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If I were to connect a bettery to the two sides of a container that contain only hydrogen gas, and its not dense (it's pressure is one atmosphere), what would it's resistance over the length of the container be?

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  • \$\begingroup\$ What's the container made of? ;-) \$\endgroup\$
    – Oli Glaser
    Jul 14 '12 at 14:32
  • \$\begingroup\$ a non conductive metirial, I want the electrons to only move through the hydrogen \$\endgroup\$ Jul 14 '12 at 15:00
  • \$\begingroup\$ Yep, I was just kidding around... \$\endgroup\$
    – Oli Glaser
    Jul 14 '12 at 16:02
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Depends on the container's dimensions. Specific resistance is often expressed in Ω cm, which means that a cube with 1 cm sides of a 1 Ω cm material will have a 1 Ω resistance between opposite sides.

At 1 atmosphere a gas is mostly empty space, with here and there a molecule. I don't think that hydrogen under those conditions will be much different from other gases (possible with the exception of noble gases). I expect a value of at least several 100 MΩ cm, probably a couple of GΩ cm.

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  • \$\begingroup\$ so my only parameters are the length of the container (the distance between the cathode and the anode) and the pressure? by your answer I can assume that the length affects the resistance alot, but how much does the pressure affects the resistance? I mean is I increase the length by 1 cm the resitance will increase by a few 100MΩ, but if I increase the pressure by 1 atmosphere, will it increase that much? \$\endgroup\$ Jul 14 '12 at 15:08
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    \$\begingroup\$ @Someonation - It's the length but also the height and width! Twice as long = twice the resistance. Twice as high or wide = half the resistance. So for a container 1 m long, 3 cm high and 5 cm you would have to calculate 100 M\$\Omega\$ cm x 100 cm / (3 cm x 5 cm). Then the unit is right: \$\Omega\$ (what they call dimensional analysis). If you increase pressure the resistance will decrease, but I have no idea how the two are related exactly. \$\endgroup\$
    – stevenvh
    Jul 14 '12 at 15:15
  • \$\begingroup\$ oh I understand... just to be clear, 100MΩ is an an assumption, if I want the exact number I need to make a cube container that is a 1cm*1cm*1cm and measure it's resistance right? \$\endgroup\$ Jul 14 '12 at 16:45
  • \$\begingroup\$ @Someonation - Right. I used it as an example for the calculation. Like I said may be much higher. At 100 V a 1 GΩ resistance (again an example) will give you 100 nA current, which is measurable as 100 mV across a 1 MΩ series resistor. Keep in mind that at these high resistances you'll have leakage currents everywhere, so make sure that you actually measure the current through the gas. \$\endgroup\$
    – stevenvh
    Jul 14 '12 at 16:57
  • \$\begingroup\$ by leakage currents you mean that the container's resistance is not much bigger than the gas's so some untrivial amount of current will go through the container? \$\endgroup\$ Jul 14 '12 at 17:05

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