A sample of 1.00 mol 0f a perfect monatomic gas with a molar heat capacity at co

Question

A sample of 1.00 mol 0f a perfect monatomic gas with a molar heat capacity at constant volume of 3/2R initially at 298K and 10l is expanded with the surroundings maintained at 298K to a final volume of 20L in three ways ways a) isothermally and reversibly b) isothermally against a constant extenal pressure 0.50 atm and c) adiabatically against a constant external pressure of 0.5 calculate deltaS , delta H and delta T for each path wher the data permis.

Explanation / Answer

(a.)isothermally and reversibly wrev=-nRT ln(Vfinal/Vinitial) = -1717.322 J * for an isothermal process either reversible or irreversible the internal energy change, Delta U, is equal to zero (dU=0) therefore: q = -w = 1717.332 J Delta S = qrev / T = 5.763 J/K If: PiVi = PfVf = nRT then: Delta(pV) = 0 Delta H = Delta U + Delta(pV) = 0 + 0 = 0 Delta T = 0 (isothermal) Delta G = Delta H -T Delta S = 1717.3144 J ( which is = q ) Delta A = Wrev = -1717.322 J b)dU = Q + W since dU = n.Cv.dT where dT = 0, then dU = 0 => Q = -W W = -pext.dV = -50650 J Q = -W = 50650J deltaS = n.R.ln(Vfinal/Vini) = (1.0)(8.314)ln(2) = 5.763J/mol A = U - TS => dA = dU -TdS => deltaA = 0 - T.deltaS = -298*5.763 = 1717J G = H - TS => dG = dH - T.dS dH = n.Cp.dT => dH = 0 for isothermal process dG = -T.dS => deltaG = -T.deltaS = deltaA = 1717J

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