# Thermodynamic Question

1. Apr 4, 2006

### Ne0

Dealing with isothermal lines. How would you go about proving if the two lines are parallel if no values are given for the Pressure or the Volume?

This is what I've got so far:
Q = -W

dU = 0, where U is internal energy.

P1V1 = nRT
P2V2 = nRT

P1V1=P2V2

Or:

P1/P2 = nRT1 / nRT2

The problem is I don't know how to start showing that the lines don't cross, i.e. the slopes are the same. Any help is greatly appreciated. Thank you.

2. Apr 4, 2006

strange.

on a P-v graph, isotherms are rectangular hyperbolas. (look like 1/x.)

they are only straight lines on a T-P or T-V graph (or a T-s graph, if you're there yet), in which case it's pretty obvious that all isotherms will be parallel.

3. Apr 4, 2006

### Skorch

I am working with Ne0 on this problem.

When he says line, he means hyperbola.

We need to show that the graphs of two isotherms on a P-V graph for the same gas at two different temperatures are parallel.

I hope this makes it a little more clear.

4. Apr 4, 2006

i didn't even know that you could say that a curve is parallel to another one!

does that mean that the tangent lines have the same slopes at the same value of V or P or something?

i don't think i can help. sorry. :/

5. Apr 4, 2006

### Physics Monkey

Skorch and Ne0,

I'm still not sure exactly what you want, but let me just throw some things out that you may find helpful. You can calculate the slope of various isotherms easily enough from the ideal gas law. Temperature appears explicitly in the result, but you can use the ideal gas law to reexpress the answer in terms of P and V only. It is then a simple matter to figure out the set of all the points (P,V) with a given slope.

6. Apr 5, 2006

### qbert

I think they're trying to get at the fact
that two isotherms don't intersect on
a P-V diagram.

if they did you'd have a system with
a single coordinate (P, V)
giving rise to two different temperatures.

7. Apr 5, 2006

### Ne0

Yes we are trying to prove using the ideal gas law that the two isotherm lines never intersect just like Gbert mentioned. I was thinking of maybe differentiating for PV=nRT and somehow show that the two isotherms have the same slope.

8. Apr 5, 2006

### Ne0

I was thinking like one point on an isothem to be (V1,P1) and another point (V2,P2) and the slope would be, P=nrt/V, (nRT/V2-nrt/V1)/(V2-V1). I am not quite sure what to do with this result because we need to somehow show from the other isothem that they are identical and therefore will never intersect. Another idea I was thinking of was differentiating PV=nrT and maybe from there show that they are the same for the two isotherms. I got VdP + PdV = nRdT but from there I am not quite sure of how to show they are the same.

9. Apr 6, 2006

### Curious3141

If all you want to do is prove they don't intersect, you can do it by a simple contradiction argument.

Let's say one isotherm is defined by $$PV = nRT_1$$
and the other is defined by $$PV = nRT_2$$ on a standard P vs V plot. $$T_1$$ and $$T_2$$ are distinct (unequal).

Let us start of by making the assumption some intersection point exists.

At that intersection point, the P and V values are identical, hence so is the product PV, so at that point $$nRT_1 = nRT_2$$ giving $$T_1 = T_2$$

But that's an impossibility since $$T_1$$ and $$T_2$$ were *defined* to be unequal. Hence a contradiction has been arrived at meaning the original assumption of an intersection is false.

Bit wordy, but that's a formal proof if you're looking for it.

10. Apr 6, 2006

### Andrew Mason

The two isotherms are not parallel but, as Curious has shown, they do not intersect.

To show that they are not parallel, draw two arbitrary horizontal lines between them. These represent lines where P=constant. If they are parallel isotherms, these isobars would be the same length (the length representing change in volume). The change in volume of the gas along each isobar

[tex}V_1 = nRT_1/P[/tex]

[tex}V_2 = nRT_2/P[/tex]

$$V_2 - V_1 = \frac{nR}{P}(T_2-T_1)$$

which is not constant for all P.

AM