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A room contains 35 kg of dry air and 0.5 kg of water vapor. The total pressure and temperature of air in the room are 100 kPa and 25°C respectively. Given that the saturation pressure for water at 25°C, is 3.17 kPa, the relative humidity of the air in the room is
  • a)
    67%
  • b)
    55%
  • c)
    83%
  • d)
    71% 
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A room contains 35 kg of dry air and 0.5 kg of water vapor. The total ...
ma = 35 kg, mV = 0.5 kg

Pt = 100 kPa

PV = 2.238 kPa
Relative humidity
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Most Upvoted Answer
A room contains 35 kg of dry air and 0.5 kg of water vapor. The total ...
Understanding Relative Humidity
Relative humidity (RH) measures the current amount of water vapor in the air compared to the maximum amount of water vapor that air can hold at a given temperature.
Given Data
- Mass of dry air = 35 kg
- Mass of water vapor = 0.5 kg
- Total pressure = 100 kPa
- Temperature = 25°C
- Saturation pressure of water at 25°C = 3.17 kPa
Calculating Partial Pressure of Water Vapor
To find the relative humidity, we first need the partial pressure of the water vapor:
- The molar mass of dry air is approximately 29 g/mol, and the molar mass of water vapor is about 18 g/mol.
- Calculate the number of moles of air:
Number of moles of dry air = Mass / Molar mass = 35,000 g / 29 g/mol ≈ 1207 moles
- Calculate the number of moles of water vapor:
Number of moles of water vapor = 500 g / 18 g/mol ≈ 27.78 moles
- Using the ideal gas law, the partial pressure of water vapor (P_w) can be approximated as:
P_w = (n_w / (n_a + n_w)) * P_total
Where:
- n_w = moles of water vapor
- n_a = moles of dry air
- P_total = total pressure
Using the values:
P_w = (27.78 / (1207 + 27.78)) * 100 kPa ≈ 2.27 kPa
Calculating Relative Humidity
Now, we can calculate the relative humidity using the formula:
RH = (P_w / P_sat) * 100%
Substituting the known values:
RH = (2.27 kPa / 3.17 kPa) * 100% ≈ 71%
Conclusion
Thus, the relative humidity of the air in the room is approximately 71%, which corresponds to option 'D'.
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Community Answer
A room contains 35 kg of dry air and 0.5 kg of water vapor. The total ...
ma = 35 kg, mV = 0.5 kg

Pt = 100 kPa

PV = 2.238 kPa
Relative humidity
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A room contains 35 kg of dry air and 0.5 kg of water vapor. The total pressure and temperature of air in the room are 100 kPa and 25°C respectively. Given that the saturation pressure for water at 25°C, is 3.17 kPa, the relative humidity of the air in the room isa)67%b)55%c)83%d)71%Correct answer is option 'D'. Can you explain this answer?
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A room contains 35 kg of dry air and 0.5 kg of water vapor. The total pressure and temperature of air in the room are 100 kPa and 25°C respectively. Given that the saturation pressure for water at 25°C, is 3.17 kPa, the relative humidity of the air in the room isa)67%b)55%c)83%d)71%Correct answer is option 'D'. 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 A room contains 35 kg of dry air and 0.5 kg of water vapor. The total pressure and temperature of air in the room are 100 kPa and 25°C respectively. Given that the saturation pressure for water at 25°C, is 3.17 kPa, the relative humidity of the air in the room isa)67%b)55%c)83%d)71%Correct answer is option 'D'. 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 A room contains 35 kg of dry air and 0.5 kg of water vapor. The total pressure and temperature of air in the room are 100 kPa and 25°C respectively. Given that the saturation pressure for water at 25°C, is 3.17 kPa, the relative humidity of the air in the room isa)67%b)55%c)83%d)71%Correct answer is option 'D'. Can you explain this answer?.
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