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A $\pu{19.3 g}$ mixture of oxygen and argon is found to occupy a volume of $\pu{16.2 L}$ when measured at $\pu{675.9 mmHg}$ and $\pu{43.4 ^\circ{}C}$. What is the partial pressure of oxygen in this mixture?

The answer is $\pu{436 mmHg}$, but I didn't manage to get to it.

I know that it is found by the formula $$P_i=X_i\cdot P_\text{total},$$ where $P_i$ is the partial pressure and $X_i$ is the mole fraction.

$P_\text{total}$ is given to find $X_i$. I should divide the amount of $\ce{O2}$ by the total amount of substances of the mixture $n_{\ce{O2}}/n_\text{total}$.

$n_\text{total}$ is easy to find with $n=PV/RT$, but my problem is, that I can't find the amount of substance of $\ce{O2}$. The mass in the problem is given for the mixture, so how to extract only the mass of $\ce{O2}$, or in other words, how to get the amount of substance for $\ce{O2}$?

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How is the mixture composed? Find expressions for mass, volume and amount of substance. \begin{align} \tag1\label1 V_\text{total} &= V_{\ce{O2}} + V_{\ce{Ar}}\\ \tag3\label2 n_\text{total} &= n_{\ce{O2}} + n_{\ce{Ar}}\\ \tag3\label3 m_\text{total} &= m_{\ce{O2}} + m_{\ce{Ar}} \end{align}

Transform $V_\text{total}$ with the ideal gas and solve for $n_{\ce{Ar}}$ \begin{align} \tag4\label4 V_\text{total} &= \frac{RT}{p_\text{total}}(n_{\ce{O2}} + n_{\ce{Ar}})\\ \tag5\label5 n_{\ce{Ar}} &= \frac{p_\text{total}V_\text{total}}{RT} - n_{\ce{O2}} \end{align}

Use total mass \eqref{3} and transform it with the with molar mass to amount of substance equation ($m = nM$), then plug in \eqref{5}: \begin{align} \tag3 m_\text{total} &= m_{\ce{O2}} + m_{\ce{Ar}}\\ \tag6\label6 m_\text{total} &= n_{\ce{O2}}M_{\ce{O2}} + n_{\ce{Ar}}M_{\ce{Ar}}\\ \tag7\label7 m_\text{total} &= n_{\ce{O2}}M_{\ce{O2}} + \left(\frac{p_\text{total}V_\text{total}}{RT} - n_{\ce{O2}}\right)M_{\ce{Ar}} \end{align}

Now there is only one variable left, which you do not know:

\begin{align} \tag8\label8 m_\text{total} &= n_{\ce{O2}}\left(M_{\ce{O2}} - M_{\ce{Ar}}\right) + \frac{p_\text{total}V_\text{total}}{RT}\cdot M_{\ce{Ar}}\\ \tag9\label9 n_{\ce{O2}} &= \left(m_\text{total} - \frac{p_\text{total}V_\text{total}}{RT} \cdot M_{\ce{Ar}}\right)\cdot \left(M_{\ce{O2}} - M_{\ce{Ar}}\right)^{-1} \end{align}

We need the following data (It would be by far better to only use SI [derived] units):

\begin{align} M_\ce{Ar} &= \pu{39.9 g//mol}\\ M_\ce{O2} &= \pu{32.0 g//mol}\\ R &= \pu{8.314 L kPa //K mol}\\ \pu{1 mmHg} &= \pu{133.322 Pa}\\ p_\text{total} &\approx \pu{90 kPa}\\ T &= \pu{316.5 K}\\ m_\text{total} &= \pu{19.3 g}\\ V_\text{total} &= \pu{16.2 L} \end{align}

Plugging this into \eqref{9}: \begin{align} \tag9 n_{\ce{O2}} &= \left(m_\text{total} - \frac{p_\text{total}V_\text{total}}{RT} \cdot M_{\ce{Ar}}\right)\cdot \left(M_{\ce{O2}} - M_{\ce{Ar}}\right)^{-1}\\ n_{\ce{O2}} &= \left(\pu{19.3 g} - \frac{\pu{90 kPa}\cdot\pu{16.2 L}}{\pu{8.314 L kPa //K mol}\cdot\pu{316.5 K}}\pu{39.9 g//mol}\right)\cdot \left(\pu{32.0 g//mol} - \pu{39.9 g//mol}\right)^{-1}\\ n_{\ce{O2}} &= \pu{0.36 mol} \end{align}

Using the ideal gas with the data provided yields: \begin{align} n_\text{total} &=\frac{pV}{RT}\\ n_\text{total} &=\frac{\pu{90 kPa}\times\pu{16.2 L}}{\pu{8.314 L kPa K-1 mol-1}\times\pu{316.5 K}}\\ n_\text{total} &=\pu{0.55 mol} \end{align}

Therefore the mole fraction is $$x_\ce{O2} = \frac{n_\ce{O2}}{n_\text{total}} \approx 0.64.$$

And the partial pressure is therefore $\pu{57.7 kPa}$ (or $\pu{433 mmHg}$ which is close enough to the solution, given that I rounded the medieval witchcraft units).

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Another visualization:

  • Assume $\pu{19.3 g}$ of pure oxygen; use the ideal gas law to find the volume $V_1$

  • Assume $\pu{19.3 g}$ of pure argon; use the ideal gas law to find the volume $V_2$, smaller than $V_1$ since argon is heavier than oxygen

  • Hopefully, the actual volume will be somewhere in between $V_1$ and $V_2$.

  • How far along the gap from $V_2$ to $V_1$ is the actual volume? That's how far you are from zero oxygen to 100% oxygen. With this fraction and the total pressure, the partial pressure is simple.

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