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

1. Nov 1, 2009

### 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.

2. Nov 1, 2009

### mikelepore

Here's a guess: pick any one 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.

3. Nov 2, 2009

### 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....

4. Nov 2, 2009

### 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?

Last edited: Nov 2, 2009
5. Nov 2, 2009

### clamtrox

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).

Last edited: Nov 2, 2009
6. Nov 2, 2009

### 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 therfore the line is not horizontal. I don't see how could I find the temperature with the information I have, using your formula....

Last edited: Nov 2, 2009
7. Nov 2, 2009

### clamtrox

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 $$k_1 / k_2 = T_1/T_2$$.

8. Nov 2, 2009

### 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.