Unknown Mass in Dalton's Law of Partial Pressures

An air with mass of $0.454\ \mathrm{kg}$ and an unknown mass of $\ce{CO2}$ occupy an 85 liters tank at $2068.44\ \mathrm{kPa}$. If the partial pressure of the $\ce{CO2}$ is $344.74\ \mathrm{kPa}$, determine its mass.


Answer is $0.138\ \mathrm{kg}$


Ok lets setup Dalton's Law: 


$$\begin{align}
p\_\text{total} &= p\_\text{air} + p\_{\ce{CO2}} \\
2068.44 &= \left(mRT/V\right)\_\text{air} + 344.74 \\
m\_\text{air} &= 0.454\ \mathrm{kg} \\
R\_\text{air} &= 0.287\ \mathrm{kJ/(kg\ K)} \\
T\_\text{air}/V\_\text{air} &= 18847.73126
\end{align}$$


Ok so I need to get something out of:


$$\begin{align}
T\_\text{air}/V\_\text{air} &= 18847.73126 \\
V\_\text{total} &= V\_\text{air} + V\_{\ce{CO2}} = 0.085\ \mathrm{m^3}
\end{align}$$


Ok so im basically stuck there. only 2 equations for 3 unknowns.


Any hint? or am I totally approaching this the wrong way?




Assuming that the total pressure $p\_\text{total}=2068.44\ \mathrm{kPa}$ is equal to the sum of the partial pressures of the individual gases in accordance with Dalton’s law, the partial pressure of air $p\_\text{air}$ can be calculated as follows:


$$\begin{align}
p\_\text{total}&=p\_\text{air}+p\_{\ce{CO2}}\\[6pt]
p\_\text{air}&=p\_\text{total}-p\_{\ce{CO2}}\\[6pt]
&=2068.44\ \mathrm{kPa}-344.74\ \mathrm{kPa}\\[6pt]
&=1723.70\ \mathrm{kPa}
\end{align}$$


Since molar mass $M$ is defined as


$$M=\frac mn$$


where $m$ is mass and $n$ is amount of substance, the ideal gas law 


$$p\cdot V = n\cdot R\cdot T$$


where $V$ is volume, $R$ is the molar gas constant, and $T$ is temperature, can be rewritten as


$$\begin{align}
p\cdot V&=\frac{m\cdot R\cdot T}M\\[6pt]
\frac V{R\cdot T}&=\frac m{M\cdot p}
\end{align}$$


This equation applies to air
$$\frac V{R\cdot T}=\frac {m\_\text{air}}{M\_\text{air}\cdot p\_\text{air}}$$
where $m\_\text{air}=0.454\ \mathrm{kg}$ is the given mass of air and $M\_\text{air}=28.97\ \mathrm{g\ mol^{-1}}$ is the average molar mass of air, as well as to $\ce{CO2}$
$$\frac V{R\cdot T}=\frac {m\_{\ce{CO2}}}{M\_{\ce{CO2}}\cdot p\_{\ce{CO2}}}$$
where $m\_{\ce{CO2}}$ is the unknown mass of $\ce{CO2}$ and $M\_{\ce{CO2}}=44.01\ \mathrm{g\ mol^{-1}}$ is the molar mass of $\ce{CO2}$.


Therefore,


$$\begin{align}
\frac {m\_{\ce{CO2}}}{M\_{\ce{CO2}}\cdot p\_{\ce{CO2}}}&=\frac {m\_\text{air}}{M\_\text{air}\cdot p\_\text{air}}\\[6pt]
m\_{\ce{CO2}}&=\frac {M\_{\ce{CO2}}\cdot p\_{\ce{CO2}}\cdot m\_\text{air}}{M\_\text{air}\cdot p\_\text{air}}\\[6pt]
&=\frac {44.01\ \mathrm{g\ mol^{-1}}\times344.74\ \mathrm{kPa}\times0.454\ \mathrm{kg}}{28.97\ \mathrm{g\ mol^{-1}}\times1723.70\ \mathrm{kPa}}\\[6pt]
&=0.138\ \mathrm{kg}
\end{align}$$


Note that the given value for the volume $V=85\ \mathrm l$ is not required for the calculation of $m\_\text{air}$. Thus, the uncertainty of the volume $V=85\ \mathrm l$ (with only two significant digits) does not affect the uncertainty of the result.


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