Answer 2:
A balloon is an elastic body filled with gas.
To a first approximation, the gas (usually air)
in the balloon can be modeled as an ideal gas
(this model assumes that the atoms or molecules
with in the gas do not interact with each
other). The ideal gas equation relates the
Pressure (P), Volume (V), amount (n usually
measured in moles), and Temperature (T measured
on an absolute scale in Kelvin) to the gas law
constant R in the following manner:
P*V=n*R*T.
If you blow up a balloon and
tie it off, you have enclosed some amount of gas
(n) at some pressure (Pi), volume (Vi), and
Temperature (Ti) such that:
(Pi*Vi)/(n*Ti)=R.
If
you now heat the gas up (Tf is greater
than Ti),
the new
pressure (Pf) and volume (Vf) can be related to
the gas constant as:
(Pf*Vf)/(n*Tf)=R
where n is
the same since the balloon is tied off and you
did not change the amount of gas inside the
balloon. Now, the gas constant is just a number
and does not change so you can equate the two
equations as:
(Pi*Vi)/(n*Ti)= (Pf*Vf)/(n*Tf)
which can be rearranged by algebra to give:
(Pi*Vi)/(Pf*Vf)=Ti/Tf
and the n drops out because n/n=1.
Since Ti is less than Tf,then that means
Pi*Vi
is less than Pf*Vf.
Now, since we know that as the pressure
increases, so too must the volume (in other
words, when you blow up a balloon (increase the
pressure), the balloon gets bigger (volume
increases), we then know that Pi is less
than Pf
and Vi is less than Vf.
Since the volume of the balloon is getting
bigger (Vf is less thanVi) then we know
that the balloon
must have expanded meaning that the
circumference of the balloon must also have
increased. In summary: as you heat a gas, it
expands. In a balloon, this expansion would
increase the circumference of the balloon. Also,
by the same logic the reverse is true: as you
cool a gas, it contracts. In a balloon, cooling
the gas inside would cause the circumference to
decrease.
