568 Temperature(b)Solving the ideal gas law equation for Vand using T=+=273 15 15 00288 15...K , we findVnRTP==⋅×=×−0523 4538 314288 151013 108166 10.....molJ mol KKPam3afbg.Then, ρ××=−−mV100 1012232kg8.166 10mkg m33(c)The 100 g sample must have an appropriate molar mass to yield n0moles of gas: that isMairg3.453 molgmol10029 0*P19.67Consider a spherical steel shell of inner radius rand much smaller thickness t, containing helium atpressure P. When it contains so much helium that it is on the point of bursting into twohemispheres, we have Prrtππ285102Nm2ej. The mass of the steel isρρπsssPaVrtr4410229Pr. For the helium in the tank, PVnRT=becomesPrnRTmMRTV4313=HeHeballoonatm.The buoyant force on the balloon is the weight of the air it displaces, which is described by1433atmballoonairairVmMRTPr. The net upward force on the balloon with the steel tank hanging
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