Ideal gas Law
Most recent answer: 12/05/2012
Q:
hello i want some clarifications. suppose i have an inflated balloon, obviously it is filled with air particles. and i have learn that particles colliding with walls of a container constitutes a pressure. now suppose i warm the balloon. the air particles will gain more kinetic energy and will more frequently collide with walls of the balloon thereby constituting greater pressure and ALSO the balloon's volume will increase. BUT this does not apply to Boyle's law which states that pressure is indirectly proportional to volume.Therefore since pressure increases,volume must decrease!What is wrong here?. please brighten my mind. thanks
- Bob (age 20)
Mauritius
- Bob (age 20)
Mauritius
A:
Yes, Boyle's law just describes how ideal gases behave at constant temperature, T. For your case, where T is changing, you need the full ideal gas law:
pV=NkT
where p is pressure, V is volume, N is the number of molecules, T is the absolute temperature, and k is a constant (Boltzmann's).
As a first approximation here, p in the balloon is just fixed at atmospheric pressure. N is also fixed, since the balloon doesn't leak much. So you just get V increasing proportional to T. (Remember that's absolute T.)
Most balloons also pull in more as they stretch more. That makes p inside a little bigger than in the atmosphere. As you heat up the balloon that might make V expand a little less than simply proportional to T, since p will go up a bit too.
Mike W.
pV=NkT
where p is pressure, V is volume, N is the number of molecules, T is the absolute temperature, and k is a constant (Boltzmann's).
As a first approximation here, p in the balloon is just fixed at atmospheric pressure. N is also fixed, since the balloon doesn't leak much. So you just get V increasing proportional to T. (Remember that's absolute T.)
Most balloons also pull in more as they stretch more. That makes p inside a little bigger than in the atmosphere. As you heat up the balloon that might make V expand a little less than simply proportional to T, since p will go up a bit too.
Mike W.
(published on 12/05/2012)