566_Physics ProblemsTechnical Physics

566_Physics ProblemsTechnical Physics - 568 Temperature(b...

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568 Temperature (b) Solving the ideal gas law equation for V and using T =+ = 273 15 15 00 288 15 . . . K , we find V nRT P == × 0 5 2 3 453 8 314 288 15 1013 10 8166 10 .. . . . mol J mol K K Pa m 3 af bg . Then, ρ × × = m V 100 10 122 3 2 kg 8.166 10 m kg m 3 3 (c) The 100 g sample must have an appropriate molar mass to yield n 0 moles of gas: that is M air g 3.453 mol gmo l 100 29 0 *P19.67 Consider a spherical steel shell of inner radius r and much smaller thickness t , containing helium at pressure P . When it contains so much helium that it is on the point of bursting into two hemispheres, we have Pr r t ππ 28 51 0 2 Nm 2 ej . The mass of the steel is ρρ π ss s Pa Vr tr 44 10 22 9 Pr . For the helium in the tank, PV nRT = becomes Prn R T m M RT V 4 3 1 3 = He He balloon atm . The buoyant force on the balloon is the weight of the air it displaces, which is described by 1 4 3 3 atm balloon air air V m M RT P r . The net upward force on the balloon with the steel tank hanging
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