# 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. – Karthik Apr 12 '19 at 7:53
• It is not. I just gave an example. – TVSuchty Apr 12 '19 at 8:29
• Neither process is isothermal. – Chet Miller Apr 12 '19 at 11:55

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.
• $T=PV/nr$, please review the ideal gas equations. – Karthik Apr 12 '19 at 8:41