= Argon detection in a binary gas mixture
{title2=$x=\frac{\rho/\rho_N-1}{M_A/M_N-1}$}
Consider an original simplified model containing only <nitrogen> and <argon>, with <argon> <mole fraction> $x$. Let their <molar masses> be $M_N$ and $M_A$. Adding the component <masses> gives the mixture's mean <molar mass>
$$
\overline M=(1-x)M_N+xM_A.
$$
For an <ideal gas> mixture at the same <pressure> and <temperature> as a pure-<nitrogen> reference, the <gas density> ratio is
$$
r=\frac{\rho}{\rho_N}=\frac{\overline M}{M_N}
=1+x\left(\frac{M_A}{M_N}-1\right),\qquad
\boxed{x=\frac{r-1}{M_A/M_N-1}.}
$$
The inversion requires known identities, the binary-mixture assumption and $M_A\ne M_N$. A density ratio cannot uniquely recover an arbitrary multicomponent <mixture>. Nor should $x$ be interpreted as the <argon> fraction of untreated air: the model sample has only the two specified components.
Choose round model values $M_N=28\,\mathrm{g\,mol^{-1}}$ and $M_A=40\,\mathrm{g\,mol^{-1}}$. For $x=0.010$, the calculated relative density excess is
$$
\boxed{r-1=0.010\left(\frac{40}{28}-1\right)
=0.00428571\ldots\simeq0.429\%.}
$$
Thus a one-percent <mole fraction> does not mean a one-percent <gas density> change. In a chosen $2.00\,\mathrm L$ globe at $100\,\mathrm{kPa}$ and $300\,\mathrm K$, the <ideal gas law> gives approximately $2.24508\,\mathrm g$ for pure <nitrogen> and $2.25471\,\mathrm g$ for this mixture: only $9.62\,\mathrm{mg}$ separates the weighings. These are model calculations with rounded <molar masses>, not historical measurements.
\Image[/argon-mixture-density-and-temperature-confounding.png]
{title=A small admixture changes gas density, but temperature can imitate the signal}
{description=Calculated binary-mixture model with stipulated <molar masses> $28$ and $40\,\mathrm{g\,mol^{-1}}$. The left panel shows the linear <gas density> contrast. The right compares three hypothetical globe fillings: a reference, one-percent <argon> at the same <temperature>, and pure <nitrogen> cooled enough to give the identical <mass>. No historical data are plotted.}
A direct control calculation explains why the filling conditions matter. At fixed <pressure>, pure <nitrogen> at
$$
T'=\frac{300\,\mathrm K}{1+0.00428571\ldots}
\simeq298.720\,\mathrm K
$$
has the same <gas density> as that mixture at $300\,\mathrm K$. A roughly $1.28\,\mathrm K$ mismatch can therefore imitate the entire stipulated composition signal. For small changes, the <ideal gas law> gives
$$
\frac{d\rho}{\rho}=\frac{dp}{p}+\frac{d\overline M}{\overline M}-\frac{dT}{T}.
$$
An independently measured <pressure> and <temperature>, matched weighing procedure, and checks for residual vapour distinguish composition from these alternative causes. The model explains the need for controls; it does not supply the complete corrections for a real gas-weighing instrument.
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