Given data
Mass of air = 1 kg
Initial temperature T1 = 150 + 273 = 423 K
Initial pressure P1 = 4.8 bar
Final temperature T2 = 200 + 273 = 473 K
The container is rigid (volume is constant)
From the ideal gas equation
T1P1V1=T2P2V2
P2=P1×T1T2
=4.8barx(423473)
= 5.37 bar
For isochoric process
Entropy change S=Cvln(T1T2)
=0.718kJ/kgK×ln(423473)
= 0.08 kJ/kgK
Average temperature T = 2(473+423)=448K
Heat supplied = T×S
=448Kx0.08kJ/kgK=35.9kJ/kg
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