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Answer:
139.97 kPa
Explanation:
Assuming propane is an ideal gas
PV = nRT
n₁ ( number of mole f the first gas) of the 1 m³ rigid tank = P₁V₁ / RT₁ R gas constant = 8.314 J/mol
n₁ = 100kPa × 1 m³ / (8.314 J/mol × 300 K) = 0.0401 mol
n₂ = 0.5 m³ × 250 kPa / ( 8.314J/mol × 400 K) = 0.0376 mol where n₂ is the number of mole of the second gas
n total = n₁ + n₂ = 0.0401 mol + 0.0376 mol = 0.0777 mol
PV = nRT
P final pressure = nRT / V where V = V₁ + V₂ = 1 m³ + 0.5 m³ = 1.5 m³
P final pressure = 0.0777 mol × 8.314 J/mol × 325 K / 1.5 m³ = 139.97 kPa
Answer:
Mass flow rate= 7.53 kg/sec
Exit Area= 0.108m^2
Explanation:
The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.
Answer:
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Answer:
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