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What do you mean by 1 mol gas occupies 22.4 litres 
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#1
Apr3012, 07:23 AM

P: 260

Avagadro law states that 1 mol of any gas( ideal) occupies 22.4 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 think it spreads out in the whole room. So what does the law mean? 


#2
Apr3012, 08:01 AM

P: 186

This law states at STP (101325 pa and 273K) one mole of an ideal gas (such as oxygen) will occupy a volume of 22.4L ;)



#3
Apr3012, 08:27 AM

P: 3,014

a) cool the room to 273.15 K;
b) hermetically seal the room; c) create a high vacuum in it by pumping all the air out; d) Add 1 mole of oxygen. The pressure in the room should be given by: [tex] 1 \, \mathrm{atm} \frac{22.4 \, \mathrm{dm}^3}{V} [/tex] 


#4
Apr3012, 04:31 PM

P: 163

What do you mean by 1 mol gas occupies 22.4 litres
Or, another way of looking at it, take an ideal empty large balloon. Fill that balloon with 1 mole of any gas, and place balloon in a room at STP. The inside volume of that balloon will be 22.4 liters...



#5
Apr3012, 06:53 PM

P: 3,014




#6
Apr3012, 07:34 PM

Mentor
P: 22,297

True if its a rubber balloon, but even still the pressure caused by the rubber will tend to be vanishingly small.



#7
Apr3012, 08:46 PM

P: 3,014

How did you estimate that the pressure difference is negligible? The Young's modulus for rubber is between 0.010.1 GPa. If we have 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 (ten 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{10 \times 10^{3} \, \mathrm{in}}{5 \, \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. 


#8
Apr3012, 08:48 PM

P: 163

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 still get close to 22.4l in any case. If it makes people happier, change 'balloon' to 'plastic garbage bag' with a nominal 50 liter capacity... as you'd only fill it to 22.4l, there will be no strain on the material and P(inside) = P(outside).



#9
Apr3012, 10:08 PM

Mentor
P: 22,297




#10
Apr3012, 10:55 PM

P: 3,014

[tex] 1000 \, \mathrm{kg} \cdot 9.81 \, \frac{\mathrm{m}}{\mathrm{s}^2} \cdot 0.97 \, \mathrm{m} = 9.52 \times 10^3 \, \mathrm{Pa} \times \frac{1 \, \mathrm{atm}}{101325 \, \mathrm{Pa}} = 9.4 \times 10^{2} \, \mathrm{atm} [/tex] That is 9.4% above atmospheric pressure. Also, see: http://www.youtube.com/watch?v=fwhi0WB_bQ As they inflate the balloon, the pressure is pretty constant around 810 mm Hg, eventually increasing to 840 mm Hg. Now, standard atmospheric pressure corresponds to: [tex] \frac{101325 \, \mathrm{Pa}}{13,600 \, \mathrm{kg} \cdot \mathrm{m}^{3} \times 9.81 \, \mathrm{m} \cdot \mathrm{s}^{2}} = 0.759 \, \mathrm{m} = 759 \, \mathrm{mm} \, \mathrm{Hg} [/tex] Thus, this is excess pressure of: [tex] \frac{810  840}{759} \, \mathrm{atm} = 1.07  1.107 \, \mathrm{atm} [/tex] I guess it is a matter of circumstance whether 10% is considered negligible or not. 


#11
May112, 02:11 AM

Admin
P: 23,578

If anything, 810760 mmHg or 0.066 atm, or about 7%. But I agree that negligibility of this number is disputable. Thank you for digging this video up. 


#12
May212, 09:31 AM

P: 3,014

There are at least 3 issues with the calculation:
Regardless, even a 9.4% increase in pressure would lead to a [itex]1  1/1.0094 = 0.0093 = 9.3 \%[/itex] decrease in molar volume. 9.3% out of 22.4 is 2.08. Thus, one cannot report the result with 3 significant figures, but, instead, one would obtain something around [itex] \approx 20 \, \mathrm{dm}^3/\mathrm{mol}[/itex] 


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