Answer:
Explanation:
Given that the tank is insulated and rigid, heat transfer to the surrounding is zero and there is no change in volume of system(container).
p1=400kpa, T1= 40°c= 313k, p2=800kpa
volume of container = volume of cube= a³=1³=1m³
Air is acting as an ideal gas, hence, applying ideal gas equatin [tex]P_1V_1=MRT_1\\\\M=\frac{p_1v_1}{RT_1}=\frac{400\times 10^3 \times 1}{287\times 313}=4.453kg[/tex] R= 287J/kg.k for air
a.) Process 1-2 constant volume process v1=v2
P∝T=[tex]\frac{P_1}{P_2}=\frac{T_1}{T_2}=\frac{400}{800}=\frac{313}{T_2}\\\\T_2=626K =353^oC[/tex]
b.) According to first law of thermodynamics
[tex]\deltaQ=dv+d\omega[/tex]
δQ=0(insulated container)
[tex]0=dv+d\omega\\d\omega=-dv=-mc_v\delta T[/tex]
[tex]c_v=\frac{1}{r-1}R[/tex]
where r= specific heat ratio; for air, r=1.4, R= 0.287KJ/kg.k
[tex]c_v=\frac{1}{r-1}R=\frac{1}{1.4-1}0.287=0.7175KJ/kg.k[/tex]
Then
[tex]d\omega=-dv=-mc_v\delta T=-4.453\times 0.7175\tims (626-313)=-1000.043KJ[/tex]
The negative sign means work done of system is in reverse.
c.) Entropy generation = change in system + change in surrounding
change in surrounding is zero
change in system=[tex]mc_vln(\frac{T_2}{T_1})+ mrln(\frac{v_2}{v_1})[/tex]
v1=v2; v2/v1=1; ln(1)=0
entropy gen= 4.453*0.7175*ln(626/313) + 0 =2.215KJ/k
