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I'd start the solution with this equation:

$n_{Ar}d_{Ar}/(n_{Ar} + n_{He}) + n_{He}d_{He}/(n_{Ar} + n_{He}) = d_{avg}$

where I emphasize that the density of the gas is a molar average value, and rewrite it as

$\chi_{Ar} d_{Ar} + (1-\chi_{Ar})d_{He} = d_{avg}$

where $\chi_{Ar} = n_{Ar}/n$ is the mole fraction of argon in the gas mixture and the $d_{i}$ are densities computed assuming all of the gas corresponds to He or Ar, for instance

$d_{Ar} = M_{Ar}n/V = M_{Ar}P/RT$$d_{Ar} = M_{Ar}n/V = M_{Ar}/V_{m}$

I add a mea culpa here, as this was massively botched in a previous answer but stands corrected nowwhere $V_{m}$ is the molar volume at STP (22.414 $m^3/kg mol$).

It follows that

$ d_{avg} = \chi_{Ar}M_{Ar}P/RT + (1-\chi_{Ar})M_{He}P/RT = (\chi_{Ar}M_{Ar} + (1-\chi_{Ar})M_{He})P/RT$$ d_{avg} = \chi_{Ar}M_{Ar}/V_{m} + (1-\chi_{Ar})M_{He}/V_{m} = (\chi_{Ar}M_{Ar} + (1-\chi_{Ar})M_{He})/V_{m} = M_{avg}/V_{m} $

Thiswhich can be rearranged assolved for $ \chi_{Ar} $:

$ \chi_{Ar} = ((d_{avg}RT/P)-M_{He})/(M_{Ar}-M_{He})$$ \chi_{Ar} = ((d_{avg}V_{m})-M_{He})/(M_{Ar}-M_{He})$

Finally, the molar percentage argon is computed as $f_{Ar}=100\chi_{Ar}$.

For your particular problem I get $f_{Ar}$= 5045.14%00%.

I'd start the solution with this equation:

$n_{Ar}d_{Ar}/(n_{Ar} + n_{He}) + n_{He}d_{He}/(n_{Ar} + n_{He}) = d_{avg}$

where I emphasize that the density of the gas is a molar average value, and rewrite it as

$\chi_{Ar} d_{Ar} + (1-\chi_{Ar})d_{He} = d_{avg}$

where $\chi_{Ar} = n_{Ar}/n$ is the mole fraction of argon in the gas mixture and the $d_{i}$ are densities computed assuming all of the gas corresponds to He or Ar, for instance

$d_{Ar} = M_{Ar}n/V = M_{Ar}P/RT$

I add a mea culpa here, as this was massively botched in a previous answer but stands corrected now.

It follows that

$ d_{avg} = \chi_{Ar}M_{Ar}P/RT + (1-\chi_{Ar})M_{He}P/RT = (\chi_{Ar}M_{Ar} + (1-\chi_{Ar})M_{He})P/RT$

This can be rearranged as

$ \chi_{Ar} = ((d_{avg}RT/P)-M_{He})/(M_{Ar}-M_{He})$

Finally, the molar percentage argon is computed as $f_{Ar}=100\chi_{Ar}$.

For your particular problem I get $f_{Ar}$= 50.14%.

I'd start the solution with this equation:

$n_{Ar}d_{Ar}/(n_{Ar} + n_{He}) + n_{He}d_{He}/(n_{Ar} + n_{He}) = d_{avg}$

where I emphasize that the density of the gas is a molar average value, and rewrite it as

$\chi_{Ar} d_{Ar} + (1-\chi_{Ar})d_{He} = d_{avg}$

where $\chi_{Ar} = n_{Ar}/n$ is the mole fraction of argon in the gas mixture and the $d_{i}$ are densities computed assuming all of the gas corresponds to He or Ar, for instance

$d_{Ar} = M_{Ar}n/V = M_{Ar}/V_{m}$

where $V_{m}$ is the molar volume at STP (22.414 $m^3/kg mol$).

It follows that

$ d_{avg} = \chi_{Ar}M_{Ar}/V_{m} + (1-\chi_{Ar})M_{He}/V_{m} = (\chi_{Ar}M_{Ar} + (1-\chi_{Ar})M_{He})/V_{m} = M_{avg}/V_{m} $

which can be solved for $ \chi_{Ar} $:

$ \chi_{Ar} = ((d_{avg}V_{m})-M_{He})/(M_{Ar}-M_{He})$

Finally, the molar percentage argon is computed as $f_{Ar}=100\chi_{Ar}$.

For your particular problem I get $f_{Ar}$= 45.00%.

    Post Undeleted by Buck Thorn
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I'd start the solution with this equation:

$n_{Ar}d_{Ar}/(n_{Ar} + n_{He}) + n_{He}d_{He}/(n_{Ar} + n_{He}) = d_{avg}$

where I emphasize that the density of the gas is a molar average value, and rewrite it as

$\chi_{Ar} d_{Ar} + (1-\chi_{Ar})d_{He} = d_{avg}$

where $\chi_{Ar} = n_{Ar}/n$ is the mole fraction of argon in the gas mixture. Insert and the expression

$d_{Ar} = M_{Ar}n_{Ar}/V = M_{Ar}\chi_{Ar}n/V= M_{Ar}\chi_{Ar}P/RT$

for the density$d_{i}$ are densities computed assuming all of argon in the mixture ($P\chi_{Ar}=P_{Ar}$ is the partial pressure of argon)gas corresponds to He or Ar, and the similar expression for helium, so that instance

$\chi_{Ar} \chi_{Ar}M_{Ar}P/RT + (1-\chi_{Ar})(1-\chi_{Ar})M_{He}P/RT = d_{avg}$$d_{Ar} = M_{Ar}n/V = M_{Ar}P/RT$

Rearranging the equationI add a mea culpa here,

$\chi_{Ar}^2(M_{Ar}+M_{He}) - 2M_{He}\chi_{Ar} +(M_{He}- d_{avg}(RT/P))=0$

or

$\chi_{Ar}^2 +\alpha\chi_{Ar} +\beta=0$

where as this was massively botched in a previous answer but stands corrected now.

$\alpha = - 2M_{He}/(M_{Ar}+M_{He})$ It follows that

$\beta = (M_{He}- (d_{avg}RT/P))/(M_{Ar}+M_{He})$$ d_{avg} = \chi_{Ar}M_{Ar}P/RT + (1-\chi_{Ar})M_{He}P/RT = (\chi_{Ar}M_{Ar} + (1-\chi_{Ar})M_{He})P/RT$

Finally, solving the quadratic equation forThis can be rearranged as $\chi_{Ar}$:

$\chi_{Ar} = -alpha/2+(1/2)(alpha^2-4\beta)^{1/2}$$ \chi_{Ar} = ((d_{avg}RT/P)-M_{He})/(M_{Ar}-M_{He})$

Finally, the molar percentage argon is computed as $f_{Ar}=100\chi_{Ar}$.

For your particular problem I get $f_{Ar}$= 7350.79%14%.

I'd start the solution with this equation:

$n_{Ar}d_{Ar}/(n_{Ar} + n_{He}) + n_{He}d_{He}/(n_{Ar} + n_{He}) = d_{avg}$

where I emphasize that the density of the gas is a molar average value, and rewrite it as

$\chi_{Ar} d_{Ar} + (1-\chi_{Ar})d_{He} = d_{avg}$

where $\chi_{Ar} = n_{Ar}/n$ is the mole fraction of argon in the gas mixture. Insert the expression

$d_{Ar} = M_{Ar}n_{Ar}/V = M_{Ar}\chi_{Ar}n/V= M_{Ar}\chi_{Ar}P/RT$

for the density of argon in the mixture ($P\chi_{Ar}=P_{Ar}$ is the partial pressure of argon), and the similar expression for helium, so that

$\chi_{Ar} \chi_{Ar}M_{Ar}P/RT + (1-\chi_{Ar})(1-\chi_{Ar})M_{He}P/RT = d_{avg}$

Rearranging the equation,

$\chi_{Ar}^2(M_{Ar}+M_{He}) - 2M_{He}\chi_{Ar} +(M_{He}- d_{avg}(RT/P))=0$

or

$\chi_{Ar}^2 +\alpha\chi_{Ar} +\beta=0$

where

$\alpha = - 2M_{He}/(M_{Ar}+M_{He})$

$\beta = (M_{He}- (d_{avg}RT/P))/(M_{Ar}+M_{He})$

Finally, solving the quadratic equation for $\chi_{Ar}$:

$\chi_{Ar} = -alpha/2+(1/2)(alpha^2-4\beta)^{1/2}$

Finally, the molar percentage argon is computed as $f_{Ar}=100\chi_{Ar}$.

For your particular problem I get $f_{Ar}$= 73.79%.

I'd start the solution with this equation:

$n_{Ar}d_{Ar}/(n_{Ar} + n_{He}) + n_{He}d_{He}/(n_{Ar} + n_{He}) = d_{avg}$

where I emphasize that the density of the gas is a molar average value, and rewrite it as

$\chi_{Ar} d_{Ar} + (1-\chi_{Ar})d_{He} = d_{avg}$

where $\chi_{Ar} = n_{Ar}/n$ is the mole fraction of argon in the gas mixture and the $d_{i}$ are densities computed assuming all of the gas corresponds to He or Ar, for instance

$d_{Ar} = M_{Ar}n/V = M_{Ar}P/RT$

I add a mea culpa here, as this was massively botched in a previous answer but stands corrected now.

It follows that

$ d_{avg} = \chi_{Ar}M_{Ar}P/RT + (1-\chi_{Ar})M_{He}P/RT = (\chi_{Ar}M_{Ar} + (1-\chi_{Ar})M_{He})P/RT$

This can be rearranged as

$ \chi_{Ar} = ((d_{avg}RT/P)-M_{He})/(M_{Ar}-M_{He})$

Finally, the molar percentage argon is computed as $f_{Ar}=100\chi_{Ar}$.

For your particular problem I get $f_{Ar}$= 50.14%.

    Post Deleted by Buck Thorn
    Post Undeleted by Buck Thorn
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I'd start the solution with this equation:

$n_{Ar}d_{Ar}/(n_{Ar} + n_{He}) + n_{He}d_{He}/(n_{Ar} + n_{He}) = d_{avg}$

where I emphasize that the density of the gas is ana averagemolar average value, and rewrite it as

$\chi_{Ar} d_{Ar} + (1-\chi_{Ar})d_{He} = d_{avg}$

where $\chi_{Ar} = n_{Ar}/n$ is the mole fraction of argon in the gas mixture. Insert the expression

$d_{Ar} = M_{Ar}n_{Ar}/V = M_{Ar}\chi_{Ar}n/V= M_{Ar}\chi_{Ar}P/RT$

for the density of argon in the mixture ($P\chi_{Ar}=P_{Ar}$ is the partial pressure of argon), and the similar expression for helium, so that

$\chi_{Ar} \chi_{Ar}M_{Ar}P/RT + (1-\chi_{Ar})(1-\chi_{Ar})M_{He}P/RT = d_{avg}$

Rearranging the equation,

$\chi_{Ar}^2(M_{Ar}+M_{He}) - 2M_{He}\chi_{Ar} +(M_{He}- d_{avg}(RT/P))=0$

or

$\chi_{Ar}^2 +\alpha\chi_{Ar} +\beta=0$

where

$\alpha = - 2M_{He}/(M_{Ar}+M_{He})$

$\beta = (M_{He}- (d_{avg}RT/P))/(M_{Ar}+M_{He})$

Finally, solving the quadratic equation for $\chi_{Ar}$:

$\chi_{Ar} = -alpha+(1/2)(alpha^2-4\beta)^{1/2}$$\chi_{Ar} = -alpha/2+(1/2)(alpha^2-4\beta)^{1/2}$

Finally, the molar percentage argon is computed as $f_{Ar}=100\chi_{Ar}$.

For your particular problem I get $f_{Ar}$= 8273.89%79%.

I'd start the solution with this equation:

$n_{Ar}d_{Ar}/(n_{Ar} + n_{He}) + n_{He}d_{He}/(n_{Ar} + n_{He}) = d_{avg}$

where I emphasize that the density of the gas is an average value, and rewrite it as

$\chi_{Ar} d_{Ar} + (1-\chi_{Ar})d_{He} = d_{avg}$

where $\chi_{Ar} = n_{Ar}/n$ is the mole fraction of argon in the gas mixture. Insert the expression

$d_{Ar} = M_{Ar}n_{Ar}/V = M_{Ar}\chi_{Ar}n/V= M_{Ar}\chi_{Ar}P/RT$

for the density of argon in the mixture ($P\chi_{Ar}=P_{Ar}$ is the partial pressure of argon), and the similar expression for helium, so that

$\chi_{Ar} \chi_{Ar}M_{Ar}P/RT + (1-\chi_{Ar})(1-\chi_{Ar})M_{He}P/RT = d_{avg}$

Rearranging the equation,

$\chi_{Ar}^2(M_{Ar}+M_{He}) - 2M_{He}\chi_{Ar} +(M_{He}- d_{avg}(RT/P))=0$

or

$\chi_{Ar}^2 +\alpha\chi_{Ar} +\beta=0$

where

$\alpha = - 2M_{He}/(M_{Ar}+M_{He})$

$\beta = (M_{He}- (d_{avg}RT/P))/(M_{Ar}+M_{He})$

Finally, solving the quadratic equation for $\chi_{Ar}$:

$\chi_{Ar} = -alpha+(1/2)(alpha^2-4\beta)^{1/2}$

Finally, the molar percentage argon is computed as $f_{Ar}=100\chi_{Ar}$.

For your particular problem I get $f_{Ar}$= 82.89%.

I'd start the solution with this equation:

$n_{Ar}d_{Ar}/(n_{Ar} + n_{He}) + n_{He}d_{He}/(n_{Ar} + n_{He}) = d_{avg}$

where I emphasize that the density of the gas is a molar average value, and rewrite it as

$\chi_{Ar} d_{Ar} + (1-\chi_{Ar})d_{He} = d_{avg}$

where $\chi_{Ar} = n_{Ar}/n$ is the mole fraction of argon in the gas mixture. Insert the expression

$d_{Ar} = M_{Ar}n_{Ar}/V = M_{Ar}\chi_{Ar}n/V= M_{Ar}\chi_{Ar}P/RT$

for the density of argon in the mixture ($P\chi_{Ar}=P_{Ar}$ is the partial pressure of argon), and the similar expression for helium, so that

$\chi_{Ar} \chi_{Ar}M_{Ar}P/RT + (1-\chi_{Ar})(1-\chi_{Ar})M_{He}P/RT = d_{avg}$

Rearranging the equation,

$\chi_{Ar}^2(M_{Ar}+M_{He}) - 2M_{He}\chi_{Ar} +(M_{He}- d_{avg}(RT/P))=0$

or

$\chi_{Ar}^2 +\alpha\chi_{Ar} +\beta=0$

where

$\alpha = - 2M_{He}/(M_{Ar}+M_{He})$

$\beta = (M_{He}- (d_{avg}RT/P))/(M_{Ar}+M_{He})$

Finally, solving the quadratic equation for $\chi_{Ar}$:

$\chi_{Ar} = -alpha/2+(1/2)(alpha^2-4\beta)^{1/2}$

Finally, the molar percentage argon is computed as $f_{Ar}=100\chi_{Ar}$.

For your particular problem I get $f_{Ar}$= 73.79%.

    Post Deleted by Buck Thorn
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