Thermodynamics - mass of gas expelled

Question

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.

Answer 1

I think your reasoning is solid and you have a correct approach.

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 $$V \propto T$$

Now another way of reading the intent of the question setter is related to Gay Lussac's law of gas volumes for reaction...."The ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers." ... and therefore the volume of gas is proportional to the 'quantity of gas' (at constant pressure) and the volumne of gas is proportional to the mass of gas. Looking at the wikipedia page of about Gay Lussac's law it is very closely linked with Avagadro's Law that equal volumes of gas contain the same number of particles at the same pressure and temperature.