A1-m3 tank containing air at 10°c and 350 kpa is connected through a valve to another tank containing

A1-m3 tank containing air at 10°c and 350 kpa is connected through a valve to another tank containing 3 kg of air at 35°c and 150 kpa. now the valve is opened, and the entire system is allowed to reach thermal equilibrium with the surroundings, which are at 19.5°c. determine the volume of the second tank and the final equilibrium pressure of air. the gas constant of air is r = 0.287 kpa·m3/kg·k.

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  1. V₂=1.76 m³

    P=222.03 KPa

    Explanation:

    Given that

    For tank 1

    V₁=1 m³

    T₁= 10°C = 283 K

    P₁=350 KPa

    For tank 2

    m₂=3 kg

    T₂=35°C = 308 K

    P₂=150 KPa

    We know that for air

    P V = m R T

    P=pressure ,V= Volume,R= gas constant ,T= temperature ,m =mass

    for tank 2

    P₂ V₂ = m₂ R T₂

    By putting the values

    150 x V₂ = 3 x 0.287 x 308

    V₂=1.76 m³

    Final mass = m₁+m₂

    m =m₁+m₂

    The final volume V= V₂+V₁

    V= 1.76 + 1 m³

    V= 2.76 m³

    The final temperature T= 19.5°C

    T= 292.5 K

    [tex]m=\dfrac{PV}{RT}[/tex]

    [tex]m_1=\dfrac{P_1V_1}{RT_1}[/tex]

    [tex]m_1=\dfrac{350\times 1}{0.287\times 283}[/tex]

    [tex]m_1=4.3\ kg[/tex]

    m =m₁+m₂

    m =4.3 + 3 = 7.3 kg

    Now at final state

    P V = m R T

    P x 2.76 = 7.3 x 0.287 x 292.5

    P=222.03 KPa

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