text: another ideal gas law problem, please help :(? this seems so simple but i don't see how height factors into any of this.   a vertical cylinder of cross-sectional area 0.047 m2 is fitted with a tight-fitting, frictionless piston of mass 5.0 kg (fig. p10.50). if there are 3.0 mol of an ideal gas in the cylinder at 400 k, determine the height, h, at which the piston will be in equilibrium under its own weight.   here is a link to the figure:  http://i136.photobucket.com/albums/q200/physics2006/physics.gif here, as the gas is in equilibrium,the pressure exerted by the piston should equal the pressure exerted by the gas..                    ppiston=mass/area =5g/0.047pa. use the ideal gas law now..  v=nrt/p and so,area*ht(=v)=nrt/p this implies ht.h=nrt/ap and use your units carefully..i think the answer will be around 203.67502 km