Question
Say there is carbon dioxide under a pressure of 15 atm in thefire extinguisher on the wall. Do the first law workup for theexpansion of the gas ( E, H, T, q and work) undereach of the following conditions. Cp,m for CO2 is 37.11 J/Kmole A) the expansion is isothermal and reversible B) the expansion is isothermal against the fixed externalpressure of 1 atm C) the expansion is quick enough that there is not chance forany heat exchange with the surroundings, that is , adiabatic,against the fixed external pressure of 1 atm D) the adiabatic expansion in which the greatest possibleamount of work is done I am really lost on this question and I have give it a trysince Thursday but to no avail! Your help is really and urgentlyneeded here and I will appreciate that. Thank you Say there is carbon dioxide under a pressure of 15 atm in thefire extinguisher on the wall. Do the first law workup for theexpansion of the gas ( E, H, T, q and work) undereach of the following conditions. Cp,m for CO2 is 37.11 J/Kmole A) the expansion is isothermal and reversible B) the expansion is isothermal against the fixed externalpressure of 1 atm C) the expansion is quick enough that there is not chance forany heat exchange with the surroundings, that is , adiabatic,against the fixed external pressure of 1 atm D) the adiabatic expansion in which the greatest possibleamount of work is done I am really lost on this question and I have give it a trysince Thursday but to no avail! Your help is really and urgentlyneeded here and I will appreciate that. Thank you
Explanation / Answer
Given that some amount of CO2 is excerting apressure of 15atm on hte walls of a fire extinguisher. First law : The internal energy of the of an isolated systemis constant Or The increase in the internal energy of a system is equal to theamount of energy added by heating the system, minus the amount lostas a result of the work done by the system on itssurroundings. mathematical form : U = q + w --------------------- (a) In isothermal reversible expansion ofgas for an isothemal process, U = 0 When temperature do not change (isothermal process) , enthalpydoes not chnage so H = 0 for isothermal process, T = 0 work done, w = -nRTln(Vf / Vi) Whare Vf, Vi are fianl and initial volumes of the gas; whenVf>Vi then w <0 otherwise w >0 q = U - w ------------- (b) When expansion is isothermal againstexternal pressure then U =0 (temeprature does not change inisothermal process so that intrenal energy do not change) T = 0 H = CpT = Cp*0 = 0 w = -pex V q = U -w = 0 -(-pex V) ------------------------- (c) In adiabatic process, The change in internal energy of a perfect gas whentemperature is changed from Ti to Tf and the volume is changedfrom Vi to Vf can be expressed as sum of two steps. In the first step, only the volume changes and the temperatureis held constant at its initial value. However, because theinternal energy of a perfect gas is independent of the volume themolecules occupy, the overall change in internal energy arises solely from the second step, the change in temperatureat constant volume. Provided the heat capacity is independent of temperature, this change is U = CV (Tf -Ti)=CV T = n,CV,m T = n(Cp,m - R)T Because the expansion is adiabatic , we know that q= 0;because U = q+w, it then fallows that U =wad wad = n(Cp,m - R)T ------------------ (d) Discussion in part c may helps you. -------------------- In the data noumber of moles of gas was not specified so Ihave given only qualitative analysis. I hope it helps you , I welcome any further clarification ifyou need. wad = n(Cp,m - R)T ------------------ (d) Discussion in part c may helps you. -------------------- In the data noumber of moles of gas was not specified so Ihave given only qualitative analysis. I hope it helps you , I welcome any further clarification ifyou need.