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An air-water vapour mixture is contained in a rigid closed vessel with a volume of 35 m3 at 1.5 bar, 120°C and Φ = 10%. The specific volume of water vapour (in m3/kg) in air is______.
(At 120°C, Psat = 1.985 bar)
    Correct answer is '9.145'. Can you explain this answer?
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    Most Upvoted Answer
    An air-water vapour mixture is contained in a rigid closed vessel wit...
    Pv = Φ x Psat
    = 0.1 x 1.985
    = 0.1985 bar
    = 19.85 kPa
    V = R/Mv x T/Pv = (8.314/18) X (393/19.85 ) = 9.145 m3/kg
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    An air-water vapour mixture is contained in a rigid closed vessel wit...
    Given:
    - Volume of the vessel (V) = 35 m^3
    - Pressure (P) = 1.5 bar
    - Temperature (T) = 120°C
    - Relative humidity (Φ) = 10%
    - Saturation pressure at 120°C (Psat) = 1.985 bar

    To find:
    The specific volume of water vapor in air.

    Assumptions:
    - The air-water vapor mixture behaves as an ideal gas.
    - The air is a mixture of oxygen, nitrogen, and other gases, with water vapor being the only condensable component.

    Solution:

    Step 1: Calculate the specific volume of dry air
    1. Convert the pressure from bar to Pa (1 bar = 100,000 Pa)
    P = 1.5 bar = 1.5 * 100,000 Pa = 150,000 Pa

    2. Convert the temperature from °C to K (K = °C + 273.15)
    T = 120°C + 273.15 = 393.15 K

    3. Use the ideal gas equation to calculate the specific volume of dry air
    The ideal gas equation is given by: Pv = RT
    Where:
    - P is the pressure in Pa
    - v is the specific volume in m^3/kg
    - R is the specific gas constant for air (287.1 J/kg·K)
    - T is the temperature in K

    Rearranging the equation: v = RT/P
    v = (287.1 * 393.15) / 150,000
    v = 0.7514 m^3/kg

    Step 2: Calculate the specific volume of water vapor
    1. Calculate the partial pressure of water vapor (Pv) using the relative humidity (Φ) and the saturation pressure (Psat) at the given temperature
    Pv = Φ * Psat
    Pv = 0.10 * 1.985 bar = 0.1985 bar = 19,850 Pa

    2. Use the ideal gas equation to calculate the specific volume of water vapor
    v = RT/P
    v = (461.5 * 393.15) / 19,850
    v = 9.145 m^3/kg

    Answer:
    The specific volume of water vapor in air is 9.145 m^3/kg.
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    An air-water vapour mixture is contained in a rigid closed vessel with a volume of 35 m3 at 1.5 bar, 120°C and Φ = 10%. The specific volume of water vapour (in m3/kg) in air is______.(At 120°C, Psat = 1.985 bar)Correct answer is '9.145'. Can you explain this answer?
    Question Description
    An air-water vapour mixture is contained in a rigid closed vessel with a volume of 35 m3 at 1.5 bar, 120°C and Φ = 10%. The specific volume of water vapour (in m3/kg) in air is______.(At 120°C, Psat = 1.985 bar)Correct answer is '9.145'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about An air-water vapour mixture is contained in a rigid closed vessel with a volume of 35 m3 at 1.5 bar, 120°C and Φ = 10%. The specific volume of water vapour (in m3/kg) in air is______.(At 120°C, Psat = 1.985 bar)Correct answer is '9.145'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An air-water vapour mixture is contained in a rigid closed vessel with a volume of 35 m3 at 1.5 bar, 120°C and Φ = 10%. The specific volume of water vapour (in m3/kg) in air is______.(At 120°C, Psat = 1.985 bar)Correct answer is '9.145'. Can you explain this answer?.
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