# 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?

• How about 344.74$kPa*V=nRT$? – LDC3 Jul 6 '15 at 5:18
• This gas mixture is stored in the same container. So they should have the same volume - which is the volume of the container. Unlike liquids or solid state, the gases have ample of free space. So the fact their volumes overlap isn't an issue. So don't you think, V_total = V_air + V_co2 should be, V_total = V_air = V_co2 ? – bonCodigo Jul 6 '15 at 7:32
• @bonCodigo If $V_{total} = V_{air} + V_{co2}$ and $V_{total} = V_{air} = V_{co2}$, then the only solution is $0m^3$. If $V_{air} = 1m^3$ and $V_{co2} = 1m^3$, then $V_{total} =2m^3$, but the third equation says it should be $1m^3$. So, no, $V_{CO2} \neq 85L$. – LDC3 Jul 7 '15 at 2:03
• @LDC3 I did not mean Vtotal=Vair+Vco2 but Vco2 = Vair = 85L. Partial pressures ratio is proportional to the mole fraction. I seem to be crashing here... care to give a heads up? – bonCodigo Jul 7 '15 at 8:57
• @bonCodigo Even if $V_{air} = V_{CO2} = 42.5L$, since $V_{total}$ needs to be 85L, it won't give you the correct answer. – LDC3 Jul 8 '15 at 1:13

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.