Question

Calculate the pressure in cm H2O, exerted in the closed end of a pressure volume apparatus if the atmospheric pressure is 720.0 torr and the heigh of the water column is 22.3 cm

Answer 1

Given that

Pressure p₀ = 720.0 torr = 720.0 mm-Hg
                 = ( 720.0 mm-Hg ) [ 1.013 × 10 ⁵ Pa / ( 760 mm-Hg ) ]
                   = 9.596842 × 10⁴ P

Height H = 22.3 cm = 22.3 × 10 ⁻ ²

Weight of water column W = ρ V g = ρ ( A • H ) g
= ( 1000 kg / m ³ ) A (22.3 × 10 ⁻ ² m ) ( 9.8 m / s ² )
= ( 2.1854 × 10 ³ Pa ) A
Since p₀ A = W + p A

Force F supporting the weight of water column must be

F = ( p₀ – p ) A = ( 2.1854 × 10 ³ Pa ) A

     p₀ – p = 2.1854 × 10 ³ Pa

p = p₀ – ( 2.1854 × 10 ³ Pa )
= ( 9.596842 × 10 ⁴ Pa ) – ( 2.1854 × 10 ³ Pa )
= 9.378302 × 10 ⁴ Pa
= ( 9.378302 × 10 ⁴ Pa ) [ 760 mm-Hg / ( 1.013 × 10 ⁵ Pa ) ]
= 704 mm-Hg
= 704 torr

P = 704 mm-Hg = 704 torr = 70.4 cm Hg