0
$\begingroup$

I have an spherical ballon filled with an ideal gas such that the difference of pressure between the interior $(P)$ and the exterior $(P_0)$ is proportional to the radius of the balloon: $$P-P_0 = \xi r$$ Where $\xi$ is a constant associated to the characteristics of the balloon. I can assume that the external pressure is constant.

How can I relate a small change in the equilibrium temperature with a small change in the radius?. In particular, I want to find $\displaystyle \frac{\Delta r}{\Delta T}$ with $\Delta r, \Delta T$ small (something like $\frac{dr}{dT}$). For example, if this ratio is constant, that means that $dr$ is proportional to $dT$. The problem here is that I don't know how to proceed. I used $PV=nRT$ assuming that the radius is dependent of the temperature and I differentiated both sides with respect to $T$ after replacing $P$ and $V$ with the information that I have, but I just got some ugly relationship that I dont know if its correct $\Big($I found something like $\frac{dr}{dT} \propto \frac{1}{r^3+r^2}$$\Big)$. The doubt is that I assumed that the radius is dependant of the temperature, but if this isn't the case I don't know how to find the ratio. Any hints or confirmation of my results are appreciated.

$\endgroup$
0
$\begingroup$

well for the gas inside the balloon you can use the ideal gas equation for sure hence we have : $$ P_{gas}\, V = nRT$$ for our system the quantity of gas is constant so let us assume that the moles of gas are constant and equal to $n_o$ and from the relation that you mentioned for the pressure we have $P_{gas} = P_{o} + \xi r $ also we know very well that $\,V = \frac{4}{3} \pi r^3$ ; this is what we all know and this yields us the following result : $$ (P_o + \xi r)\,\frac{4}{3} \pi r^3 = n_o R T$$

hence from the above equation we can see that the radius is indeed a function of temperature , if this wasn't the case then radius could have only been a constant and the ratio $ \gamma = \frac {\Delta\, r}{\Delta\,T} = 0$

let us however work the results for the previous case, well if radius is indeed a function of temperature then we can indeed use the derivative with respect to T : $$ (3P_o r^2 + 4\xi r^3) \frac{dr}{dt} = \frac{3n_oR}{4}$$ $$(A r^2 + B r^3)\;\frac{dr}{dt}\; = \;1 \,$$ $$A = \frac{4P_o}{n_oR} \;\;\; and \;\;\; B = \frac{16\xi}{3n_oR}$$

also from our calculus courses we know that we can relate $\Delta r $ & $\Delta T$ as follows :

$$r \, - \, r_o = \frac{dr}{dT} (T-T_o)$$ $$ \Delta r = \frac{dr}{dT} \, \Delta T$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.