The problem statement, all variables and given/known data

There is a perfectly spherical balloon with surface painted black. It is placed in a perfect vacuum. It is gently inflated with an ideal mono-atomic gas at Kelvin temperature $T_i$, slowly enough so that it reaches thermal equilibrium with the gas, and then it is sealed off. It has radius $r_i$ at this time and contains N atoms. The vacuum is large, so radiation from its walls can be ignored.

a) Show that $\frac{T}{T_i}= \left(\frac{r}{r_i}\right)^3$ if the pressure inside the balloon is independent of its radius.

b) How much energy does the balloon radiate per second when it is at radius $r$? Express your answer in terms of $r$ and constants.

c) What is the rate of change of the internal energy of the gas? Express you answer in terms of $r$, $r˙$, and constants.

Relevant equations:

$$PV = nRT$$

$$J * A = σT^4 (4\pi r^2)$$

The attempt at a solution:

I already showed a), so I don't need help with that. In part b), I wrote pretty much $J * A = σT^4 (4\pi r^2)$ but I am unsure how to get a speed of radiation. I also have no idea how to do part c), so any help on parts b) or c) is welcome. Thanks!


closed as too localized by Qmechanic Feb 15 '13 at 15:39

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.


I don't know what J*A is, so i'm not exactly sure which form of Stefan's law you're using. I think that if you use the following form, you should be fine: $$\frac{\mathrm dQ_b}{\mathrm dt}=A\sigma T^4$$ Here, $\mathrm Q_b$ is the radiant energy emitted by a perfectly black body in a small time $\mathrm dt$, so $\frac{\mathrm dQ_b}{\mathrm dt}$ is the rate of emission of energy. $A$ is the surface area of balloon.

I think that b and c become pretty easy once you use this formula. Fortunately, we have a vacuum, so we don't have to consider energy absorbed or work done against vacuum.

  • $\begingroup$ Just out of interest, Is $J$ some sort of heat-current-density? It looks like emissive power to me.. $\endgroup$ – Manishearth Mar 13 '12 at 5:15
  • $\begingroup$ To clarify, Q is the energy? Not the heat? Then what I wrote for part b is basically correct? Also for part c, the internal energy shouldn't be the same as the emitted energy right? $\endgroup$ – Eric Mercer Mar 13 '12 at 5:20
  • $\begingroup$ @EricMercer: By first law of thermodynamics, $Q=U+w$. $Q$ is heat, $U$ is internal energy. $w$ is work, but it is zero here because its a vacuum. So here, heat lost=decrease in internal energy. So they're the same here. $\endgroup$ – Manishearth Mar 13 '12 at 5:28
  • $\begingroup$ I guess what you did for b is correct. $dQ/dt$ is the amount of energy radiated per second. For c, since $Q=U$, their corresponding rates are the same as well (there may be a $\pm$, though). $\endgroup$ – Manishearth Mar 13 '12 at 5:29
  • $\begingroup$ If it helps, the $Q$ used in the first law of thermoD is $-Q_b$ here (since I took energy emitted) $\endgroup$ – Manishearth Mar 13 '12 at 5:31

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