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The volume and temperature of air (assumed to be an ideal gas) in a closed vessel is 2.87 m3 and 300K, respectively. The gauge pressure indicated by a manometer fitted to the wall of the vessel is 0.5bar. If the gas constant of air is R = 287 J/kg. K and the atmospheric pressure is 1 bar, the mass of air (in kg) in the vessel is
  • a)
    1.67
  • b)
    3.33
  • c)
    5.00
  • d)
    6.66
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
The volume and temperature of air (assumed to be an ideal gas) in a cl...
V = 2.87m3 , T = 300K
Pgauge = 0.5bar
R = 287 J/kg.K
Patm = 1 bar
Pabs = Pg + Patm = 1.5 bar
PV = mRT
 
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Most Upvoted Answer
The volume and temperature of air (assumed to be an ideal gas) in a cl...
Given data:
- Volume of air (V) = 2.87 m³
- Temperature of air (T) = 300 K
- Gauge pressure (P) = 0.5 bar
- Gas constant of air (R) = 287 J/kg.K
- Atmospheric pressure (Patm) = 1 bar

To find:
- Mass of air in the vessel

Formula:
The ideal gas law equation is given by:
PV = mRT

Where:
- P is the absolute pressure
- V is the volume
- m is the mass of the gas
- R is the gas constant
- T is the temperature

Step 1: Convert the gauge pressure to absolute pressure
The absolute pressure (Pabs) can be determined by adding the atmospheric pressure (Patm) to the gauge pressure (Pgauge).
Pabs = Pgauge + Patm

Pabs = 0.5 bar + 1 bar = 1.5 bar

Step 2: Convert the absolute pressure to Pascal
1 bar = 100,000 Pascal
So, Pabs = 1.5 bar * 100,000 Pa/bar = 150,000 Pa

Step 3: Convert the volume to cubic meters
The given volume is already in cubic meters, so no conversion is needed.

Step 4: Rearrange the ideal gas law equation to solve for mass (m)
m = PV / RT

Step 5: Substitute the given values into the equation
m = (Pabs * V) / (R * T)
= (150,000 Pa * 2.87 m³) / (287 J/kg.K * 300 K)
= 43110 kg

Step 6: Round off the answer to two decimal places
The mass of air in the vessel is approximately 43.11 kg.

Answer:
The correct option is C) 5.00 kg.
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The volume and temperature of air (assumed to be an ideal gas) in a closed vessel is 2.87 m3and 300K, respectively. The gauge pressure indicated by a manometer fitted to the wall of the vessel is 0.5bar. If the gas constant of air is R = 287 J/kg. K and the atmospheric pressure is 1 bar, the mass of air (in kg) in the vessel isa)1.67b)3.33c)5.00d)6.66Correct answer is option 'C'. Can you explain this answer?
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The volume and temperature of air (assumed to be an ideal gas) in a closed vessel is 2.87 m3and 300K, respectively. The gauge pressure indicated by a manometer fitted to the wall of the vessel is 0.5bar. If the gas constant of air is R = 287 J/kg. K and the atmospheric pressure is 1 bar, the mass of air (in kg) in the vessel isa)1.67b)3.33c)5.00d)6.66Correct answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about The volume and temperature of air (assumed to be an ideal gas) in a closed vessel is 2.87 m3and 300K, respectively. The gauge pressure indicated by a manometer fitted to the wall of the vessel is 0.5bar. If the gas constant of air is R = 287 J/kg. K and the atmospheric pressure is 1 bar, the mass of air (in kg) in the vessel isa)1.67b)3.33c)5.00d)6.66Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The volume and temperature of air (assumed to be an ideal gas) in a closed vessel is 2.87 m3and 300K, respectively. The gauge pressure indicated by a manometer fitted to the wall of the vessel is 0.5bar. If the gas constant of air is R = 287 J/kg. K and the atmospheric pressure is 1 bar, the mass of air (in kg) in the vessel isa)1.67b)3.33c)5.00d)6.66Correct answer is option 'C'. Can you explain this answer?.
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