# What do you mean by 1 mol gas occupies 22.4 litres

by jd12345
Tags: litres, occupies
 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?
 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 ;)
 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: $$1 \, \mathrm{atm} \frac{22.4 \, \mathrm{dm}^3}{V}$$
 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...
P: 3,014
 Quote by RocketSci5KN 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...
No, it won't. The pressure inside an inflated balloon is larger than the exterior pressure, due to the elastic tension in the material from which the balloon is made of.
 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.
 P: 3,014 How did you estimate that the pressure difference is negligible? The Young's modulus for rubber is between 0.01-0.1 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 $\sigma$: $$\sigma = \frac{R}{t} \, \Delta p \Leftrightarrow \Delta p = \frac{t}{R} \, \sigma$$ 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. $\epsilon \sim 1$. Therefore: $$\Delta p \sim \frac{10 \times 10^{-3} \, \mathrm{in}}{5 \, \mathrm{in}} \, \left( 0.01 - 0.1 \, \mathrm{GPa} \right) = 0.02 - 0.2 \, \mathrm{MPa} = 0.2 - 2 \, \mathrm{atm}$$ I think this is exactly of the same order of magnitude as the atmospheric pressure.
 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).
Mentor
P: 22,297
 Quote by Dickfore How did you estimate that the pressure difference is negligible?
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 have had!), the max pressure we can generate is about 1psi, so you're an order of magnitude high with your calculation: http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1501025/
P: 3,014
 Quote by russ_watters 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 have had!), the max pressure we can generate is about 1psi, so you're an order of magnitude high with your calculation: http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1501025/
 In men, maximum expiratory pressure increased with volume from 63 to 97 cmH2O and maximum inspiratory pressure decreased with volume from 97 to 39 cmH2O.
97 cm H20 refers to a pressure difference of:
$$1000 \, \mathrm{kg} \cdot 9.81 \, \frac{\mathrm{m}}{\mathrm{s}^2} \cdot 0.97 \, \mathrm{m} = 9.52 \times 10^3 \, \mathrm{Pa} \times \frac{1 \, \mathrm{atm}}{101325 \, \mathrm{Pa}} = 9.4 \times 10^{-2} \, \mathrm{atm}$$

That is 9.4% above atmospheric pressure.

Also, see:

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:
$$\frac{101325 \, \mathrm{Pa}}{13,600 \, \mathrm{kg} \cdot \mathrm{m}^{-3} \times 9.81 \, \mathrm{m} \cdot \mathrm{s}^{-2}} = 0.759 \, \mathrm{m} = 759 \, \mathrm{mm} \, \mathrm{Hg}$$
Thus, this is excess pressure of:
$$\frac{810 - 840}{759} \, \mathrm{atm} = 1.07 - 1.107 \, \mathrm{atm}$$

I guess it is a matter of circumstance whether 10% is considered negligible or not.
P: 23,578
 Quote by Dickfore That is 9.4% above atmospheric pressure.
Which is way lower than your initial estimate

 Quote by Dickfore Thus, this is excess pressure of: $$\frac{810 - 840}{759} \, \mathrm{atm} = 1.07 - 1.107 \, \mathrm{atm}$$

If anything, 810-760 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.
P: 3,014
 Quote by Dickfore That is 9.4% above atmospheric pressure.
 Quote by Borek Which is way lower than your initial estimate
 Quote by Dickfore How did you estimate that the pressure difference is negligible? The Young's modulus for rubber is between 0.01-0.1 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 $\sigma$: $$\sigma = \frac{R}{t} \, \Delta p \Leftrightarrow \Delta p = \frac{t}{R} \, \sigma$$ 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. $\epsilon \sim 1$. Therefore: $$\Delta p \sim \frac{10 \times 10^{-3} \, \mathrm{in}}{5 \, \mathrm{in}} \, \left( 0.01 - 0.1 \, \mathrm{GPa} \right) = 0.02 - 0.2 \, \mathrm{MPa} = 0.2 - 2 \, \mathrm{atm}$$ I think this is exactly of the same order of magnitude as the atmospheric pressure.
My lower estimate is only about twice as big as the other result. I would say that is not bad for an order of magnitude estimate :SMILE:

There are at least 3 issues with the calculation:
• There might be a numerical factor in the equation relating latteral radial pressure difference, and sheer stress. I am not sure what this factor is.
• The thickness I used ($10 \, \mathrm{mil} = 254 \, \mathrm{\mu m}$) was taken from some ad for ballons used to isolate clean rooms! It might be too big!
• Finally, I used the quoted values for Young's modulus in Wikipedia. Rubber behaves in an atypical fashion that is not in exact coincidence with Hooke's Law. That is why the range of values.

Regardless, even a 9.4% increase in pressure would lead to a $1 - 1/1.0094 = 0.0093 = 9.3 \%$ 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 $\approx 20 \, \mathrm{dm}^3/\mathrm{mol}$

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