A steel container with a volume of 500. 0 mL is evacuated, and 25.0 g of CaCO3 i

Question

A steel container with a volume of 500. 0 mL is evacuated, and 25.0 g of CaCO3 is added. The container and contents are then heated to 1500 K, causing the CaCO3 to decompose completely, according to the equation

Using the ideal gas law and ignoring the volume of any solids remaining in the container, calculate the pressure inside the container at 1500 K

Now make a more accurate calculation of the pressure inside the container. Take into account the volume of solid CaO (density = 3.34 g/mL) in the container, and use the vander Waals equation to calculate the pressure. The vander Waals constants for CO2(g) are: a=3.59 (L2atm)/mol2and b=0.0427 L/mol.

Explanation / Answer

CaCO3(s) -----------> CaO(s) + CO2(g)

Molar mass of CaCO3 = 100 g/mole ; CaO = 56 g/mole ; CO2 = 44 g/mole

Now, moles of CaCO3 decomposing = mass/molar mass = 25/100 = 0.25

Thus, moles of CO2 formed = moles of CaCO3 decomposing = 0.25

Now, applying Ideal Gas EQuation, i.e., P*V = n*R*T ; where P = pressure in atm , V = volume of the container in litres , R = universal Gas Constant = 0.0821 atm-litre/mole/K , n = moles of the gas & T = temperature in kelvin, we get

P*0.5 = 0.25*0.0821*1500
or, P = 61.575 atm

Now, After taking CaO volume into consideration:-

Volume, V available for CO2 = Total volume of container - volume occupied by CaO
Now, Volume occupied by CaO = mass of CaO/density of CaO = (moles of CaO*molar mass of CaO)/density of CaO = (0.25*56)/3.34 = 4.192 ml

Hence , volume available for CO2 = 495.808 ml = 0.496 litres

Now, Applying Vander Waal's Equation, i.e., (P - (n2a/V2))*(V - nb) = n*R*T

or, (P - (0.252*3.59))*(0.496 - 0.25*0.0427) = 0.25*0.0821*1500

or, P = 63.212 atm

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