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An open end mercury manometer was constructed from a u-shaped tube. in a particular measurement, the level in the end connected to the gas manifold, on which the experiment was being conducted, measured 76.2 cm above the u-neck, while the level in the open end (to the atmosphere) was 23.8 cm above the u-neck. the outside air pressure in the laboratory was measured as 754 torr. what is the pressure in the gas manifold?

Respuesta :

barnuts

We can solve this problem using the formula:

P2 – P1 = ρ g (h2 – h1)

where,

P2 = pressure of the gas manifold = ?

P1 = pressure of the outside air = 754 torr = 100,525 Pa

ρ = density of mercury = 13560 kg/m^3

h2 = height in gas manifold = 76.2 cm = 0.762 m

h1 = height in air = 23.8 cm = 0.238 m

 

So w can directly find for P2:

P2 = 13560 * 9.81 * (0.762 – 0.238) + 100,525

P2 = 170,229.37 Pa

 

Convert back to torr:

P2 = 170,229.37 Pa = 1276.83 torr              (ANSWER)

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