From Newsgroup: sci.physics
Christopher Howard wrote:
Hi, I tried to work something out on paper, but the result seems counter-intuitive, so I'm wondering if my mathematical formulation is
wrong, or just my understanding of the concepts.
In the way of Linus Torvalds:
Talk is cheap. Show your calculations.
Assume we have pumped some mass of an ideal gas into a very strong
container, i.e., no change in volume possible.
OK.
And assume the container is such a good heat insulator that loss of heat
to the environment is negligible.
OK.
But say that we are able to add heat to the container, maybe
through an electric heating element inside.
OK.
And lets say I apply a steady amount of heat transfer, say 80 watts.
So the increase of heat is 80 J per second.
The pressure will increase.
True. With constant volume, the temperature will increase according to
reaQ = m c_V reaT rco reaT = reaQ/(m c_V),
where Q is heat; m is the mass of the gas, and c_V is its specific heat capacity at constant volume.
And since the equation of state (EOS) of an ideal gas, also known as "Ideal
Gas Law", can be written
p V = N k_B T,
if the volume V and number of particles N are constant, and the absolute temperature T increases, then the pressure p will increase proportionally to the change in T, therefore to the change in Q:
reap = (N k_B/V) reaT reY reaT reY reaQ.
(k_B is the Boltzmann constant, named after Ludwig Boltzmann, which we need only to convert between energy and absolute temperature.)
My question: will the rate of pressure change, over time, be dependent
on how great a mass of gas I have originally pumped into the container?
Yes, as you can see above, given a constant volume, the greater the mass of
the gas, the less the temperature will change if the heat changes by the
same amount.
You can understand this by considering that the temperature of a substance
is a measure that we invented for the /average/ kinetic energy of a particle
of that substance (without necessarily knowing that at the time). For an
ideal gas, the conversion is proportional to the degrees of freedom of a gas particle because by the equipartition theorem the kinetic energy is
distributed evenly among the degrees of freedom:
ri-E_kri- = (f/2) k_B T rco T = 2 ri-E_kri-/(f k_B) rcA reaT = 2 reari-E_kri-/(f k_B).
[Atomic or elementary particle (e.g. photon, electron) gas:
3 translations rcA f = 3;
diatomic gas: 3 translations; 3 possible, but only 2 relevant
rotations; 1 possible way of vibration rcA f = 5 or 6;
triatomic gas: 3 translations, 3 rotations, up to 2 ways of vibration
rcA f = 7 or 8.]
If the mass of the gas is larger than some other mass, then there are two possibilities that can be combined: there are more particles or the mass of each particle is larger.
If there are more particles of the same gas at *the same* temperature, then
the same additional energy needs to be distributed among more particles
(this happens by them colliding with each other or, actually, repelling each other electromagnetically before they can collide), so the /average/ kinetic energy of a particle does not increase as much.
I.e., will my pressure gauge needle swing more slowly if I have 10 kg of
the gas in there, as opposed to 1 kg?
Yes, because with the same gas, for a larger mass you need more particles,
so you need to add more energy/heat than before so that the average kinetic energy of a particle increases as much to correspond to an equal change in absolute temperature which, in turn, with a constant value is proportional
to an equal change in pressure.
See also:
<
https://www.feynmanlectures.caltech.edu/I_01.html>
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PointedEars
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