From Newsgroup: sci.physics

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.

Assume we have pumped some mass of an ideal gas into a very strong
container, i.e., no change in volume possible. And assume the container
is such a good heat insulator that loss of heat to the environment is negligible. But say that we are able to add heat to the container, maybe through an electric heating element inside.

And lets say I apply a steady amount of heat transfer, say 80 watts. The pressure will increase. 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? 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?

I've been assuming also that the specific heat of the gas remains the
same throughout the whole process, which I believe is a justifiable
assumption under practical circumstances (...?)
--
Christopher Howard
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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>
--
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.
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From Newsgroup: sci.physics

Christopher Howard writes:

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? 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.
--
John Hasler
john@sugarbit.com
Dancing Horse Hill
Elmwood, WI USA
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From Newsgroup: sci.physics

Thomas 'PointedEars' Lahn <PointedEars@web.de> writes:

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.

Intuitively, it makes sense to me that if I have a greater mass of gas involved, therefore the pressure will increase more slowly with the same
amount of applied heat.

I can see certainly that temperature will increase more slowly, because,
as you point out, reaT = reaQ/(m c_V).

However, in the equation p V = N k_B T, the number of particles, N, does
change with an increase in mass, correct? I think if we assumed we were
dealing with air, with a molar mass of about 30 g/mol, or 3x10rU+-# kg/mol, then the equation converts to this, correct?:

M = N |u 3x10rU+-# kg/mol
N = M / (3x10rU+-# kg/mol)

+op = (M / (3x10rU+-# kg/mol)) |u (R / V) |u +oT
+op = (M / (3x10rU+-# kg/mol)) |u (R / V) |u (reaQ / M c_air)

And therefore the mass terms (M) cancel out, correct...?
--
Christopher Howard
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From Newsgroup: sci.physics

Christopher Howard writes:

And therefore the mass terms (M) cancel out, correct...?

Leaving you with mols.
--
John Hasler
john@sugarbit.com
Dancing Horse Hill
Elmwood, WI USA
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From Newsgroup: sci.physics

Christopher Howard <christopher@librehacker.com> wrote or quoted:

However, in the equation p V = N k_B T, the number of particles, N, does >change with an increase in mass, correct? I think if we assumed we were >dealing with air, with a molar mass of about 30 g/mol, or 3x10rU+-# kg/mol, >then the equation converts to this, correct?:

So, the ideal gas law is,

pV=nRT (1).

The total mass m is the number of moles n times the molar mass M,

m = nM (2) or
n = m/M (2').

(2') inserted into (1) gives:

pV=m/M RT (3) or
p=m/(MV) RT (3') or
p/T=m/(MV) R (3'').

The definition of the molar heat capacity at constant volume C_v is,

dQ = n C_v dT (4) or
dQ = m/M C_v dT (4' by 2') or
dT = dQ/(C_v m/M) (4'').

Taking the derivative of (3') with respect to time gives,

dp/dt = m/(MV) R dT/dt (5)

On the other hand, (4) gives by differentiation with respect to t,

dQ/dt = n C_v dT/dt (6) and
dT/dt = dQ/dt/(n C_v) (6').

(5) with (6') gives,

dp/dt = m/(MV) R dT/dt (5)
= m/(MV) R dQ/dt/(n C_v) (7) using (6')
= m/(MV) R dQ/dt/(m/M C_v) (7') using (2')
= 1/V R dQ/dt/(C_v)
= R/(V C_v) dQ/dt (7''),

meaning the pressure rises with the heat with no dependency on m!

I started with (1), (2), and (4) and did some calculations,
arriving at (7''). But since I am not an expert in this field,
I am now not able to judge whether my result (7'') makes sense!


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From Newsgroup: sci.physics

ram@zedat.fu-berlin.de (Stefan Ram) wrote or quoted:

I started with (1), (2), and (4) and did some calculations,
arriving at (7''). But since I am not an expert in this field,
I am now not able to judge whether my result (7'') makes sense!

Ok, maybe one can make sense this way:

If the mass difference is due to more particles,
they hit the walls less hard, but more often.

If the mass difference is due to heavier particles,
they hit the walls slower, but with more mass.


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