Physics 3 (P3) - Gas Law (explained)

Physics 3 (P3) - Gas Law (explained); inc. calculation examples.

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Physics - P3Sophie Mitchell
Physics ­ Gas Law
Gay-Lusacs law, part of the combined ideal gas law ((P1V1/T1) = (P2V2/T2) ­ where V1=V2=V), states
that temperature is relative to pressure (in a constant volume) in the sense that as temperature (in
Kelvin (K)) goes down, pressure goes down ( ). They relate to the movement of particles inside
of the container (given volume) ­ and the energy that they gain from heat. As the heat energy is
absorbed by a particle, it is converted into kinetic energy, increasing the particles velocity. These
particles, moving in random motion, collide with the walls of the container, exerting a force with
every collision - this force will be greater or lesser due to the velocity of the particle (F=m x a). It is
this force against the walls of the container that affect the pressure (pressure = force/area); as force
is increased, and/or the area the force is acting upon is decreased, pressure inside a constant volume
will increase.
( ^ A diagram to show temperatures effect on the rate of collisions per given time changes with
temperature and pressure)
On a graph showing its proportional relation, data can be extrapolated to the origin of the graph,
where temperature would be 0 Kelvin (K) which is equal to -273oc ­ (all temperatures relating to
Boyles law and in fact the ideal gas law are measured in K ­ Kelvin is just 273 more than oc, eg: 0oc =
273K) , At this temperature, pressure would theoretically be around or exactly 0.
In a container of helium gas, at room temperature, (around 21 oc which is 294K), the molecules of
helium within will be moving around in random motion, absorbing 39881.1105 joules of heat energy
(1oc = 1899.1005 joules) , which when converted into KE, is equal. We will assume forces exerted on
the inside of the container are equal to the reaction forces given by normal air pressure ­ 100kPa.
Volume is constant; we'll say V = 100cm3.
As the container is cooled, there will be less heat energy to be converted into kinetic energy to the
particles, which reduces the rate of collisions between the inside of the container and the particles,
and the force exerted by each impact. In turn, this reduces the pressure. At 283K (10oc) pressure
would be worked out like so;
(P1V1/T1) = (P2V2/T2) ­ where V1=V2=V.
P1 = 100kPa P2 = ?
T1 = 294K T2 = 283K

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Physics - P3Sophie Mitchell
(100 x V1)/294 = (P2V2)/283
P2 = (100 x 283)/294 = 96.2585034 =96.26 kPa(2dp)
As it is cooled further, the particles will convert less energy due to availability, and therefore have a
slower velocity (V = (KE/½M)) reducing the force on impact and the rate of the collisions over a
given time. A gradual decrease in pressure is shown with the decrease in temperature, as continued
below, shown also when looking at the graph show its proportional relationship.…read more

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Physics - P3Sophie Mitchell
We could use this data to extrapolate the line of best fit to the origin, to theoretically show that; jJust
like previously mentioned, if the container was cooled to absolute zero ­ it would technically have no
pressure, as the particles would have no heat energy to transfer into kinetic energy, and therefore
no velocity ((V = (0/½M) = 0))
0K -
(P1V1/T1) = (P2V2/T2) ­ where V1=V2=V.…read more

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