Answer:
1) the entropy generated is Δs= 0.0363 kJ/kg K
2) the minimum theoretical work is w piston = 201.219 kJ/kg
Explanation:
1) From the second law of thermodynamics applied to an ideal gas
ΔS = Cp* ln ( T₂/T₁) - R ln (P₂/P₁)
and also
k= Cp/Cv , Cp-Cv=R → Cp*( 1-1/k) = R → Cp= R/(1-1/k)= k*R/(k-1)
ΔS = k*R/(k-1)* ln ( T₂/T₁) - R ln (P₂/P₁)
where R= ideal gas constant , k= adiabatic coefficient of air = 1.4
replacing values (k=1.4)
ΔS = k*R/(k-1)* ln ( T₂/T₁) - R ln (P₂/P₁)
ΔS = 1.4/(1.4-1) *8.314 J/mol K * ln( 600K/300K) - 8.314 J/mol K * ln (10 bar/ 1bar)
ΔS = 1.026 J/ mol K
per mass
Δs = ΔS / M
where M= molecular weight of air
Δs = 1.026 J/ mol K / 28.84 gr/mol = 0.0363 J/gr K = 0.0363 kJ/kg K
2) The minimum theoretical work input is carried out under a reversible process. from the second law of thermodynamics
ΔS =∫dQ/T =0 since Q=0→dQ=0
then
0 = k*R/(k-1)* ln ( T₂/T₁) - R ln (P₂/P₁)
T₂/T₁ = (P₂/P₁)^[(k-1)/k]
T₂ = T₁ * (P₂/P₁)^[(k-1)/k]
replacing values
T₂ = 300K * ( 10 bar/1 bar)^[0.4/1.4] = 579.2 K
then from the first law of thermodynamics
ΔU= Q - Wgas = Q + Wpiston ,
where ΔU= variation of internal energy , Wgas = work done by the gas to the piston , Wpiston = work done by the piston to the gas
since Q=0
Wpiston = ΔU
for an ideal gas
ΔU= n*Cv*(T final - T initial)
and also
k= Cp/Cv , Cp-Cv=R → Cv*( k-1) = R → Cv= R/(k-1)
then
ΔU= n*R/(k-1)*(T₂ - T₁)
W piston = ΔU = n*R/(k-1)*(T₂ - T₁)
the work per kilogram of air will be
w piston = W piston / m = n/m*R/(k-1)*(T₂ - T₁) = (1/M*) R/(k-1)*(T₂ - T₁) ,
replacing values
w piston = (1/M*) R/(k-1)*(T₂ - T₁) = 1/ (28.84 gr/mol)* 8.314 J/mol K /0.4 * ( 579.2 K - 300 K) = 201.219 J/gr = 201.219 kJ/kg
