- Joined
- Apr 17, 2016
- Sythe Gold
- 121
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I was pumping my bicycle tire the other day when I came to think of how much energy I would save if I used helium instead of air whilst biking. I decided to work it out.
The idea here is that you need to figure out what volume of gas held at STP conditions is needed in order for the tire to have a volume of 1.35 L at 8.5 atm and 20.0∘C.
Since pressure, temperature, and volume change, you can use the combined gas law equation to find the volume of gas at STP.
The combined gas law equation looks like this:
P1V1/T1 = P2V2/T2, where P1V1T1 = the pressure, volume, and absolute temperature of the gas at an initial state
and P2V2/T2 the pressure, volume, and absolute temperature of a gas at a final state.
So, STP conditions are defined as a pressure of 100 kPa and a temperature of 0∘C. To convert the pressure to atm and the temperature to Kelvin, use the conversion factors.
1 atm = 101.325 kPa
T[K]=t[∘C]+273.15
You're starting with the gas under STP conditions, then changing its temperature to 20.0∘C and its pressure to 8.5 atm.
Rearrange the combined gas law equation to solve for V1
P1V1T1=P2V2T2⇒V1=P2P1⋅T1T2⋅V2
Plug in your values to get:
8.5 atm/(100/101.325 atm) *(273.15K + 0) K / (273.15 + 20.0) K * 1.35 L = V1 = 10.834 L, which we round up to 11 L.
1 m^3 ofweighs 1.3 kg, therefore 1 litre of air weighs 1.3g ==> 1 dm^3 = 1 l = 1 m^3/1000
1.3g * 22 = 28,6 g
1 m^3 of helium weighs 0.1785 kg, therefore 1 litre of helium weighs 0.1785 g -II-
0.1785 * 22 = 3,927 g
28,6 - 3,927 * 10^-3 = 0.024673 kg
0.024673 kg * 9.82 = 0,24228886 N
I bike 3 km everyday. Let's see how much I'd save biking a whole year.
W = F * s
3000 * 0,24228886 N = 726,9 J
726,9 J * 365 = 265 kJ
In conclusion: I'd save 265 thousand Joule if I were to use helium instead of air in my tires. This is of course a rough estimate.
What do you guys say? Should I start using helium in stead of air?
The idea here is that you need to figure out what volume of gas held at STP conditions is needed in order for the tire to have a volume of 1.35 L at 8.5 atm and 20.0∘C.
Since pressure, temperature, and volume change, you can use the combined gas law equation to find the volume of gas at STP.
The combined gas law equation looks like this:
P1V1/T1 = P2V2/T2, where P1V1T1 = the pressure, volume, and absolute temperature of the gas at an initial state
and P2V2/T2 the pressure, volume, and absolute temperature of a gas at a final state.
So, STP conditions are defined as a pressure of 100 kPa and a temperature of 0∘C. To convert the pressure to atm and the temperature to Kelvin, use the conversion factors.
1 atm = 101.325 kPa
T[K]=t[∘C]+273.15
You're starting with the gas under STP conditions, then changing its temperature to 20.0∘C and its pressure to 8.5 atm.
Rearrange the combined gas law equation to solve for V1
P1V1T1=P2V2T2⇒V1=P2P1⋅T1T2⋅V2
Plug in your values to get:
8.5 atm/(100/101.325 atm) *(273.15K + 0) K / (273.15 + 20.0) K * 1.35 L = V1 = 10.834 L, which we round up to 11 L.
1 m^3 ofweighs 1.3 kg, therefore 1 litre of air weighs 1.3g ==> 1 dm^3 = 1 l = 1 m^3/1000
1.3g * 22 = 28,6 g
1 m^3 of helium weighs 0.1785 kg, therefore 1 litre of helium weighs 0.1785 g -II-
0.1785 * 22 = 3,927 g
28,6 - 3,927 * 10^-3 = 0.024673 kg
0.024673 kg * 9.82 = 0,24228886 N
I bike 3 km everyday. Let's see how much I'd save biking a whole year.
W = F * s
3000 * 0,24228886 N = 726,9 J
726,9 J * 365 = 265 kJ
In conclusion: I'd save 265 thousand Joule if I were to use helium instead of air in my tires. This is of course a rough estimate.
What do you guys say? Should I start using helium in stead of air?
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