# Does the property of additivity for the amounts of different substances hold up for ideal gas law?

A mixture contains $$\pu{8.00 g}$$ each of $$\ce{O2},$$ $$\ce{CO2},$$ and $$\ce{SO2}$$ at STP. Calculate the volume of this mixture. Which of the gases would exert the greatest pressure and why?

Here is my work:

\begin{align} n(\ce{O2}) &= \frac{\pu{8.00 g}}{\pu{32.00 g mol-1}} = \pu{0.250 mol}\\ n(\ce{CO2}) &= \frac{\pu{8.00 g}}{\pu{44.01 g mol-1}} = \pu{0.182 mol}\\ n(\ce{SO2}) &= \frac{\pu{8.00 g}}{\pu{64.07 g mol-1}} = \pu{0.125 mol} \end{align}

$$n_\mathrm{tot} = \pu{0.250 mol} + \pu{0.182 mol} + \pu{0.125 mol} = \pu{0.557 mol}$$

$$pV = nRT \quad\implies\quad V = \frac{nRT}{p}$$

$$V = \frac{\pu{0.557 mol}\times\pu{0.0821 atm L mol^-1 K^-1}\times\pu{273 K}}{\pu{1.00 atm}} = \pu{12.5 L}$$

Considering $$p = nRT/V,$$ the gas that takes up the least amount of volume with the greatest amount of particles will exert the greatest pressure. This will be $$\ce{O2}$$ because it has the lowest molar mass/volume.

Is this how handling the amounts of substance work (can different substances be added like this)?

• You are using (intuitively) Dalton's law of partial pressure when you add the moles. Ideal gas law doesn't know the molecular formula. All it considers is the number of particles. May 23, 2020 at 3:07
• Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. May 23, 2020 at 7:24
• You are assuming a pressure of 1.00 atm. Please note that, for about 40 years, STP corresponds to a pressure of 100 kPa = 1.00 bar. Also note that the use of the unit standard atmosphere (atm) is deprecated.
– user7951
May 23, 2020 at 8:05

## 1 Answer

The question would be better phrased as which gas contributes most to the total pressure? The contribution is referred to as the partial pressure. The sum of the partial pressures is the total pressure.

The formula you would use to compute the partial pressure $$p_i$$ of gas component $$i$$ is either of the following:

$$p_i=\chi_i p = \frac{n_i}{n}p = \frac{n_i}{V}RT$$

where $$n_i$$ is the amount of component $$i$$ in the gas and $$\chi_i$$ its mole fraction. So the argument "the gas that takes up the least amount of volume with the greatest amount of particles will exert the greatest pressure" is not quite right. It suffices to say that the gas that contributes the greatest amount of particles will contribute (exert) the greatest pressure. If you mix identical masses of each gas, that will be the gas with lowest molecular mass.