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Find the density of air at $99$ kPa and $322.15$ K.

My attempt:

Let's assume that air is $78\%$ nitrogen and $22\%$ oxygen by volume.

Average molar mass of air = $0.78\times 28+0.22\times 32\approx 28.9$ g

We need to find the mass of $1\ m^3$ of air. Let this be $x$ kg.

Number of moles of air = $\displaystyle n=\frac{x}{28.9\times 10^{-3}}\approx 34.6x$ mol

Using the ideal gas equation $pV=nRT$ for air,

$99\times 10^3\times 1=34.6x\times 8.314\times 322.15$

$x\approx 1.068$ kg

So, $\rho\approx 1.068\ kg/m^3$


Is this right? In particular, is using the the ideal gas equation $pV=nRT$ for air, which is a mixture of gases, justifiable. Or should one calculate the number of moles of nitrogen and oxygen individually and then use the ideal gas equation for one of these gases?

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According to Dalton's Law, the total pressure of a mixed gas is equal to the sum of the gas component pressures. This is one simple evidence that you can consider the gases as one, since the pressures acts as a one, at a simple level.

In fact, if you perform this calculation separately for each element (using their own individual molar masses and their own fraction of the volume), then add them together, you will get the same result. I got $\rho \approx 1.067\space kg/m^3$ using this method, thus confirming your answer.

Also keep in mind that the ideal gas law is only ideal - it does not represent reality. So, looking at a gas mixture as one is sufficient.

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  • $\begingroup$ You can also simplify your calculation since you multiply by $V$ and then divide by $V$. Since $n=\frac{m}{M_r}$, where $M_r$ is the molar mass: $$PV=(\frac{m}{M_r})RT$$ So if you divide by $V$: $$P=(\frac{m}{V})\frac{RT}{M_r}$$ $$PM_r=(\rho)RT$$ Now you can directly solve for density, $\rho$: $$\therefore \space \frac{P M_r}{RT}=\rho$$ $\endgroup$ – Alaz Cig Jul 16 '19 at 19:19

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