Oxygen at 5 bar and 320 K is supplied to an insulated tank of volume 5
m3
. The tank initially contains oxygen at 1 bar and 300 K. Determine:
i) the final temperature in the tank when its pressure reaches 5 bar.
ii) the mass that has entered the tank.
Gives
P1=100kpaP_1=100kpaP1=100kpa
T1=300k
V1=P2V2×T1T2×P1V_1=\frac{P_2V_2 \times T_1}{T_2\times P_1}V1=T2×P1P2V2×T1
Put Value
V1=23.43 m3
P2=500kpaP_2=500kpaP2=500kpa
T2=320KT_2=320KT2=320K
V2=5m3V_2=5m^3V2=5m3
P1V1T1=P2V2T2\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}T1P1V1=T2P2V2
T2=T1P1V1×P2V2T_2=\frac{T_1}{P_1V_1}\times P_2V_2T2=P1V1T1×P2V2
Put value
T2=300×500Kpa×5100Kpa×23.43=T_2=\frac{300 \times500Kpa\times5}{100Kpa\times23.43}=T2=100Kpa×23.43300×500Kpa×5=320K
Part(b)
We know that
m=P1V1RT1m=\frac{P_1V_1}{RT_1}m=RT1P1V1
m=100×23.434.12×400=1.42kgm=\frac{100 \times23.43}{4.12\times400}=1.42kgm=4.12×400100×23.43=1.42kg
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