{T}=\frac{nR}{P}$$

since $P$ is constant

$$\frac{V}{T}=constant$$

Since the external force applied to the gas by the piston and atmosphere is always approximately equal to and opposite the force the piston and atmosphere apply to the gas, Newton;s third law applies.

Hope this helps.

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