Answer:
the entropy change of the gas is ΔS= 2.913 J/K
Explanation:
starting from the first law o thermodynamics for an adiabatic reversible process
ΔU= Q - W
where
ΔU = change in internal energy
Q= heat flow = 0 ( adiabatic)
W = work done by the gas
then
-W= ΔU
also we know that the ideal compression work Wcom= - W , then Wcom = ΔU. But also for an ideal gas
ΔU= n*cv* (T final - T initial)
where
n=moles of gas
cv= specific heat capacity at constant volume
T final =T₂= final temperature of the gas
T initial =T₁= initial temperature of the gas
and also from an ideal gas
cp- cv = R → cv = 7/2*R - R = 5/2*R
therefore
W com = ΔU = n*cv* (T final - T initial)
for an ideal gas under a reversible adiabatic process ΔS=0 and
ΔS= cp*ln(T₂/T₁) - R* ln (P₂/P₁) =0
therefore
T₂ = T₁* (P₂/P₁)^(R/cp) = T₁* (P₂/P₁)^(R/(7/2R))= T₁* (P₂/P₁)^(2/7)
replacing values T₁=25°C= 298 K
T₂ =T₁* (P₂/P₁)^(2/7) = 298 K *(7 bar/2 bar)^(2/7) = 426.25 K
then
W com = ΔU = n*cv* (T₂- T₁)
and the real compression work is W real = 1.35*Wcom , then
W real = ΔU
W real = 1.35*Wcom = n*cv* (T₃ - T₁)
T₃ = 1.35*Wcom/n*cv + T₁ = 1.35*(T₂- T₁) + T₁ =1.35*T₂ - 0.35*T₁ = 1.35*426.25 K - 0.35 *298 K = 471.14 K
T₃ = 471.14 K
where
T real = T₃
then the entropy change will be
ΔS= cp*ln(T₃/T₁) - R* ln (P₂/P₁) = 7/2* 8.314 J/mol K *ln(471.14 K /298 K ) - 8.314 J/mol K* ln (7 bar / 2 bar) = 2.913 J/K
ΔS= 2.913 J/K
