How to obtain temperature from a PV vs P diagram using ideal gas law?

[trelek2]

As stated in the question: I have PV(T) [J/kg] vs P [Pa] plotted for 2 different temperatures. I'm to approximate the the temperatures using the lines and the ideal gas law. Note: V is the specific volume. I have no clue how to do this:
I know that PV vs P is just as if I had nRT vs P. From this the temperature should be obtainable, however the slopes of the lines are negative, so I don't see how I should approach this. The pressure varies from 0 to 10^6 Pa and the PV varies from about 310600 to 306600 J giving a slope of about -0.004. I also had been given the information that the weight is 0.018kg/mole.

 

[mikelepore]

Here's a guess: pick anyone point, out of the infinite number of point available along one of the isotherms, read off the values of P and V from the graph, and then calculate T=PV/nR.

 

[trelek2]

I don't think its that easy: The y-axis corresponds to different values of PV. That would mean that if I take any two different points and calculate the temperature this way, I'll get a different value of T for each point along the isotherm which is clearly wrong...

 

[Gerenuk]

Not sure if I understood this correctly. The problem is that pV_m is not constant along the line that's supposed to be the isotherm? Then the only solution is, that the ideal gas law doesn't apply?
Maybe it can be fitted to a Van der Waals gas?
I'll think about that...

 

[clamtrox]

The pressure varies from 0 to 10^6 Pa and the PV varies from about 310600 to 306600 J giving a slope of about -0.004.

So your temperature drops as you increase pressure? This just means that you have actual real world data... Take the ideal gas law, pV=NT. Then

\frac{d}{dp} (pV) =T \frac{dN}{dp}.

So, as you increase your isothermal systems pressure, unless it is perfectly sealed, you expect to see a small downward slope on the pV-p graph as your molecules are leaking outside (dN/dp < 0).

 

[trelek2]

Hey, you're right. I just got the hint that "all gases behave as ideal gases in the low pressure limit". Do you think I should take the point for the lowest pressure and approximate the temperature from that?

clamtrox: Are you sure about this? Or is it what Grenuk pointed out: The gas doesn't behave exactly as an ideal gas and therefore the line is not horizontal. I don't see how could I find the temperature with the information I have, using your formula...

 

[clamtrox]

clamtrox: Are you sure about this?

I certainly am not, I just said the first thing that popped into my head :D But you can check: just assume that dN/dp is independent of T (as it probably is). Then the slopes on the graphs should be proportional to the respective temperatures, so you can just check if k_1 / k_2 = T_1/T_2.

 

[trelek2]

That's not really what I get:
For the highter temperature the slope is -0.003, and taking the point at lowest pressure (0.1MPa) I get the temp = 772 K = 500C.
For the lower temp the slope is -0.004 and I get the temperature to be 672K = 400C...
These values are pretty reasonable considering we're dealing with superheated steam. Maybe I should just say I had taken the point measured at lowest pressure and at that point we can assume that the gas behaved as an ideal gas.