# Find the density of air at the given temperature and pressure [closed]

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?

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.
• 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, $\rho$: $$\therefore \space \frac{P M_r}{RT}=\rho$$ – Alaz Cig Jul 16 '19 at 19:19