A steel container with a volume of 500.0 mL is evacuated, and 25.0 g of CaCO3 is added. The cont

step 1 This problem gives us a reaction that produces a gas at the end. And it's asking us to use the i

A steel container with a volume of 500.0 mL is evacuated,...

A steel container with a volume of 500.0 $\mathrm{mL}$ is evacuated, and 25.0 $\mathrm{g}$ of $\mathrm{CaCO}_{3}$ is added. The container and contents are then heated to 1500 $\mathrm{K}$ , causing the $\mathrm{CaCO}_{3}$ to decompose completely, according to the equation $\mathrm{CaCO}_{3}(s) \rightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$ . (a) Using the ideal gas law and ignoring the volume of any solids remaining in the container, calculate the pressure inside the container at 1500 $\mathrm{K}$ . (b) Now make a more accurate calculation of the pressure inside the container. Take into account the volume of solid CaO (density $=3.34 \mathrm{g} / \mathrm{mL}$ ) in the container, and use the van der Waals equation to calculate the pres- sure. The van der Waals constants for $\mathrm{CO}_{2}(g)$ are: $a=3.59\left(\mathrm{L}^{2} \cdot \mathrm{atm}\right) / \mathrm{mol}^{2},$ and $b=0.0427 \mathrm{L} / \mathrm{mol} .$

A steel container with a volume of 500.0 $\mathrm{mL}$ is evacuated, and 25.0 $\mathrm{g}$ of $\mathrm{CaCO}_{3}$ is added. The container and contents are then heated to 1500 $\mathrm{K}$ , causing the $\mathrm{CaCO}_{3}$ to decompose completely, according to the equation $\mathrm{CaCO}_{3}(s) \rightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$ .
(a) Using the ideal gas law and ignoring the volume of any solids remaining in the container, calculate the pressure inside the container at 1500 $\mathrm{K}$ .
(b) Now make a more accurate calculation of the pressure inside the container. Take into account the volume of solid CaO (density $=3.34 \mathrm{g} / \mathrm{mL}$ ) in the container, and use the van der Waals equation to calculate the pres- sure. The van der Waals constants for $\mathrm{CO}_{2}(g)$ are:
$a=3.59\left(\mathrm{L}^{2} \cdot \mathrm{atm}\right) / \mathrm{mol}^{2},$ and $b=0.0427 \mathrm{L} / \mathrm{mol} .$

Key Concepts

Mole Concept and Molar Mass

The mole concept is a central idea in chemistry that provides a bridge between the mass of a substance and the number of its constituent particles, using molar mass as the conversion factor. This concept is essential when quantifying reactants and products in chemical reactions and applying gas laws, as it allows for the conversion of given masses into moles, which are directly used in thermodynamic calculations.

Decomposition Reaction

A decomposition reaction is a type of chemical reaction in which a single compound breaks down into two or more simpler substances. In thermodynamics, understanding the stoichiometry of decomposition reactions is crucial for determining the molar amounts of products formed, which subsequently influence calculations of pressure and other state variables in a closed system.

Volume Correction for Solids

Volume correction for solids involves adjusting calculations in systems where both gases and solids are present, particularly when the solids occupy a significant fraction of the volume. In such cases, the physical volume taken up by the solid must be subtracted from the total container volume, ensuring that subsequent gas behavior calculations, such as those using the ideal gas law or van der Waals equation, accurately reflect the true volume available to the gas molecules.

Ideal Gas Law

The ideal gas law, expressed as PV = nRT, is a fundamental equation in thermodynamics that relates the pressure, volume, temperature, and number of moles of a gas. It assumes that gas particles do not interact and that their individual volumes are negligible. This law is used for approximations and initial calculations in systems where gases behave ideally.

Van der Waals Equation

The van der Waals equation is an adjustment of the ideal gas law that accounts for the non-ideal behavior of real gases. It introduces two additional parameters: 'a' for intermolecular attractive forces and 'b' for the finite volume occupied by the gas molecules. This equation is particularly useful for calculating gas properties under high-pressure or low-temperature conditions where deviations from ideality become significant.

Transcript

00:01 This problem gives us a reaction that produces a gas at the end.

00:04 And it's asking us to use the ideal gas law to find the pressure at the end of the reaction.

00:09 And then it's asking us to find the real pressure using the vanderwals equation.

00:13 So, and that means that we have to take in account this solid contributing to the volume.

00:20 So first, to find the pressure using the ideal gasol, that's pb equals nrt.

00:28 Pressure is equal to nrt over v.

00:31 We need to get our units into liters, atms, moles, and kelvin.

00:36 And so the volume is 0 .5 liters.

00:40 And if we, then we need to find the number of moles.

00:43 So we need to find the number of moles of co2 gas.

00:46 I like to use a bca table, stands for before, change, and after of the reaction.

00:51 So 25 grams, we need to get this into moles.

00:54 So the molar mass of c .i .c .o .3 is 100 grams.

00:57 Per 1 mole.

00:59 So this lost 0 .25 moles and it ended at 0 .20 and co2 gained 0 .25 moles and it started at 0.

01:12 And so the ending amount of moles of gas is 0 .25 moles.

01:17 So then if we plug all these values in, we get the pressure to be equal to 61 .5 atm.

01:25 And that's the same as the partial pressure of co2, because it's the only gas...

Please give Ace some feedback