# Show that the Change in Volume is Independent of the Path

1. Feb 25, 2014

### McAfee

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

I understand what the question is asking. Both ways I should get the same answer. I'm having trouble figuring out the mathematics behind this question.

2. Feb 25, 2014

### Goddar

Hi.
You're dealing with an ideal gas: what's the equation of state? (the one relating P, V and T)

3. Feb 25, 2014

### McAfee

The equation of state would be PV=nRT.
P- pressure
V- volume
n- number of moles
T-temperature
R-universal constant

4. Feb 25, 2014

### Goddar

That was your only missing piece, now you can work the integrals.

5. Feb 25, 2014

### McAfee

So would I have to change the equation of state around and plug it into each change of V, change of T, and change of P?

Edit: I should plug them into the partial difference equation? The change in should be constants I think.

Edit(again): The only equation I really have to plug in is V=(RT)/P. I think I'm just taking different differentials followed by integration.

Last edited: Feb 25, 2014
6. Feb 25, 2014

### Goddar

Use the equation of state to determine the derivatives. Example:
PV = nRT
so that, if n is constant:
VΔP + PΔV = nRΔT
When P is constant: $\frac{\partial V}{\partial T}$ = $\frac{nR}{P}$,
When T is constant: $\frac{\partial V}{\partial P}$ = – $\frac{nRT}{P^{2}}$

Just pay attention to what's constant and what's not when following your equations. Take the time to understand what you're doing, but you have everything you need.

7. Feb 25, 2014

### McAfee

Here is the work I have done so far. I did use the equations but I was having a hard time understanding in my mind the final values to plug in specifically for the final T value. I assumed it was referring to the final state so I plugged in 400 K. Please let me know if I made any mistakes.
Thanks again for the help.

8. Feb 25, 2014

### Goddar

Ok, you have to be careful when integrating. First with:
ΔV1-2-3 = ($\frac{\partial V}{\partial T}$)P1 ΔT + ($\frac{\partial V}{\partial P}$)T2 ΔP,
You get:
∫ΔV1-2-3 = Vfinal – Vinitial = ∫($\frac{\partial V}{\partial T}$)P1 ΔT +∫($\frac{\partial V}{\partial P}$)T2 ΔP = [V(P1,T2) – V(P1,T1)] + [V(T2,P2) – V(T2,P1)]
= R($\frac{ΔT}{P_{1}}$ + $\frac{T_{2}}{P_{2}}$ – $\frac{T_{2}}{P_{1}}$)

So you see that using integrations is purely formal here: at the end you don't have to differentiate or integrate anything.
Also, be careful with your units: atm is not a S.I. unit, so the gas constant must be adjusted if you use atms...

9. Feb 25, 2014

### McAfee

Ok thanks I'm going to take another look at what I did. Will post when I finish.