Consider one mole of an ideal monatomic gas ( For a monatomic gasC V,m = 3R/2 =

Question

Consider one mole of an ideal monatomic gas ( For a monatomic gasCV,m = 3R/2 = 12.472 J/mol-K, Cp,m = 5R/2 =20.786 J/mol-K). The gas is initially at a pressure p = 4.000 atmand a temperature T = 400.0 K.Find Ssyst, Ssurr, Sunivfor the following processes:
a) An isothermal reversible compression of the gas to a finalpressure pf = 10.000 atm. b) An reversible cooling of the gas from Ti = 400.0K to Tf = 300.0 K, at a constant pressure p = 4.00atm. c) An adiabatic reversible compression of the gas to a finalpressure pf = 10.000 atm. d) An adiabatic irreversible expansion of the gas against aconstant pressure pex = pf =1.000atm.
I know it's in parts, but if you could please help me I willimmediately rate lifesaver minutes after receivinghelp.
a) An isothermal reversible compression of the gas to a finalpressure pf = 10.000 atm. b) An reversible cooling of the gas from Ti = 400.0K to Tf = 300.0 K, at a constant pressure p = 4.00atm. c) An adiabatic reversible compression of the gas to a finalpressure pf = 10.000 atm. d) An adiabatic irreversible expansion of the gas against aconstant pressure pex = pf =1.000atm.
I know it's in parts, but if you could please help me I willimmediately rate lifesaver minutes after receivinghelp.

Explanation / Answer

Given: CV,m= 3R/2 = 12.472J/mol-K            Cp,m = 5R/2 = 20.786J/mol-K            initial pressure of gas, Pi = 4.0atm            initial temperature, Ti = 400K a) S = nRln(Pi / Pf)      CV,m = CV/ n and  Cp,m = Cp /n       Cp - CV =R so n (Cp,m - CV,m) = R                                 n = R/ (Cp,m - CV,m)     Hence, Ssys = R/(Cp,m - CV,m)*Rln(Pi / Pf)                       = R2 / (Cp,m - CV,m) *ln(Pi /Pf)                       = ((8.314Jmol-1K-1 )2 /8.314Jmol-1K-1 )*ln (4/10)                       = 8.314Jmol-1K-1 *-0.916                       = -7.618Jmol-1K-1   Thus, per mole of gas Ssys =-7.618J/K       Ssurr = -qsys/T               , T is the temperature of the surroundings and in isothermalprocess the sys and sur at same temperature. S univ = S sys + S surr -------------- S = nRln(Vf / Vi)     we know pV = nRT then pVi = nRTi and  pVf = nRTf at constantpressure so Vf/ Vi = Tf/Ti then Ssys = nRln(Tf / Ti) ------------- For an adiabatic process, S sys = -nRTln(Vf/Vi)                                       S surr = 0 as there is no exchangeof heat between system andsurroundings -------------- For an adiabatic irreversible expansion of a gas the entropyis same as for the isothermal expansion so for adiabatic irreversible expansion , S = nRln(Pi / Pf)   Thus, per mole of gas Ssys =-7.618J/K       Ssurr = -qsys/T               , T is the temperature of the surroundings and in isothermalprocess the sys and sur at same temperature. S univ = S sys + S surr -------------- S = nRln(Vf / Vi)     we know pV = nRT then pVi = nRTi and  pVf = nRTf at constantpressure so Vf/ Vi = Tf/Ti then Ssys = nRln(Tf / Ti) ------------- For an adiabatic process, S sys = -nRTln(Vf/Vi)                                       S surr = 0 as there is no exchangeof heat between system andsurroundings -------------- For an adiabatic irreversible expansion of a gas the entropyis same as for the isothermal expansion so for adiabatic irreversible expansion , S = nRln(Pi / Pf)

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