A mixture of 10.0 g of Ne and 10.0 g of Ar was kept in a 1.00 L container at 765 torr. Calculate the partial pressure of Ar inside the container.
the two gases are occupying the same volume at the same temperature. according to the ideal gas law, what do you think would be the ratio of their partial pressures? if we know the ratio of partial pressures, then we can invoke (dalton's) law of partial pressures to find their actual values.
have absolutley no idea
is there something you have some idea on? we can start from there.
no i don't ...i have read the chapter, have had the hardest time figuring this stuff out and my final is tomorrow..which it's safe to say i won't be passing. At this point i'm studying just so i feel better of at least having seen the material
should we try ideal gas law? pv=nrt?
ok guess not
that's right. v and t are the same for both gases. what's p(Ar)/p(Ne)?
There are two volumes though? so how would i set up the equation?
they are in the same container. so the volume of Ar = volume of Ne. each gas fills the vessel completely.
Lets look at the situation: We define the partial pressure for a gas (ideal or real) J in a mixture as: \[pJ=x _{j}*p\] where xj is the mole fraction: \[x _{j}=\frac{ n _{j} }{ n }\] and n equals the total amount of substance: \[n=\sum_{j}^{}n _{j}\] But sense the sum of the mole fraction is equal to 1: \[\sum_{j}^{}x_{j}=1\] It most then follow that the sum of partial pressure for the single gasses in a mixture is the total pressure: \[\sum_{j}^{}p _{j}=\left( \sum_{j}^{}x_{j} \right)p=p\] In addition to that we can use Dalton's law that state: For a mixture of ideal gasses, it follows that the partial pressure of a gas in a mixture is the pressure the gas would do if it was alone in the container. So we can write: \[p _{j}=n _{j}\frac{ RT }{ V }\] This however is only for ideal gasses. Assume that neon and argon is ideal gasses and use the above laws to calculate the partial pressures.
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