I am thinking about how the heating of a room changes from 0 to 20 degrees with and without the window open.
If I heat it with the window closed, I have calculated the internal energy as follows:
$$V = const$$ $$dU = cNK_{b} dT$$ with $c = \frac{7}{2}$. So we I calculated the integral: $$\int dU = \int cNK_{b} dT$$ $$\Delta U = \int cNK_{b} dT = 70 \cdot N \cdot K_{b} [J]$$ Now my interest concerns the heating of the same room, but with the window open, so the pressure is constant, as well as the volume, what changes is the number of particles, being that when the gas (oxygen and nitrogen) expands and escapes through the window. For my part I would have chosen to use the formula: $$dS = \frac{dU}{T} + \frac{p}{T} dV - \frac{\mu}{T}dN$$ The problem is that I wouldn't know how to vary both the temperature and the number of particles...
Can anyone help me?

  • $\begingroup$ What about the cold outside air entering the room. Have you included that effect in your calculation? $\endgroup$ – my2cts Mar 21 at 10:20
  • $\begingroup$ I have not taken this into account for this calculation, as my interest is in the particles that escape from the chamber with the heating of the chamber, the external environment is not relevant. $\endgroup$ – Albano Tabacchi Mar 21 at 10:22
  • $\begingroup$ Is the room airtight when the window is closed. Even if it is, does it expand with pressure? $\endgroup$ – my2cts Mar 21 at 10:36
  • $\begingroup$ With the window open the air volume is not constant when heated. $\endgroup$ – my2cts Mar 21 at 10:39
  • $\begingroup$ The volume of the chamber and the pressure are constant, you cannot expand the volume of the chamber, so the particles escape through the window $\endgroup$ – Albano Tabacchi Mar 21 at 10:40

You should use the ideal gas law. This is a relation between pressure, volume, temperature and number of moles. You appear to keep pressure and volume constant so the law reduces to a relation between temperature and molar density, which gives you the number of moles in the fixed volume at fixed pressure at different temperatures. The bedroom case is rather more complicated ☺️.

  • $\begingroup$ So basically we'll end up in this situation $$ PV=const.$$ $$N_{1}K_{b}T_{1} = N_{2}K_{b}T_{2}$$ then my temperature difference is 20K, from 273.15K to 293.15K, so I should integrate over this range. The problem then is how to handle the number of moles... $\endgroup$ – Albano Tabacchi Mar 21 at 13:32
  • $\begingroup$ There is no integration. Just calculate for the two temperatures and compare. $\endgroup$ – my2cts Mar 21 at 15:03
  • $\begingroup$ Do you mean something like this:$$\frac{N_{1}}{N_{2}} = \frac{293.15}{273.15}=1.0732$$ $\endgroup$ – Albano Tabacchi Mar 21 at 17:09

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