Answer:
a) 0.759 m , 0.759 m
b) 10.1 m , 10.3 m
c) 12.3 m , 13 m
Explanation:
Vapour Pressure included:
[tex]P_{atm,vp} = y*h + P_{vp} \\\\h = \frac{P_{atm,vp}-P_{vp}}{y} ... Eq1[/tex]
mercury
[tex]h_{Hg,vp} = \frac{101*10^3-1.6*10^(-1)}{133*10^3} \\\\h_{Hg,vp} = 0.759m[/tex]
[tex]h_{Hg} = \frac{101*10^3}{133*10^3} \\\\h_{Hg} = 0.759m[/tex]
water
[tex]h_{H2O,vp} = \frac{101*10^3-1.77*10^3}{9.8*10^3} \\\\h_{H2O,vp} = 10.1m[/tex]
[tex]h_{H2O} = \frac{101*10^3}{9.8*10^3} \\\\h_{H20} = 10.3m[/tex]
Ethyl Alcohol
[tex]h_{EA,vp} = \frac{101*10^3-5.9*10^3}{7.74*10^3} \\\\h_{EA,vp} = 12.3m[/tex]
[tex]h_{EA} = \frac{101*10^3-5}{7.74*10^3} \\\\h_{EA} = 13m[/tex]
For mercury barometers the effects of vapour pressure are negligible and required height for mercury barometer is reasonable as compared to that of water and ethyl alcohol.