A gymnasium is 16 m high. By what percent is the air pressure at the floor greater than the air pressure at the ceiling? Assume that pressure at the bottom is 1 atm and the density of air stays constant. The density of air is 1.29 kg/m3
Δp=ρg(h2−h1)=1.29⋅9.81⋅16≈202 (Pa)\Delta p=\rho g(h_2-h_1)=1.29\cdot9.81\cdot16\approx202\ (Pa)Δp=ρg(h2−h1)=1.29⋅9.81⋅16≈202 (Pa)
h1=101 325 (Pa)h_1=101\ 325\ (Pa)h1=101 325 (Pa)
h2=101 123 (Pa)h_2=101\ 123\ (Pa)h2=101 123 (Pa)
e=100−101123⋅100/101325=0.2 (%)e=100-101123\cdot100/101325=0.2\ (\%)e=100−101123⋅100/101325=0.2 (%) . Answer
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