Question
Consider a system consisting of 2.0 mol CO2 (g), initially at25C and 10 atm and confirmed to a cylinder of cross section10.0cm2. It is allowed to expandadiabatically against a constant externalpressure of 1 atm until the piston has moved outwardsthrough 20 cm. Assume that carbon dioxide may be consideredas a perfect gas with the temperature independent molar heatcapacity Cm,V = 28.8 J K mol and calculate: q (heat), v(volume change), w (work), U (internal energy change),T (temperature change), final temperature (Tf) and S(entropy change) R = 8.31451 J K mol R = 8.20578 X 10^-2 L atm K mol R = 8.31451 X 10^-2 L bar K mol R = 8.31451 L kPa K mol Please help in solving this problem. This is a practiceexample for an exam I will be having this week. Thankyou. R = 62.364 L Torr K mol Consider a system consisting of 2.0 mol CO2 (g), initially at25C and 10 atm and confirmed to a cylinder of cross section10.0cm2. It is allowed to expandadiabatically against a constant externalpressure of 1 atm until the piston has moved outwardsthrough 20 cm. Assume that carbon dioxide may be consideredas a perfect gas with the temperature independent molar heatcapacity Cm,V = 28.8 J K mol and calculate: q (heat), v(volume change), w (work), U (internal energy change),T (temperature change), final temperature (Tf) and S(entropy change) R = 8.31451 J K mol R = 8.20578 X 10^-2 L atm K mol R = 8.31451 X 10^-2 L bar K mol R = 8.31451 L kPa K mol Please help in solving this problem. This is a practiceexample for an exam I will be having this week. Thankyou. R = 62.364 L Torr K mol
Explanation / Answer
Moles of CO2 = 2.0 mol Molar heat capacity Cm,V = 28.8 J K^-1 mol^-1 Initial volume of CO2 =(2.0mol)(0.0821Latmmol^-1)K^-1(25+273)K / 10atm = 4.89L After the expansion, volume of gas = 4.89L +10.0cm^2 *20cm =4.89L +0.2L = 5.09L 10.0cm^2 * 20cm = 200cm^3 = 200cm^3(0.1dm / 1cm)^3 = 200x10^-3 dm^3 = 0.2L w = -Pext* V = -1atm (101.325x10^3 Pa 1atm )* (5.09 - 4.89)L = -20.265kJ For adiabatic process , U = w =20.265kJ H = nCp,mT Tf = Ti *{(Cv,m + Pext *R / Pi) /(Cv,m + Pext *R / Pf)} T = Tf -Ti = Tf - 298K