what does one mole of h2o correspond to

jd12345

Avagadro law states that i mol of any gas( ideal) occupies 22.four litres.
I dont really understand it. You take one mole of oxygen and leave it in a room - will it go in a corner and occupy 22.4litres? I call up it spreads out in the whole room.
And so what does the police force hateful?

miniradman · Apr thirty, 2012

This police force states at STP (101325 pa and 273K) 1 mole of an ideal gas (such as oxygen) volition occupy a book of 22.4L ;)

Dickfore

a) cool the room to 273.15 K;

b) hermetically seal the room;

c) create a high vacuum in information technology by pumping all the air out;

d) Add together ane mole of oxygen.

The pressure in the room should exist given by:
[tex]
one \, \mathrm{atm} \frac{22.iv \, \mathrm{dm}^iii}{V}
[/tex]

RocketSci5KN

Or, another style of looking at it, accept an ideal empty large balloon. Fill up that balloon with 1 mole of any gas, and place balloon in a room at STP. The inside volume of that balloon volition be 22.four liters...

Dickfore

Or, some other way of looking at information technology, accept an platonic empty large airship. Make full that balloon with 1 mole of any gas, and place balloon in a room at STP. The inside volume of that balloon will be 22.four liters...

No, it won't. The force per unit area inside an inflated balloon is larger than the exterior pressure, due to the elastic tension in the material from which the airship is made of.

russ_watters

True if its a rubber balloon, merely even still the pressure level caused past the rubber will tend to be vanishingly pocket-size.

Dickfore

How did you estimate that the pressure deviation is negligible? The Young's modulus for rubber is between 0.01-0.1 GPa. If nosotros have a sphere with radius R, and thickness t, then a pressure deviation on both sides of the sphere translates to a strain [itex]\sigma[/itex]:
[tex]
\sigma = \frac{R}{t} \, \Delta p \Leftrightarrow \Delta p = \frac{t}{R} \, \sigma
[/tex]

A typical thickness of a balloon is ten mil (10 thousandths of an inch), and a typical inflated radius of a balloon is five inches. But, the balloons are often inflated twice their undeformed length, i.eastward. [itex]\epsilon \sim 1[/itex]. Therefore:
[tex]
\Delta p \sim \frac{10 \times 10^{-3} \, \mathrm{in}}{five \, \mathrm{in}} \, \left( 0.01 - 0.1 \, \mathrm{GPa} \right) = 0.02 - 0.2 \, \mathrm{MPa} = 0.2 - two \, \mathrm{atm}
[/tex]

I call back this is exactly of the same order of magnitude every bit the atmospheric pressure.

RocketSci5KN

I said 'ideal' balloon - this implies zero constraining forces from the material itself. P(inside) = P(outside)... Yes, in a normal balloon P(inside) slightly larger than P(outside). You'll nevertheless get shut to 22.4l in whatever example. If it makes people happier, change 'balloon' to 'plastic garbage purse' with a nominal 50 liter capacity... as you'd only fill it to 22.4l, there volition be no strain on the material and P(inside) = P(outside).

russ_watters · Apr 30, 2012

How did yous gauge that the pressure level difference is negligible?

I base of operations that on the fact that I tin can blow one up with my lungs! Based on this link (and an everyday experience nosotros all have had!), the max pressure nosotros can generate is about 1psi, so y'all're an order of magnitude high with your adding: http://www.ncbi.nlm.nih.gov/pmc/manufactures/PMC1501025/

Dickfore

I base that on the fact that I can blow one up with my lungs! Based on this link (and an everyday experience we all accept had!), the max pressure level we can generate is about 1psi, then you lot're an order of magnitude loftier with your calculation: http://world wide web.ncbi.nlm.nih.gov/pmc/articles/PMC1501025/

From the link you had provided:

In men, maximum expiratory pressure increased with book from 63 to 97 cmH2O and maximum inspiratory pressure decreased with book from 97 to 39 cmH2O.

97 cm H20 refers to a pressure difference of:
[tex]
1000 \, \mathrm{kg} \cdot 9.81 \, \frac{\mathrm{m}}{\mathrm{due south}^two} \cdot 0.97 \, \mathrm{m} = nine.52 \times 10^3 \, \mathrm{Pa} \times \frac{ane \, \mathrm{atm}}{101325 \, \mathrm{Pa}} = 9.4 \times 10^{-ii} \, \mathrm{atm}
[/tex]

That is 9.4% above atmospheric pressure.

Also, see:
http://world wide web.youtube.com/picket?v=fwh-i0WB_bQ

As they inflate the balloon, the pressure is pretty constant around 810 mm Hg, eventually increasing to 840 mm Hg. At present, standard atmospheric pressure level corresponds to:
[tex]
\frac{101325 \, \mathrm{Pa}}{13,600 \, \mathrm{kg} \cdot \mathrm{m}^{-3} \times 9.81 \, \mathrm{k} \cdot \mathrm{due south}^{-2}} = 0.759 \, \mathrm{chiliad} = 759 \, \mathrm{mm} \, \mathrm{Hg}
[/tex]
Thus, this is excess force per unit area of:
[tex]
\frac{810 - 840}{759} \, \mathrm{atm} = i.07 - 1.107 \, \mathrm{atm}
[/tex]

I guess it is a matter of circumstance whether ten% is considered negligible or not.

Borek

That is 9.4% above atmospheric pressure.

Which is way lower than your initial guess

Thus, this is backlog force per unit area of:
[tex]
\frac{810 - 840}{759} \, \mathrm{atm} = 1.07 - 1.107 \, \mathrm{atm}
[/tex]

If anything, 810-760 mmHg or 0.066 atm, or about 7%.

But I agree that negligibility of this number is disputable. Thank you lot for digging this video up.

Dickfore

That is 9.4% higher up atmospheric pressure level.

Which is style lower than your initial gauge

How did you guess that the pressure difference is negligible? The Young's modulus for safe is between 0.01-0.ane GPa. If we take a sphere with radius R, and thickness t, then a pressure difference on both sides of the sphere translates to a strain [itex]\sigma[/itex]:
[tex]
\sigma = \frac{R}{t} \, \Delta p \Leftrightarrow \Delta p = \frac{t}{R} \, \sigma
[/tex]
A typical thickness of a balloon is 10 mil (x thousandths of an inch), and a typical inflated radius of a balloon is 5 inches. But, the balloons are often inflated twice their undeformed length, i.e. [itex]\epsilon \sim 1[/itex]. Therefore:
[tex]
\Delta p \sim \frac{x \times x^{-iii} \, \mathrm{in}}{five \, \mathrm{in}} \, \left( 0.01 - 0.1 \, \mathrm{GPa} \right) = 0.02 - 0.2 \, \mathrm{MPa} = 0.2 - 2 \, \mathrm{atm}
[/tex]

I think this is exactly of the same order of magnitude as the atmospheric pressure.

My lower guess is only almost twice every bit big as the other result. I would say that is not bad for an social club of magnitude estimate :SMILE:

At that place are at to the lowest degree 3 problems with the calculation:

At that place might be a numerical factor in the equation relating latteral radial pressure divergence, and sheer stress. I am not sure what this cistron is.
The thickness I used ([itex]10 \, \mathrm{mil} = 254 \, \mathrm{\mu thousand}[/itex]) was taken from some ad for ballons used to isolate make clean rooms! Information technology might be too big!
Finally, I used the quoted values for Young's modulus in Wikipedia. Safety behaves in an atypical fashion that is not in exact coincidence with Hooke'southward Police. That is why the range of values.

Regardless, even a 9.4% increase in pressure would lead to a [itex]1 - 1/1.0094 = 0.0093 = nine.three \%[/itex] decrease in molar book. ix.three% out of 22.iv is ii.08. Thus, one cannot written report the result with 3 significant figures, merely, instead, 1 would obtain something around [itex] \approx twenty \, \mathrm{dm}^three/\mathrm{mol}[/itex]

what does one mole of h2o correspond to

What do you mean past 1 mol gas occupies 22.iv litres

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