Calculate the Internal Energy of Monoatomic Gas

[themagiciant95]

Homework Statement

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.

Homework Equations

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 don't know why does it works...

 

[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.

 

[BvU]

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

just integrate

 

[themagiciant95]

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

 

[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 ?

 

[themagiciant95]

\gamma ln V + ln P = lnC

 

[BvU]

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

 

[BvU]

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

 

[themagiciant95]

How is it possible to integrate independent variables ? I always integrated functions. I am sorry , but i can't 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...

 

[Delta2]

How is it possible to integrate independent variables ? I always integrated functions. I am sorry , but i can't 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...

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 $\int \frac{V'(t)}{V(t)}dt=lnV(t)+C$.

 

[themagiciant95]

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

 

[Delta2]

Sorry I can't recommend English books, I just use my textbooks from my undergrad studies 20 years ago, which are written in greek :D...

 

[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.