# What kind of ideal gas process a positively sloped line in a $pV$-diagram represents?

I've come across an ideal gas process which can be represented by a positively sloped line in the $$pV$$ diagram. I found something about this on KhanAcademy, but it's simply a worked example on how to compute the work done by such a process. My question is what kind of a physical process does this represent? It's clearly not isochoric, isobaric, isothermal or adiabatic and it doesn't seem to appear in any basic heat engines.

Here's an image of the process (from KhanAcademy):

• This probably isn't very common, but you could get this behavior if you heated the gas while the piston was pushed down by a spring. As the volume goes up, so does the spring force and hence the pressure. – knzhou Nov 29 '18 at 16:08

It does not have a name, and I am not sure how practical is to do it, it is just an expansion in which the pressure increases linearly with the volume. In order to do that you need to put the gas in contact with a variable thermal source whose temperature will augment quadratically with the volume: $$T=(a+bV)V/nR$$

• "In order to do that you need to put the gas in contact with a variable thermal source whose temperature will augment quadratically with the volume: $T=(a+bV)V/nR$" Why? This is not obvious to me. – Drew Nov 29 '18 at 15:55
• @Drew All he is saying is that the temperature is rising in a specific way along the process path. From the ideal gas law, T=PV/nR, so for any combination of P and V along the process path, you can calculate a corresponding temperature. – Chet Miller Nov 29 '18 at 16:11
• @Drew yes, just as Chester Miller said – Wolphram jonny Nov 29 '18 at 16:13
• @Drew The equation of the straight line in this case is $P = bV + a$, compare with the general equation of a straight line $y=mx+c$, and the equation of state is $PV=nRT$ – Farcher Nov 29 '18 at 17:38

Any path on the $$PV$$ graph (or on any other thermodynamic plane) represents a quasistatic process that can, in principle, be conducted experimentally. How easily, that's another question. Some paths represent simple processes, for example isothermal, isobaric, isentropic etc. Others require elaborate setups.

For the linear path on the $$PV$$ plane we might construct a temperature controller that adjusts the temperature to make the path a straight line. For an ideal gas, the controller would set the temperature according to the current volume as follows:

$$T = \frac{V \left(P_0 V-P_0 V_1-P_1 V+P_1 V_0\right)}{n R \left(V_0-V_1\right)}$$

If you plug this temperature in the ideal gas law it will produce a linear path between $$(P_0,V_0)$$ and $$(P_1,V_1)$$. That's the same quadratic formula of @Wolphram jonny, evaluated for the specific end points of the path.