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Gases Finding mass escape

  1. Jul 12, 2014 #1
    1. The problem statement, all variables and given/known data

    A cylinder containing 19 kg of compressed air at a pressure 9.5 times that of the atmosphere is kept in a store at 7 degrees Celsius. When it is moved to a workshop where the temperature is 27 degrees Celsius a safety valve on the cylinder operates, releasing some of the air. If the valve allows air to escape when its pressure exceeds 10 times that of the atmosphere, Calculate the mass of air that escapes..

    2. Relevant equations

    PV= nRT

    where P - Pressure
    V- Volume
    n- number of moles
    R- universal gas constant (8.314 J / mol. K )
    T - temperature on Kelvin

    Moles = mass / Mr

    pressure1 9.5 x 760 mmHg = 7220 mmHg

    Temperature1 7°C = 280 K

    pressure2 10 x 760 mmHg = 7660 mmHg

    Temperature 27°C = 300 K

    3. The attempt at a solution

    I assumed that the volume was constant and at STP 22.4 x 10-3 m3

    rearranging the equation PV = nRT, placing n as the subject

    With Pressure 1 and Temperature 1

    n = 7220(22.4 x 10-3) / 8.314(280)

    n = 0.0694 moles

    With Pressure 2 and Temperature 2

    n = 7660(22.4 x 10-3) / 8.314(300)

    n = 0.0687 moles

    Rearranging the equation for moles to find Mr, using the initial mass and moles from pressure 1 and temperature 1

    Mr = 19 kg / 0.0694 moles

    Mr = 274

    using the Mr value calculated to find mass from pressure 2 and temperature 2

    274 x 0.0687 = 18.82 kg

    subtracting the above value from the initial mass 19

    19 - 18.82= 0.18 kg


    Im not sure if i worked this out right...or if it makes any sense at all but it was as close i could of gotten in attempting this problem... please any assistance of guidance will be grateful....
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Jul 12, 2014
  2. jcsd
  3. Jul 12, 2014 #2

    Simon Bridge

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    Not bad - modelling the air as an ideal gas ... the ideal gas equation takes form:
    I think you did a lot more work than you needed to. It helps t do all the algebra before you plug the numbers in.

    The trick with this sort of problem is to realize that you can divide the ideal gas equations for each state.

    Step 1. lynchpin: write out PV=nRT in terms of the total mass of gas present M: M=nm: m=molar mass of the gas.

    Thus: ##PV=(R/m)MT##

    This is useful because you have the amount of gas as a total mass, and writing it out this way avoids having to convert moles a couple of times. Since R/m is a constant, it will cancel out in the next step so we don't even need to know the values of R and m.

    step 2. if state 2 is the final state, state 1 is the initial state: divide the ideal gas equations out like this: $$\frac{P_2V_2=(R/m)M_2T_2}{P_1V_1=(R/m)M_1T_1}$$... you should be able to simplify the equation, remember that ##V_2=V_1## and ##[P]_{atmos}\propto [P]_{mmHg}## ... and you are not actually looking for ##M_2## yu are looking for the mass lost.
     
    Last edited: Jul 12, 2014
  4. Jul 12, 2014 #3
    :redface: ooo my... thats embarrassing...i always end up doing extra work...or over thinking it....thanks soo much, followed your guidance and got M2 to be 18.67kg... by subtraction from 19 .. Mass escape worked out to be 0.33kg
     
  5. Jul 12, 2014 #4

    Simon Bridge

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    Many students feel more comfortable with numbers but the algebra is your friend.
    It's too easy to make rounding off errors or lose track if you don't do the algebra first.
    That's the main lesson here.
     
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