Ideal Gas Frenzy

Relevant wiki: Ideal Gas Laws

Typing takes too much time, so I'll just post this:

With the given information, the formula most appropriate to be used here is the ideal gas law . Therefore:

$PV = nRT$

However, the container holds a certain number of ideal gas, $\alpha$ . The equation of ${ \alpha }_{ x }$ would be:

${ P }_{ x }{ V }_{ x } = { n }_{ x }RT$

Considering the ideal gas a whole, as opposed to just individual multiple ideal gas, would require Dalton's law as that allows us to add all the ideal gasses together in respect to their pressure, thus, rearranging the above equation:

${ P }_{ x } = \frac{{ n }_{ x }RT}{ { V }_{ x } }$

According to Dalton's law and since R is the total pressure in the container,

(It is important to note that the R on the LHS is the value of pressure, and the R on the RHS is the Avogadro constant . Their values are the same, but their units are different.)

$R=\sum _{ x=1 }^{ a }{ { P }_{ x } } =\sum _{ x=1 }^{ a }{ \frac { { n }_{ x }RT }{ { V }_{ x } } } =\alpha (\frac { { n }_{ total }RT }{ V } )$

Therefore,

$\frac { V }{ T } =\frac { \alpha { n }_{ total }R }{ R } =\alpha { n }_{ total }$

The final variable that has to be calculated is $n$ , which represents the amount of gas measured in moles.

Since $M = \frac{m}{n}$ , where M and m represents molar mass and mass respectively,

$n = \frac{m}{M}$

Therefore,

${ n }_{ total } = \frac{{ m }_{ total }}{{ M }_{ total }}$

To calculate ${ n }_{ total }$ , one would need to first calculate ${ m }_{ total }$ and ${ M }_{ total }$ . The answer to this lies in the question about rate of diffusion and the proportionality between mass and molar mass.

Since the rate of diffusion or effusion of every succeeding gas forms a geometric sequence with a common ratio $\beta$ , and according to Graham's law ,

$\frac { { Rate }_{ 1 } }{ { Rate }_{ 2 } } = \sqrt { \frac { { M }_{ 2 } }{ { M }_{ 1 } } }$

Since $\frac { { Rate }_{ 2 } }{ { Rate }_{ 1 } } = \beta$ ( $common ratio = \frac{{ T }_{ n+1 }}{{ T }_{ n }}$ ),

$\frac { 1 }{ \beta } =\sqrt { \frac { { M }_{ 2 } }{ { M }_{ 1 } } }$

Squaring both sides,

$\frac { { M }_{ 2 } }{ { M }_{ 1 } } = \frac { 1 }{ { \beta }^{ 2 } }$

Or in more general terms,

$\frac { { M }_{ n+1 } }{ { M }_{ n } } = \frac { 1 }{ { \beta }^{ 2 } }$

Therefore, the common ratio of $M$ is $\frac { 1 }{ { \beta }^{ 2 } }$ .

We can now find the mass from the equation ${ m }_{ n } = { \beta }^{ 3 }{ M }_{ n+1 }$ as given by the second paragraph of the problem.

Since ${ M }_{ n+1 } =\frac { { M }_{ n } }{ { \beta }^{ 2 } }$ , we can substitute that into ${ m }_{ n } = { \beta }^{ 3 }{ M }_{ n+1 }$ ,

${ m }_{ n } = { \beta }^{ 3 }\frac { { M }_{ n } }{ { \beta }^{ 2 } }$

${ m }_{ n } = \beta{ M }_{ n }$

Therefore, the common ratio of ${ m }_{ n }$ is the same as ${ M }_{ n }$ , and has all of its values except each one of them are multiplied by $\beta$ .

Let $a$ be the first term of the geometric series of ${ M }_{ n }$ , hence the first term of the geometric series of ${ m }_{ n }$ is $a\beta$ , thus:

$M(n)=a\cdot { ({ \beta }^{ -2 }) }^{ n-1 }$ $m(n)=a\beta\cdot { ({ \beta }^{ -2 }) }^{ n-1 }$

We can also find their geometric sum:

${ M }_{ total } = \frac { a[{ (\beta ) }^{ -2n }-1] }{ { \beta }^{ -2 }-1 }$ ${ m }_{ total } = \frac { a\beta[{ (\beta ) }^{ -2n }-1] }{ { \beta }^{ -2 }-1 }$

Therefore:

${ n }_{ total } = \frac{{ m }_{ total }}{{ M }_{ total }} = \frac{\frac { a\beta[{ (\beta ) }^{ -2n }-1] }{ { \beta }^{ -2 }-1 }}{\frac { a[{ (\beta ) }^{ -2n }-1] }{ { \beta }^{ -2 }-1 }} = \frac{a\beta}{a} = \beta$

Now, for the penultimate of this solution, substituting $\beta$ for ${ n }_{ total }$ in $\frac { V }{ T } = \alpha { n }_{ total }$ , thus:

$\frac { V }{ T } = \alpha \beta$

Quod Erat Demonstrandum.