IGCSE Physics Question About Fluid Pressure

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Hi, very quick question... It does like this: Helium in a cylinder (Which will be used to inflate a number of balloons) has a volume of (6*10-3) cubic meters and a pressure of (2.75*106)Pa. The pressure in each of the helium balloons is (1.1*105)Pa, and the volume of each balloon (3*10^-3) cubic meters.The temperature of helium does not change. How many balloons were inflated?

-How i solved it: Pressure * Volume = Constant so.... (6.0*10^-3)*(2.75*10^6)=V2*1.1*10^5... V2= 0.15m3... Now what i did here was divide the total volume of helium at 1.1*105 by the volume of helium in each balloon and so.....0.15/3*10-3.... My answer was 50 balloons.

-Mark scheme has the following answer: (0.15-(6*10-3))/3*10-3... Can someone tell me where i went wrong?
Thanks in advance.
 

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Hi, very quick question... It does like this: Helium in a cylinder (Which will be used to inflate a number of balloons) has a volume of (6*10-3) cubic meters and a pressure of (2.75*106)Pa. The pressure in each of the helium balloons is (1.1*105)Pa, and the volume of each balloon (3*10^-3) cubic meters.The temperature of helium does not change. How many balloons were inflated?

-How i solved it: Pressure * Volume = Constant so.... (6.0*10^-3)*(2.75*10^6)=V2*1.1*10^5... V2= 0.15m3... Now what i did here was divide the total volume of helium at 1.1*105 by the volume of helium in each balloon and so.....0.15/3*10-3.... My answer was 50 balloons.

-Mark scheme has the following answer: (0.15-(6*10-3))/3*10-3... Can someone tell me where i went wrong?
Thanks in advance.
You forgot to subtract the equivalent volume remaining in the tank when the pressures equilibrate.
 

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