You are given an p(V) - Diagram like this:


Source: How to calculate the efficiency from a $p$-$v$ diagram?

There you see two isothermal processes and two isochor processes. For every given isothermial process I can exactly calculate the heat supplied by:

$\Delta U = \Delta W + \Delta Q = 0 \implies \int p(V)dV=- \Delta Q$

Now I assume an isobar process (not seen in the Diagram): Again I know:

$\Delta U = \Delta W + \Delta Q \implies \int p(V)dV= \Delta U - \Delta Q$

But can I make a more precise statment? Can I calculate the heat supplied? Same for isochor processes, where:

$\Delta U = \Delta Q$

I am wondering because I suppose that given this diagram (the circle), there should be an equilibrium of Energy supplied and Energy absorbed...

Thank you.

  • $\begingroup$ How is this an isobaric process? I mean you say so in the title. I don't see how it is. $\endgroup$ – Karthik Apr 12 '19 at 7:53
  • $\begingroup$ It is not. I just gave an example. $\endgroup$ – TVSuchty Apr 12 '19 at 8:29
  • $\begingroup$ Neither process is isothermal. $\endgroup$ – Chet Miller Apr 12 '19 at 11:55

The linked source has an accepted answer that explains what you have to do very well. All you have to do is to calculate the temperature at the four points in your diagram by using the ideal gas equation.

As you can see, the first deviation ($PV^{1.1} ...$) is a polytropic process as it is of the form $PV^n = k$, where $k$ is some constant. The heat required to do such a polytropic process is given in that very link.

You can find the heat in that isochoric process by using the formula $Q=C_v \Delta T$, where $C_v$ is the specific heat capacity and $\Delta T$ is the temperature difference.

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  • $\begingroup$ Thank you. But if I do not know the temperatures? I do not see the solution... $\endgroup$ – TVSuchty Apr 12 '19 at 8:30
  • $\begingroup$ $T=PV/nr$, please review the ideal gas equations. $\endgroup$ – Karthik Apr 12 '19 at 8:41
  • $\begingroup$ No problems! :) $\endgroup$ – Karthik Apr 12 '19 at 16:33

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