Gas Leak

Chemistry Level 2

Three identical boxes each contain a different gas at the same temperature and pressure. The first box contains oxygen gas, the second box contains carbon dioxide, and the third box contains water vapor. At the same time, a small hole is pricked through each of these boxes so that the gases inside them begin to leak out. Which box will be the first to completely leak out?

The second box Two of the boxes at the same time The third box All of the boxes at the same time The first box

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2 solutions

Steven Yuan
Jan 11, 2018

Graham's law of effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. In simple terms, the lower the molar mass of the gas, the faster the gas will leak out. Among the three gases in the boxes, water vapor has the lowest molar mass at around 18 grams per mole, compared to oxygen's 32 grams per mole and carbon dioxide's 44 grams per mole. Therefore, the third box will be the first to leak out completely.

Graham's law of effusion is a chemistry law no classical mechanics.

You should change the topic of problem into chemistry.

Matin Naseri - 3 years, 5 months ago

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I'm debating whether to make it physics or chemistry, since technically the law can be derived from kinetic energy. I'm gonna change it to chemistry until someone objects.

Steven Yuan - 3 years, 5 months ago

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Temporary solution.

Matin Naseri - 3 years, 5 months ago

As they all have same pressure p p and temperature T , T, if we consider them as perfect gases, the only variables that can change in their state equation are volume V , V, mass m , m, and molar mass M . M. According to the state equation for perfect gases the density of the gas will be

p V = m M R T m V = ρ = p R T M = α M . pV = \frac{m}{M} RT \Rightarrow \frac{m}{V} = \rho = \frac{p}{RT} M = \alpha M.

If we prick a hole through the box and the external pressure is p 0 p_0 (also the same for all situations, I'm assuming), then according to Bernoulli's formula

p = p 0 + 1 2 ρ 0 v 2 , p = p_0 + \frac{1}{2} \rho_0 v^2,

where v v is the velocity the gas will scape the box and ρ 0 \rho_0 is the density of the gas outside the box. The density outside the box may also be found by using the perfect gas state equation. For a particular gas p V = n R T , pV = nRT, leads us to p = β ρ . p = \beta \rho. So

p 0 ρ 0 = p ρ ρ 0 = ρ p p 0 = M α p p 0 . \frac{p_0}{\rho_0} = \frac{p}{\rho} \Rightarrow \rho_0 = \rho \frac{p}{p_0} = M \frac{\alpha p}{p_0}.

So for the velocity we'll get

v = 2 M ( p p 0 ) α ( p / p 0 ) . v = \sqrt{\frac{2}{M} \frac{(p - p_0)}{\alpha (p/p_0)}}.

Therefore the gas with least molar mass will be the first to completely leak out. That would be water vapor.

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