Estimate the work done during the decomposition of 1.0 mol CaCo3 (s) under 1 bar

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CaCO3 (s) -> CaO (s) + CO2 (g) So, at consant pressure of 1 bar, work done = P*(volume of product-volume of reactant) We know that, volume of gas is much larger than volume of solid., and only CO2 is in gaseous form => we can neglect the volume of solids Hence, work done = P(Volume of CO2) Now, 1 mole of CaCO3 gives 1 mole of CO2 => volume of CO2 = (1*8.314*(273+25))/(1*10^5) = 0.02477 m^3 => work done = (1*10^5)*(0.02477) = 2477 J = 2.48 KJ

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