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Gas Pressure Law

  1. Jan 25, 2009 #1
    the intiial pressure of a mass of dry gas in a rigid vessel is 4.5*10^5Pa. The volume of the vessel is 0.1m^3. It is connected to a vacuum pump, whih consists of a cylinder fitted with a piston. The pump cylinder has a volume of 0.005m^3. i am supposed to determine the number of strokes necessary to reduce the gas pressure to 1*10^5Pa

    i used the formula: P_1*V_1 = P_2*(0.1 + n(0.005)), where n is the number of strokes
    P_1 = 4.5*10^5, V_1 = 0.1, P_2 = 1*10^5.

    the answer i got is 70 strokes, which according to the book is incorrect. The answer on the book is 31. however, when i substiute 31 instead of n, P_2 is not equal to 1*10^5 Pascals, it is however when i substitute n=70. I'm not sure whether it's a mistake in the book, but i think it is. the first part of the question asks me to find the pressure after 2 strokes. when i substitute 2 in my equation, i get the pressure that he asked for so I'm pretty sure the equation is not wrong.

    can someone show me where i'm mistaken please?
     
  2. jcsd
  3. Jan 25, 2009 #2

    Doc Al

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    Staff: Mentor

    The book is correct. Your equation is mistaken--it's only correct for a single stroke. (As long as the number of strokes is small it's approximately correct, which explains you getting an OK answer for 2 strokes.)

    Answer this: With every stroke the pressure is reduced by what fraction?
     
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