Calculate the work done during an isothermal reversible expansionof a van der wa

Question

Calculate the work done during an isothermal reversible expansionof a van der waals gas. Account physically for the way inwhich th ecoefficients a and b appear in the finalexpression. Plot on the same graph the indicator diagrams forthe isothermal reversible expansion of a) a perfect gas, b) a vander waals gas in which a=0 and b=5.11x10-2dm3mol-1, and c) a=4.2dm6atmmol-2 and b=0. The valuesselected exaggerate the imperfections but give rise to significanteffects on the indicator diagrams. Take vi = 1.0dm3, n=1.0 mol, and T=298 K.

Explanation / Answer

For VDW P = nRT(V-nb) - an2/V2 W = -integral [ P*dV ] = -integral [ dV*nRT/(V-nb) ] +integral [ dV*an2/V2] W = -nRT * ln [(Vf- nb)/(Vi - nb)] - an2*[1/ Vf - 1 Vi] a) perfect gas,   W = -integral [ nRT*dV/V] =-nRT*ln(Vf/Vi) W = -(1mol * 8.314 J/mol-K * 298 K) *ln(Vf / 1.0 dm3) W = -2477.572*ln(Vf/ 1.0dm3) plot W vs Vf         by theabove relation,   (Vf is in dm3 and W is in J) b) W = -nRT * ln [(Vf- nb)/(Vi - nb)] - an2*[1/ Vf - 1 Vi ] a=0 and b=5.11x10-2 dm3mol-1 W = -(1 mol *8.314 J/mol-K * 298K) * ln ( Vf - 1 mol * 5.11e-2dm3/mol)/ (1.0dm3 - 1mol *5.11e-2 dm3/mol) - 0 last term is due to a = 0 W = -2477.572*ln[(Vf - 5.11e-2 dm3)/(0.9489dm3) ] c) c) a=4.2 dm6atmmol-2 and b=0 W = -nRT * ln [(Vf- nb)/(Vi - nb)] - an2*[1/ Vf - 1 Vi ] W = -nRT*ln(Vf/Vi) - an2*[ 1/ Vf - 1 Vi ] convert a = 4.2 dm6*atm/mol2 * ( 1 m / 10 dm)3 *(101325 Pa/ 1atm) = 425.56500 dm3 *m3*Pa/mol2 recall m3*Pa =1 J a = 425.565 J*dm3 /mol2 W = -2477.572*ln(Vf/ 1.0dm3) - (425.56500J*dm3/mol2)*(1mol)2 * (1/Vf - 1/ 1.0 dm3) W = -2477.572*ln(Vf/ 1.0dm3) - 425.56 dm3 *(1/ Vf -dm-3) Plot W vs. Vf   for W in J and Vf in dm3

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