Question

Consider 1 mol of aluminum oxide that is compressed reversibly at 298 K. Estimate the pressure increase that is required to impart 1 J of mechanical work during the isothermal process.
Take the starting pressure (Pi) to be 1 atm, the molar volume of aluminum oxide (Vm) to be 25.715 cm3 mol-1 and the volume coefficient of compressibility of aluminum oxide (β) to be 8.0×10-7 atm-1. With the increasing pressure, the volume change of the aluminum oxide is not zero but it is negligibly small (i.e., the ratio of the volume change to the initial volume, ΔV/Vi <<1).

Answer 1

Given T =298 K Work = 1 J Pi = 1 atm=101325 Pa Vm=V1 = 25.715 *10^-6 m^3   

   volume coefficient of compressibility = 8.0 *10^-7 atm^-1

   work done in a isothermal process W = 2.303RTlog(V1/V2)

1 = 2.303 (8.314 )(298)log((25.715*10^-6)/V2)

1=5705.84log (25.715*10^-6/V2)

   V2 = (25.715 *10^-6)/e^(1/5705.84)

V2 =2.571*10^-5 m^3

but we know   = 8.0 *10^-7 atm^-1

   = (V2-V1)/(V2(P2-P1))

   8.0 *10^-7 = (0.144*10^-5)/((2.571*10^-5)(P2-101325))

   P2-101325 = 0.056/8*10^-7

   P2 =70000+101325

P2 =171325 Pa

P2= 1.69 atm