Four gas balloons A, B, C, D of equal volumes containing $\ce{H2}$, $\ce{N2O}$, $\ce{CO}$, $\ce{CO2}$ respectively prickled with a needle and immsersed in a tank containing $\ce{CO2}$. Which of them will shrink after some time?

a) A
b) B
c) C
d) Both A and C

My attempt: I think that the ballons filled with the gases which will react with $\ce{CO2}$ will be the ones that will shrink. Now $\ce{H2}$ reacts with $\ce{CO2}$ to form $\ce{CH4}$, so the amount of $\ce{H2}$ will decrease and the balloon will shrink. Also I know $\ce{CO}$ cannot react with $\ce{CO2}$, so the answer has to be (a). I need hep determining the correct solution.

  • $\begingroup$ This is a rather unclear question (your test question, not your question so much). If they're pricked, I would expect gas to be released. If it weren't for that detail, I would assume this was about ideal gas vs. nonideal gas. Are you sure that bit was in the question? You wrote "pickled" but I am suspecting it's a typo. Also, the options are A/B/C not the gases. $\endgroup$ – user7652 Nov 18 '15 at 9:33
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    $\begingroup$ $\ce{H2}$ will not react with $\ce{CO2}$ for two reasons: (1) such a reaction requires pretty high temperatures and (2) $\ce{H2}$ is inside the balloon, while $\ce{CO2}$ is outside. And if we consider that a gas can flow out of a pierced balloon (as our common sense says), then it will be the case with all four of them, so all balloons will shrink. $\endgroup$ – Ivan Neretin Nov 18 '15 at 10:39
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    $\begingroup$ pickled balloons? Like this? $\endgroup$ – Jason B. Nov 18 '15 at 13:10
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    $\begingroup$ I have a feeling this is a poorly dressed up question asking about the relative effusivity between the gasses. $\endgroup$ – Nicolau Saker Neto Nov 18 '15 at 13:26

The relative molecular mass ($M_r$) of the gas determines the rate of diffusion of the gases from a region of higher concentration to a region of lower concentration.

The smaller the $M_r$ of the gas, the greater the rate of diffusion (movement of gas molecules).

$M_r$ of $\ce{CO2}$ = 40

$M_r$ of $\ce{H2}$ = 2 (Balloon A)
Since hydrogen gas has a smaller $M_r$ than carbon dioxide gas, more hydrogen gas diffuse out of the balloon than carbon dioxide gas into the balloon per unit time. Hence, this balloon will shrink.

$M_r$ of $\ce{N2O}$ = 44 (Balloon B)
Since nitrous oxide gas has a greater $M_r$ than carbon dioxide gas, more carbon dioxide gas diffuse into the balloon than nitrous oxide gas out of the balloon per unit time. Hence, this balloon will inflate

$M_r$ of $\ce{CO}$ = 28 (Balloon C)
This balloon will shrink (same logic as $\ce{H2}$).

$M_r$ of $\ce{CO2}$ = 40 (Balloon D)
Since the balloon and the tank both contains $\ce{CO2}$ (same $M_r$ of gases), there is no net diffusion of gases in/out of the balloon, thus the balloon will not inflate/shrink (theoretically).


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