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2. In the middle of a horizontally positioned cylinder of length 1 = 60 cm is a piston of mass m = 20 kg. The lower part of t
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Answer #1

Let mo be mass of Oz net force on poiston = 0 Poxa + mg - PH2XA 0 Poz T PV=nRT = mox 8.31x300 32 xha 0.6m Oz PH2 = .08x8.31x3

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