I know it's a dumb question but I am having a little misunderstanding with Boyle's Law. Shouldn't pressure be inversely proportional with Volume , or this is meant for the pressure outside of gases outside the balloon,can you use another example for this relation .

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    $\begingroup$ Boyle's law is assuming that the quantity of gas (in moles or atoms) remains constant. Is this precondition met in the case you are considering? $\endgroup$ – dmckee --- ex-moderator kitten Sep 29 '18 at 18:52
  • $\begingroup$ Oh surely its not ,Sorry I forgot about this $\endgroup$ – user597368 Sep 29 '18 at 18:55

The general relation for an ideal gas is $$PV=nRT$$ where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles of the gas particles, $R\approx 8.314 \mbox{ J/ mol}\cdot\mbox{K}$ is the gas constant and $T$ is the temperature in Kelvin.

As you inflate a balloon at constant temperature, $T$ remains constant while $n$ increases because you are adding gas with your lungs, and $V$ increases because it is inflating. Your question is what happens to the pressure $P$?

The answer is... the gas equations doesn't give you enough information to figure this out. Since $RT$ is constant you have $PV/n=\mbox{const.}$. But both $P$ and $V$ are allowed to vary so you need more information to answer this than just the gas law alone.

However, you know that once you've blown into the balloon, the system will come to equilibrium, which means that forces on the balloon are balanced. You have pressure inside the balloon pushing out, and this must be balanced by the outside air pressure pushing in PLUS the force of the elastic which is also pushing in. The more the balloon is inflated, the more force the elastic will apply (think Hooke's law). Outside air pressure can be taken as constant. Therefore pressure will increase as you inflate the balloon, because the force from the elastic grows as the balloon is inflated.

No need to invoke the gas law at all. In fact, the gas law can't save you here!

  • $\begingroup$ Yes I get it now. this case is dependent of numbers to know exactly what will happen, so a real example for Boyle's would be that the balloon already has a certain amount of gas and I increase outer pressure then its volume will decrease but There is a question if I do increase the pressure of gases inside the balloon somehow its volume will increase ,wouldn't it? $\endgroup$ – user597368 Sep 29 '18 at 19:12
  • $\begingroup$ It's a little hard to increase the pressure of a system without changing anything else (the gas law says that if $V$, $n$, and $T$ all remain constant, then so does $P$). But either way, $P$ and $V$ go like inverses so a rise in pressure (assuming $T$ and $n$ are fixed) corresponds to a drop in volume. This makes perfect sense, if I squeeze my balloon (shrink the volume) - the pressure will go up. $\endgroup$ – bRost03 Sep 29 '18 at 19:20
  • $\begingroup$ Perhaps you are thinking about when the balloon is out of equilibrium, so the pressure in the balloon is "too high". Like if we let go after squeezing the. When we let go, the pressure in the balloon is "too high" and the volume will increase. But when $V$ increases, $P$ will decrease - precisely because the pressure was "too high". It's wants to get to lower pressure, and it can do this by increasing it's volume. Or if you like, it could also do it by lowering the temperature (this is the basis for refrigerators), or ditching some molecules (like when you crack open a soda). $\endgroup$ – bRost03 Sep 29 '18 at 19:28
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    $\begingroup$ Blowing into a balloon is hardest at the start. And after that it gets easier and easier. It must be related to decreasing thickness of material as it streches. $\endgroup$ – physicsguy19 Sep 29 '18 at 19:37
  • $\begingroup$ It is hardest at the start because you are changing the surface area the most rapidly, thus the force from the air pressure is growing very quickly. The first lungful of air might triple the balloons surface area, while the next lungful may only increase it by 50% (totally made up numbers, but the point stands). It is not related to the stretching of the rubber. Try holding a deflated balloon and just stretching it with your hands, easiest at the beginning then harder the more you stretch it :) $\endgroup$ – bRost03 Sep 29 '18 at 19:42

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