Well allows see. Let us say girlfriend inflate theballoon at the surface ar where the push is 1 bar(105 Pa). Now in the balloon, thepressure of the gas is slightly better than onebar due to the fact that the stamin of the elastic balloon.The pressure of the gas within is slightly greaterthan 1 bar. This is obvious since if we pop theballoon the goes pop... And a press wave is sentout POP!!). Now we enable this balloon come rise. Letus first assume the environment is isothermal. Asthe balloon rises that encounters approximately atmosphereat lower pressure... Now since the now theequation that state claims that pV=mRT... Where p ispressure, V is the volume the the balloon, m is themass the gas in the balloon (constant) and also R isuniversal gas constant divided by mean molar wt ofair (about 28 g/mol) and T is thermodynamictemperature (T in Kelvin; T=273 +t (in degC).

You are watching: A balloon in a constant environment gets smaller and

So together the balloon rises come lowerpressure environment, the volume that the balloonwill broaden so the pV stays continuous as the mustfor an isothermal atmosphere.

So i think theballoon will certainly expand... In ~ some suggest when the skinthickness reaches a crucial value (as balloonexpands the thickness the balloon skin decreases)it will certainly burst because the pressure of the gasinside will certainly exceed the strength of the elasticballoon material.

P.S. The best gas regulation Ihave used has been created in type to explicitlyconsider the mass ,m in the balloon since that isconstant. And also one deserve to see exactly how the product pV mustalso over there fore continue to be constant.

Consider a helium balloon the is filled at sealevel. In ~ sea level, the external atmosphericpressure the the air is equal to 14.7lbs/in2 or 1.0135 bar or 1 atm (thosethree values are all equal just like 1 garden isequal come 0.9144 meters). Because the balloon"svolume is not changing, we recognize that the outsidepressure top top the balloon is well balanced by the airpressure of the air inside the balloon. Anotherway of saying this is that the outside andinternal pressure forces are balancing.

Nowrelease the balloon. The exterior air pressuredecreases as the balloon floats higher up in theair (assuming stagnant air and constanttemperature). This deserve to be defined by thefollowing example: as soon as you"re swimming from thebottom of a pool as much as the surface the waterpressure decreases once you gain to the surface ar ofthe water. Stagnant water push is led to bythe weight of the water from above pushing under onyou. Ago to the balloon example, in ~ sea level,the weight of the air in the setting is pushingon the balloon. The higher up the balloon go theless air there is to press down top top the balloon sothe push decreases.

When the balloon isreally high, over there is less air above the balloonthan there was at sea level - so the weight of theair above the balloon is less than in ~ sea level.The outside air press is advertise on the balloonless 보다 it was when the balloon was at sealevel. Therefore, the balloon will increase sincethere is less pressure being applied on it. So,the balloon should expand the greater up the floatsin the atmosphere.

Now, think about taking anempty balloon yes, really high up in the atmosphere andfilling the up through air. Would certainly its volume increaseor decrease as you lugged it back down come sealevel?

Hint: The same thing would take place ifyou take it an empty water party (that is filledwith air), placed its lid on, and also brought it to thebottom that a deep pool. The volume the a waterbottle filled v air would certainly _______ as it wasbrought down to the bottom the the pool.