# Thermodynamics - mass of gas expelled

Question:

An open vessel contains 200 mg of air at 17 degrees Celsius. What percentage of weight of air would be expelled if the vessel is heated to 117 degrees Celsius if the pressure is to remain constant:

Approach: We have $PV = nRT$

$n$ can be rewritten as $m\over{M}$, where $m$ is the weight of the gas while $M$ is the molecular mass of the gas. Thus: $$PV = \frac{m}{M}RT$$ The only two variables changing are $m$ and $T$

Substituting values, we have: $$PV = \frac{0.200}{M}R(290)$$ $$PV = \frac{m}{M}R(390)$$ Where $m$ is the mass of the gas remaining in the vessel. Dividing the two equations: $$\frac{390}{290} = \frac{0.200}{m}$$ $$m = 149 mg$$

From here, the percentage can be calculated easily. Is there any problem with this approach? Or is there any simpler approach that can be used. I am actually supposed to solve this question using Gay-Lussac's Law.

I think the question was wanting you to use the expression at constant pressure $$V \propto T$$
This would give you the same answer because the ideal gas law $$PV = nRT$$ can be rearranged to give $$V = {nR\over P}T$$ which at constant pressure gives $$V = const.~T$$ which is equivalent to $$V \propto T$$