# Calculating volume of gases in different conditions

Doctor Who has a secret laboratory on Venus where atmospheric pressure is $$92$$ times that of Earth and room temperature is $$\pu{735 K}$$. There, Doctor Who mixes $$\pu{4.00 g}$$ of $$\ce{CaCO3}$$ with $$\pu{40.00 mL}$$ of $$\pu{1.00 M}$$ $$\ce{HCl}$$. What volume of carbon dioxide gas at VSTP (Venusian Standard Temperature and Pressure) is produced?

I know that the chemical equation is

$$\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}$$

Since there are $$\pu{0.04 mol}$$ of $$\ce{CaCO3}$$ and $$\pu{0.04 mol}$$ of $$\ce{HCl}$$, $$\ce{HCl}$$ is the limiting reagent; thus, $$\pu{0.02 mol}$$ of $$\ce{CO2}$$ are produced.

This is equivalent to:

$$V_\ce{CO2} = \frac{nRT}{p} =\frac{0.02 \times 0.08205 \times 735}{92} = \pu{0.013 L}.$$

However, the answer key says $$\pu{26.2 mL}$$ of $$\ce{CO2}$$ is produced, which is double my answer. Where did I go wrong in my reasoning?

• You should include units in your calculations, but otherwise your maths seems to be correct. Does it ask for how much CO2 gas is produced, or how much gas is produced? At 735 K water would be gaseous, and assuming ideality (of course) this would lead to twice the volume of gas. – orthocresol Dec 23 '18 at 21:16
• @orthocresol thank you for checking the calculations. The question asked for the amount of carbon dioxide produced. – Cyclopropane Dec 23 '18 at 23:03
• @DrPepper The CO2 and the H2O will occupy the same volume of container. So it seems to me that orthocresol is correct. – Chet Miller Dec 23 '18 at 23:46

I haven't found any flaws in your math either, and I can only join those who recommended to include units in your calculations. By googling around I found this problem appeared in 2012 Australian Science Olympiad Exam, and $$\pu{13.1 mL}$$ was one of the options there. Maybe your source of the answer key is incorrect?

Now to the fun part. I'm pretty sure you were expected to use ideal gas law, but come on, it's Venus and no way under the pressure of $$\pu{92 atm}$$ any gas would behave even remotely as the ideal one. In fact, $$\ce{CO2}$$ on Venus with its critical temperature of $$\pu{304 K}$$ and critical pressure of $$\pu{7.39 MPa}$$ is going to be in supercritical state!

You have to use Van der Waals' equation:

$$pV_\mathrm{m} = \frac{RT}{V_\mathrm{m} - b} - \frac{a}{V_\mathrm{m}^2},$$

where $$p$$ – pressure; $$V_\mathrm{m}$$ – molar volume; T – temperature; R – molar gas constant; $$a$$ and $$b$$ – characteristic parameters related to the critical temperature $$T_\mathrm{cr}$$ and pressure $$p_\mathrm{cr}$$ (reduced form):

$$a = \frac{27R^2T_\mathrm{cr}^2}{64p_\mathrm{cr}}; \qquad b = \frac{RT_\mathrm{cr}}{8p_\mathrm{cr}}$$

Using the above-mentioned values coefficients $$a$$ and $$b$$ can be found:

$$a = \frac{27\cdot (\pu{8.314 J mol-1 K-1})^2\cdot (\pu{304 K})^2}{64\cdot\pu{7.39e6 Pa}} = \pu{3.65e-1 J2 mol-2 Pa-1}$$

$$b = \frac{\pu{8.314 J mol-1 K-1}\cdot \pu{304 K}}{8\cdot \pu{7.39e6 Pa}} = \pu{4.28e-5 J mol-1 Pa-1}$$

Now you need to plug both values in the first equation and solve quadratic equation for $$V_\mathrm{m}$$. Wolfram Alpha claims $$V_\mathrm{m} = \pu{2.56e-2 m3 mol-1}$$, so

$$V(\ce{CO2}) = \pu{2.56e-2 m3 mol-1}\cdot\pu{2e-2 mol} = \pu{5.12e-1 L},$$

which, of course, deviates quite a lot from the value found using ideal gas law.