Respuesta :
Answer:
a)P₂ =4 bar
b)W= - 1482.48 KJ
It means that work done on the system.
c)S₂ - S₁ = 3.42 KJ/K
Explanation:
Given that
T₁ = 300 K ,V₁ = 3 m³ ,P₁=2 bar
T₂ = 600 K ,V₂=V₁ 3 m³
Given that tank is rigid and insulated.It means that volume of the gas will remain constant.
Lets take the final pressure = P₂
For ideal gas P V = m R T
[tex]\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}[/tex]
[tex]P_2=\dfrac{T_2}{T_1}\times P_1[/tex]
[tex]P_2=\dfrac{600}{300}\times 2[/tex]
P₂ =4 bar
Internal energy
ΔU = m Cv ΔT
Cv=0.71 KJ/kg.k for air
[tex]m=\dfrac{PV}{RT}[/tex]
[tex]m=\dfrac{200\times 3}{0.287\times 300}\ kg[/tex]
m= 6.96 kg
ΔU= 6.96 x 0.71 x (600 - 300)
ΔU=1482.48 KJ
From first law
Q= ΔU + W
Q= 0 Insulated
W = - ΔU
W= - 1482.48 KJ
It means that work done on the system.
Change in the entropy
[tex]S_2-S_1=mC_v\ \ln\dfrac{T_2}{T_1}[/tex]
[tex]S_2-S_1=6.96\times 0.71\ \ln\dfrac{600}{300}[/tex]
S₂ - S₁ = 3.42 KJ/K
