To suck water up a straw to a maximum height of 68 mm, what minimum gauge pressure must be produced in the lungs? Note that in order to be able to suck up water, the pressure in the lungs must be lower than the pressure outside by the exact amount of pressure created by the weight of the column of water in the straw.

Question

To suck water up a straw to a maximum height of 68 mm, what minimum gauge pressure must be produced in the lungs? Note that in order to be able to suck up water, the pressure in the lungs must be lower than the pressure outside by the exact amount of pressure created by the weight of the column of water in the straw.

Answer 1

It is negative because gauge pressures are measured relative to atmospheric pressure. It has to be negative to suck water up the straw.

See
http://www.jiskha.com/display.cgi?id=1259197828
for a similar response.

Answer 2

68 mm = 0.068 m

P= pgd (where p=density)
P= (1000 kg/m^3)g(0.068 m)
P= 666 Pa

and it's negative, I don't know why.

Is this for UTSC? I have the same question, exact wording :D

Answer 3

To calculate the minimum gauge pressure needed in the lungs, we can use the concept of pressure difference between the lungs and the atmosphere.

First, let's determine the pressure created by the weight of the water column inside the straw.

The pressure created by a column of liquid is given by the equation:

Pressure = density * g * height

Where:
- density is the density of water (approximately 1000 kg/m^3)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- height is the height of the water column (68 mm or 0.068 m)

Plugging in the values:

Pressure = 1000 kg/m^3 * 9.8 m/s^2 * 0.068 m
Pressure = 67.816 Pascal (Pa)

This means the pressure difference between the lungs and the atmosphere should be at least 67.816 Pa in order to suck water up the straw to a maximum height of 68 mm.

Please note that this calculation assumes ideal conditions and does not consider factors such as air resistance or fluid velocity.

Answer 4

To determine the minimum gauge pressure required to suck water up a straw to a maximum height of 68 mm, we need to consider the pressure difference between the lungs and the outside environment.

Let's go through the steps to find the answer:

Step 1: Determine the pressure difference due to the height of the water column.
The pressure difference is caused by the weight of the water column in the straw. The pressure at a given depth in a fluid is given by the equation:

P = ρgh

Where:
P is the pressure,
ρ is the density of the fluid (water),
g is the acceleration due to gravity,
h is the height of the water column.

Step 2: Calculate the pressure difference due to the height of the water column.
Since the pressure inside the lungs needs to be lower than the pressure outside by the exact amount of pressure created by the weight of the column of water, we'll set the height to 68 mm:

P_water = ρ_water * g * h_water

Substituting the known values:
ρ_water (density of water) ≈ 1000 kg/m³
g (acceleration due to gravity) ≈ 9.8 m/s²
h_water (height of water column) = 0.068 m

P_water = 1000 kg/m³ * 9.8 m/s² * 0.068 m

Step 3: Convert pressure to gauge pressure.
The pressure we obtained from step 2 is the absolute pressure since it includes atmospheric pressure. To calculate the gauge pressure, we need to subtract atmospheric pressure.

Standard atmospheric pressure is approximately 101325 Pa.

P_gauge = P_water - atmospheric pressure

Step 4: Convert the gauge pressure to mmHg.
Since the maximum height is given in mm, we can convert the gauge pressure to mmHg using the following conversion factor:

1 mmHg = 133.322 Pa

P_gauge_mmHg = P_gauge * (1 mmHg / 133.322 Pa)

Now, let's calculate the minimum gauge pressure required:

P_water = 1000 kg/m³ * 9.8 m/s² * 0.068 m
P_gauge = P_water - 101325 Pa
P_gauge_mmHg = P_gauge * (1 mmHg / 133.322 Pa)

By performing the calculations, you'll obtain the minimum gauge pressure required in mmHg to suck water up the straw to a maximum height of 68 mm.