Solving Gas Pressure Law: 4.5*10^5Pa to 1*10^5Pa

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The discussion revolves around calculating the number of strokes needed to reduce the gas pressure from 4.5*10^5 Pa to 1*10^5 Pa in a rigid vessel connected to a vacuum pump. The initial formula used, P_1*V_1 = P_2*(0.1 + n(0.005)), is only accurate for a single stroke, leading to discrepancies in the calculated number of strokes. The user calculated 70 strokes, while the book states the answer is 31, prompting confusion about the correctness of the book's answer. The key point is that the equation must account for the cumulative effect of multiple strokes on pressure reduction. Understanding the fraction by which pressure decreases with each stroke is crucial for accurate calculations.
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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?
 
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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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