Question

Consider the complete combustion of 1.00g of glucose to carbon dioxide and water. Calculate the work of expansion when water obtained is(a)liquid, (b)vapour at 20.0°C when the external pressure is 1.20atm.For both cases, briefly comment on the meaning of the sign/result.

Answer 1

Glucose : CGH12O6 - Combustion of Glucose CCHIQ + go a 6 coat 6H20 (19) Mo = 180 (21) I gm Glucose - ( 1 ) moles Mole I mole I mole Glucose = Glucose = 6mole cog + 6 mole to 6 x t mole (cog) + 6 mole 14 180 180 180

Now - Work of expansion work against constant (2) pressure. External pressure is constant) Pext = 1.do atm Constant external pressure Work = done against - pex(AV) - Volume of Gases formed. Consider Volume of glucose negligibles liquid at Room Temperature) * Case I (a) - Water is liquid → Volume of Gras (cog only) Moles of coa = mole Volume of cop using ideal gas equation pr=MRT V= MRT mol. k. MART = Geo mole) * 254 0.082 La [Pext = Pin at equilibrium) 1.20 atm = 0.6673 litre. Work done – – 1.80 (0.6673) (atmxL) = 0.8 atm x L)
a
Here negative sign represents by the systemor system lose expansion work done energy while Case (b) when water is Vapour. Work done = -Pex CAV) Volume of Gases formed ccog + H2O) * Vases = ? (using pr=MRT Moles of (cog+H2O) logo (Pext = Pin at equilibrium V x 1.90 atm = 16 (180 - 6 mole x 0.088 L-atm 293 molek V = 12 mol x 180 0.0 8 2 L.atm malik 293K 1.80 atm 1.3348 Litre. Work = = - done – Pext. Ar -1.90 x C1-3848) Litre. x atm -1.6 Litre at mit Here are sign represents work done by the system/Gases. !! -