What is the fractional decrease in pressure when a barometer is raised 35m?

  • Thread starter ZoeGab
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In summary, when a barometer is raised 35 meters to the top of a building, the pressure decreases by 0.4%.
  • #1
ZoeGab
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Help! What is the fractional decrease in pressure when a barometer is raised 35m to the top of a building? (assume that the density of air is constant over that distance)
 
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  • #2
You must post some attempt before anyone at PF can help you (forum rules).

What I can do though is ask you, how would find the change in pressure at the height of 35m?
 
  • #3
I have tried to work this problem in different ways and I have gotten different answers. This was my last attempt but it still does not seem right.

35 m of air at approx 29 gm/mole at 24.1 liter per mole at room temperature weighs 4.2 gram per cm squared.

The air pressure at sea level is 1035 grams per cm squared.

Raising the barometer 35 meters lowers the pressure 4.2/1035 or 0.4%.
 
  • #4
ZoeGab said:
I have tried to work this problem in different ways and I have gotten different answers. This was my last attempt but it still does not seem right.

35 m of air at approx 29 gm/mole at 24.1 liter per mole at room temperature weighs 4.2 gram per cm squared.

The air pressure at sea level is 1035 grams per cm squared.

Raising the barometer 35 meters lowers the pressure 4.2/1035 or 0.4%.

0.4% looks fine to me using your method.
 
  • #5
here is another way with a different answer
change in pressure , ΔP = ρgh

where,
ρ = density of medium = 1.184 kg/m3 (for air at 250 C)
g = acc due to gravity = 9.8 m/s2
h = height change = 35 m

=> ΔP = 1.184*9.8*35 = 406.112 Pa

1 atm = 101325 Pa
=> 1 Pa = 1/101325 atm
=> 406.112 Pa = 406.112/101325 = 0.004 atm (approx)
 
  • #6
zoegab said:
here is another way with a different answer
change in pressure , Δp = ρgh

where,
ρ = density of medium = 1.184 kg/m3 (for air at 250 c)
g = acc due to gravity = 9.8 m/s2
h = height change = 35 m

=> Δp = 1.184*9.8*35 = 406.112 pa

1 atm = 101325 pa
=> 1 pa = 1/101325 atm
=> 406.112 pa = 406.112/101325 = 0.004 atm (approx)

Δp/p = 0.004 = 0.4%
 
  • #7
take a look at this; where did I go wrong this time

ρgh = 1.15kg/m3*9.8*35 = 389.4*10-5 Pa
 
  • #8
ZoeGab said:
take a look at this; where did I go wrong this time

ρgh = 1.15kg/m3*9.8*35 = 389.4*10-5 Pa

How did you get the 10-5? I got 394.8525 Pa
 
  • #9
I put it in wrong in my calculator! I just realized this. Thanks for your help!
 

1. What is pressure in physics?

Pressure in physics is defined as the amount of force applied per unit area. It is typically measured in units of pascals (Pa) or newtons per square meter (N/m^2).

2. What is Pascal's principle?

Pascal's principle, also known as the principle of transmission of fluid-pressure, states that a change in pressure at any point in a confined fluid is transmitted equally to all parts of the fluid. This principle is the basis for many hydraulic systems.

3. How does pressure affect objects?

Pressure can affect objects by exerting a force on them. This force can cause objects to change shape, move, or even break. Objects with larger surface areas are more affected by pressure than those with smaller surface areas.

4. What are some real-life applications of Pascal's principle?

Pascal's principle has many real-life applications, such as hydraulic car brakes, hydraulic lifts, and hydraulic jacks. It is also used in devices such as syringes, hydraulic presses, and hydraulic turbines.

5. How does pressure change with depth?

Pressure increases with depth in a fluid due to the weight of the fluid above pushing down. This is known as hydrostatic pressure. The equation for hydrostatic pressure is P = ρgh, where P is pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth.

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