posted by yuri .
a tank contains 26.0 kg of oxygen gas at a gauge pressure of 8.70 atm. if oxygen is replaced by helium, how many kilograms of the latter will be needed to produce a gauge pressure at 7.00 atm?
A gauge pressure of 8.70 atm means the absolute pressure is 9.70 atm. Similarly, a gauge pressur eof 7.00 atm means an absolute pressure of 8.00 atm.
Assuming the temperature remains the same, the gas mass ratio is:
M(He)/M(O2) = (4/32)(8.00/9.70) = 0.1031
M(He) = 26*0.1031 = 2.68 kg