if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

Explanation:

The air in the tube can be considered an ideal gas,

P V = nR T

In that case we have the tube in the air where the pressure is P1 = P_atm, then we introduce the tube to the water to a depth H

For pressure the open end of the tube is

P₂ = P_atm + ρ g H

Let's write the gas equation for the colon

P₁ V₁ = P₂ V₂

P_atm V₁ = (P_atm + ρ g H) V₂

V₂ = V₁ P_atm / (P_atm + ρ g h)

If the air obeys Boyle's law e; volume within the had must decrease due to the increase in pressure, if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

The main assumption is that the temperature during the experiment does not change

6 0

2 years ago

Water in a tank is pressurized by air and pressure measured using a multi-fluid manometer. Determine the gage pressure of air in

sattari [20]

Answer:

The gauge pressure of air is 110 kpa

Explanation:

Atmospheric pressure, $P_{atm}$ = 101 Kpa

$P_{gauge} + \rho_w gh_1 + \rho_o gh_2 -\rho_{Hg} gh_3 =P_{atm}$

$P_{gauge} = P_{atm} - \rho_w gh_1 - \rho_o gh_2 +\rho_{Hg} gh_3$

where;

ρw is the density of water = 1000 kg/m³

ρo is the density of oil = 800 kg/m³

ρHg is the density of mercury = 13,600 kg/m³

g is acceleration due to gravity = 9.8 m/s²

$P_{gauge} = 101,000 - (1000* 9.8*0.2) - (800* 9.8*0.3) +(13,600* 9.8*0.46)\\\\P_{gauge} = 101,000 - 1960 - 2352 + 13610.26\\\\P_{gauge} = 110,298.26 pa$

Therefore, the gauge pressure of air is 110 kpa

4 0

3 years ago

What is the difference between speed and volocity?

Sliva [168]

The average speed is the distance per time ratio. velocity is the rate at which the position changes

3 0

2 years ago