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AQA Unit 5 Nuclear/Thermal Physics Specimen Paper q2a

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    The question's about a hand pump and bottle
    The volume of the bottle is 3.5x10^(-4)
    During the upstroke of the pump, the pump chamber volume increases from zero to 6.5x10^(-4)
    Initial pressure is 99kPa
    (constant temperature)

    Calculate the pressure of the bottle after one upstroke


    I did:
    pV = 99 x 10^3 x 3.5 x 10^(-4) = 34.65

    V(after upstroke) = 34.65/ (10x10^(-4)) = 34650Pa

    actual answer is 83.5kPa, can anyone see how to get that?
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    (Original post by hchawn)
    The question's about a hand pump and bottle
    The volume of the bottle is 3.5x10^(-4)
    During the upstroke of the pump, the pump chamber volume increases from zero to 6.5x10^(-4)
    Initial pressure is 99kPa
    (constant temperature)

    Calculate the pressure of the bottle after one upstroke


    I did:
    pV = 99 x 10^3 x 3.5 x 10^(-4) = 34.65

    V(after upstroke) = 34.65/ (10x10^(-4)) = 34650Pa

    actual answer is 83.5kPa, can anyone see how to get that?
    Just doing this paper now. I think the key is definitely 'On an up-stroke of the pump, air enters the pump chamber from the bottle.' So if V1 = 3.5x10^-4 then V2 (an up-stroke) is going to be that original volume plus the amount of air it's let in. So V1xP1= 3.5x10^-4 x 99000
    V2xP2= (3.5x10^-4 +6.5x10^-5) x ?
    then just solve for P2 and you get 83934

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Updated: June 16, 2012
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