# Can you calculate the supplied heat given a p(V) - Diagram of a isobar process?

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

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.

• How is this an isobaric process? I mean you say so in the title. I don't see how it is. – KV18 Apr 12 at 7:53
• It is not. I just gave an example. – TVSuchty Apr 12 at 8:29
• Neither process is isothermal. – Chet Miller Apr 12 at 11:55

## 1 Answer

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.

• Thank you. But if I do not know the temperatures? I do not see the solution... – TVSuchty Apr 12 at 8:30
• $T=PV/nr$, please review the ideal gas equations. – KV18 Apr 12 at 8:41
• No problems! :) – KV18 Apr 12 at 16:33