# Homework Help: Calculate the Internal Energy of Monoatomic Gas

1. Dec 22, 2017

### themagiciant95

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

Re-arranging the equations of potential (internal) energy in a monoatomic gas, i get this differential equation:

$$\gamma (dV/V ) + (dP/P) = 0.$$
where V is volume and P is pressure.
I have to integrate it.

2. Relevant equations

3. The attempt at a solution

I tried, but i wasnt able to understand how (and why) is it possible to integrate an equation with different functions and different differentials in it. My intuition tells me : just integrate ! , but i dont know why does it works...

2. Dec 22, 2017

### Delta²

yes it seems to work in this form, use that $\int \frac{dV}{V}=lnV-lnV_0=ln\frac{V}{V_0}$ and similar for the other term with P. Its because the differentials and the denominators are for the same variable (V or P) so integration works and gives the logarithmic function as result.

3. Dec 22, 2017

### BvU

Delta was faster, but:
Well, what do you get when you

4. Dec 22, 2017

### themagiciant95

That's very nice.
In this case P and V are functions of which variables ?

5. Dec 22, 2017

### BvU

They are independent variables. Solving the differential equation gives you a relationship between the two.

What did you get when you 'just integrate'd ?

6. Dec 22, 2017

### themagiciant95

$$\gamma ln V + ln P = lnC$$

7. Dec 22, 2017

### BvU

Why $\log C$ and not $C$ ?

8. Dec 22, 2017

### BvU

Because it's more comfortable to write $P\;V^\gamma = C \quad$

9. Dec 23, 2017

### themagiciant95

How is it possible to integrate independent variables ? I always integrated functions. Im sorry , but i cant clearly understand.
A friend suggested me to try a "dummy variable" in order to parametrize U,V,P . But i didnt find anything explaining on the web about this...

Last edited: Dec 23, 2017
10. Dec 23, 2017

### Delta²

You can write P,V as functions of a dummy variable t, that is P=P(t) , V=V(t). Then the differentials dP and dV will become $dP=P'(t)dt$, $dV=V'(t)dt$ so at the end you ll integrate both functions $\frac{P'(t)}{P(t)},\frac{V'(t)}{V(t)}$ with respect to the dummy variable t. The end result will not change due to the chain rule of derivatives, we ll have for example $\int \frac{V'(t)}{V(t)}dt=lnV(t)+C$.

11. Dec 23, 2017

### themagiciant95

Thanks so much. Do you know where i can find some theoric math materials about "integration of independent variables" (just as this case) ?

12. Dec 23, 2017

### Delta²

Sorry I cant recommend English books, I just use my textbooks from my undergrad studies 20 years ago, which are written in greek :D...

13. Dec 23, 2017

### BvU

I hope I didn't confuse you with my post #5, I called both variables independent and then continued with 'solving the differential equation gives you a relationship between the two'.

That isn't as contradictory as it seems: it means theat there is only one degree of freedom: change P (which you can do) and V changes, and vice versa: change V (which you can also do) and P changes.

If you replace one of the two by a $y$ and the oher by an $x$ you are close to textbook material on differential equations. Your own familiar and trusted calculus book is probably the best place to freshen up your skills.