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Pressure and volume

  1. Sep 10, 2011 #1
    1. The problem statement, all variables and given/known data
    I have attached the question. I have no clue with this one. Is there a specific equation that I have to use? How do I approach a problem like this??


    2. Relevant equations
    density = mass/volume
    specific volume = volume/mass = 1/density
    mass flow rate = density x volume flow rate
    volume flow rate = cross sectional area x velocity
    density of water = 1000kg/m^3
    density of air = 1.22521kg/m^3 i think?
    total volume = mass x specific volume


    3. The attempt at a solution
    -
     

    Attached Files:

  2. jcsd
  3. Sep 10, 2011 #2
    Write down the ideal gas equation PV=nRT for each container's initially while assuming unknowns.
     
  4. Sep 10, 2011 #3
    Then we know that total no. of moles in system(container 1 + container 2) is constant.
    Then write final gas equation PV=(n1+n2)RT.

    Sufficient data is given to you.Molar mass of air may be required
     
  5. Sep 10, 2011 #4
    okay so the ideal gas law eqs: 500 x 10^3 (1) = n x 287 x (25 +273.15)------container 1.
    200 x 10^3 (V) = n x 287 x (35 +273.15)------------container 2.

    how do i find Volume knowing that it weighs 5kg. Am i suppose to add the two above equations?? then make it equal to P (V1+V2)=(n1+n2) x 287 x (20 + 273.15) ???????
     
  6. Sep 10, 2011 #5
    You know the volume of second tank using ideal gas equation for container-2.You have pressure temperature and number of moles(you need molar mass of air).
     
  7. Sep 10, 2011 #6
    Using this you get V1+v2 in final equation.There you have total no. of moles, total volume and equilibrium temperature.Get equilibrium Pressure
     
  8. Sep 10, 2011 #7
    I'm not sure i understand what you mean. Isn't the molar mass used to find R in the ideal gas law equation??? I used R as 287 is that right??? i dont get how to find n
     
  9. Sep 10, 2011 #8
    molar mass of air is 28.97 and R is 8.31J/mol*K.
    You are using wrong values of R.
     
  10. Sep 10, 2011 #9
    In the ideal gas equation for first container you have temperature, pressure, and volume calculate no. of moles for container 1.
     
  11. Sep 10, 2011 #10
    For second container you have mass of air 5kg=5000g.
    no. of moles= 5000/28.97=172.59
     
  12. Sep 10, 2011 #11
    Here is how to solve the problem:

    Use finite states to have a picture of the problem easily.

    State 1:
    V_a1 = 1 m^3
    T_a1 = 25 deg C + 273 = 323 K
    P_a1 = 500 kPa
    m_b1 = 5 kg
    T_b1 = 35 deg C + 273 = 308 K
    P_b1 = 200 kPa

    State 2:
    T_a2 = T_b2 = 20 deg C + 273 = 293 K (thermal equilibrium w/ surroundings)
    V_b2 = ?
    P_b2 = ?

    where a is the first tank, and b is the second tank.

    Solution:
    1. Obtain unknowns of each tank for easy computations at the initial state or State 1.

    1.1 Obtain the mass of the first tank using PV=mRT
    m_a1=P_a1*V_a1 / (R*T_a1), where R is 286.9 J/(kgK)

    1.2 Obtain the volume of the second tank
    V_b1= m_b1 * R * T_b1 / P_b1

    2. Analyze the transition from State 1 to State 2, taking note of the parameters in PV=mRT equation.

    2.1 When the valve is opened, the volume of the 2 tanks will not obviously change. So the volume of the second tank V_b2 is equal to V_b1.

    2.2 Since the valve is already opened, the contents of the 2 tanks are combined, that is: m_a1 + m_b1 = m_a2 = m_b2. Now, we can solve for the final pressure using PV = mRT:
    P_b2 = m_b2 * R * T_b2 / V_b2
     
  13. Sep 10, 2011 #12
    How come R is 286.9????It's 8.31 as i said earlier.
    You will need molar mass of air.
     
  14. Sep 10, 2011 #13
    8.31 im pretty sure is the universal gas constant. to find the gas constant of air it is universal gas constant divided by molar mass. R of air = universal gas constant (8.31) / molar mass of air(28.956) = 0.2869kPa = 286.9Pa
     
  15. Sep 10, 2011 #14
    anyway my working out now is:
    container 1: 500 x 10^3 (1) = m x 287 x (25 + 273.15) thus m = 5.84kg.
    container 2: 200 x 10^3 (V) = 5 x 287 x (35 + 273.15) thus V = 2.21 m^3

    so mass total is 5.84 + 5 = 10.84 kg

    therefore P ( 2.21) = 10.84 x 287 x (20 + 273.15), which means P = 412616 Pa = 412.6kPa. Is that correct?
     
  16. Sep 10, 2011 #15
    Then it's okay.However using universal gas constant seems more fundamental.
     
  17. Sep 10, 2011 #16
    is my working out correct as well?
     
  18. Sep 10, 2011 #17

    seems OK.Check for any calculation errors and if your answer matches with correct answer(if known) then fine.
     
  19. Sep 11, 2011 #18
    to find the final equilibrium pressure, what volume do i have to use? the volume of the first/second or both added together?
     
  20. Sep 11, 2011 #19
    Both added together.
     
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