An open vessel contains 200 mg …
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Govind Singh 7 years, 11 months ago
Approach: We have PV=nRT
n can be rewritten as {tex}\frac{m}{M}{/tex}, where m is the weight of the gas while M is the molecular mass of the gas. Thus:
{tex}PV = \frac{m}{M}RT{/tex}
The only two variables changing are m and T
Substituting values, we have:
{tex}PV = \frac{{0.200}}{M}R\left( {290} \right){/tex}
{tex}PV = \frac{m}{M}R\left( {390} \right){/tex}
Where m is the mass of the gas remaining in the vessel. Dividing the two equations:
{tex}\frac{{390}}{{290}} = \frac{{0.200}}{m}{/tex}
m = 149mg
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.
0Thank You