Recent content by Mafi0z

  1. M

    AP Question involving Ideal Gas Law - Thermal Physics - Pressure

    Thanks a lot for the help, LowlyPion! :D I've got the answer.. .00542 moles of air enters the can Your method did work.. P1V1=P2V2 V2 = ((82093)(0.005*0.15)) / (1x10^5) V2 = 6.157x10^-4 m^3 ΔV = V(can) - V2 ΔV = (0.005*0.15) - 6.157*10-4 ΔV = 1.343x10^-4 n=PV/RT = ((100000)(1.343x10^-4)) /...
  2. M

    AP Question involving Ideal Gas Law - Thermal Physics - Pressure

    Oh ok i got that part now(ΔV). thanks. so basically what that's asking is what change in volume will occur when the pressure increases from 82093 to 100,000, right? I forgot to subtract the overall volume, not sure why.. i did on my first attempt. Anyway, so now would I still use n = PV/RT ...
  3. M

    AP Question involving Ideal Gas Law - Thermal Physics - Pressure

    hmm not sure i understand. do you mean: P1V1=P2V2 with P1 = 82093 and P2 = 1atm? V2 = ((82093)(0.005*0.15)) / (1x10^5) V2 = 6.157x10^-4 m^3 then do the n = PV/RT with V = V2 = 6.157x10^-4 and P = 1atm and T =298 ? n= 0.0248 moles still doesn't seem right.. although i am sure i...
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    AP Question involving Ideal Gas Law - Thermal Physics - Pressure

    [SOLVED]AP Question involving Ideal Gas Law - Thermal Physics - Pressure I am having some trouble with an old ap problem(PART C ONLY), see below: Homework Statement 1996 AP PHYSICS B Free Response Question: The inside of the cylindrical can has a cross sectional area of 0.005 m^3 and...
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    Solve Hard Fluid Problem: Calculate Minimum Work & Power of Pump

    W=Fd W=(mg)h W=(rho)Vgh Note: (rho) = density - in this case of water W=1000*.35*9.8*85 W=291550 Joules P=W/t P=291550/7200 Note 2hrs = 7200s P=40.5 watts Next one you have done correctly.. A1v1=A2v2 V2 = 2.88 m/s Last one.. You must use bernoullis principle P+(.5(rho)v^2)+((rho)gh) P is...
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