Assume patmos=1.00atm. what is the gas pressure pgas? express your answer in pascals to three significant figures.
2 answers:
The gas pressure is about 9.40 × 10⁴ Pa
Let's recall Hydrostatic Pressure formula as follows:
<em>where:</em>
<em>P = hydrosatic pressure ( Pa )</em>
<em>ρ = density of fluid ( kg/m³ )</em>
<em>g = gravitational acceleration ( m/s² )</em>
<em>h = height of a column of liquid ( m )</em>
Let us now tackle the problem!
<em><u>Complete Question:</u></em>
<em>A U-shaped tube is connected to a box at one end and open to the air at the other end. The box is full of gas at pressure pgas, and the tube is filled with mercury of density 1.36 × 10⁴ kg/m³ . When the liquid in the tube reaches static equilibrium, the mercury column is h₁ = 10.5 cm high in the left arm and h₂ = 5.00 cm high in the right arm. Assume p_atmos = 1.00 atm , what is the gas pressure p_gas?</em>
<u>Given:</u>
density of mercury = ρ = 1.36 × 10⁴ kg/m³
atmospheric pressure = p_o = 1.00 atm = 76 cmHg
height difference of liquid column = Δh = 10.5 - 5.00 = 5.5 cm
<u>Asked:</u>
gas pressure = p_gas = ?
<u>Solution:</u>
<em>Because the height difference of mercury column is </em><em>5.5 cm</em><em> , then the pressure difference between the left arm and the right arm will be </em><em>5.5 cmHg.</em>
- Buoyant Force : brainly.com/question/13922022
- Kinetic Energy : brainly.com/question/692781
- Volume of Gas : brainly.com/question/12893622
- Impulse : brainly.com/question/12855855
- Gravity : brainly.com/question/1724648
Grade: High School
Subject: Physics
Chapter: Pressure
You might be interested in
Answer:
r₂ = 0.316 m
Explanation:
The sound level is expressed in decibels, therefore let's find the intensity for the new location
β = 10 log
let's write this expression for our case
β₁ = 10 log \frac{I_1}{I_o}
β₂ = 10 log \frac{I_2}{I_o}
β₂ -β₁ = 10 ( )
β₂ - β₁ = 10
log \frac{I_2}{I_1} = = 3
= 10³
I₂ = 10³ I₁
having the relationship between the intensities, we can use the definition of intensity which is the power per unit area
I = P / A
P = I A
the area is of a sphere
A = 4π r²
the power of the sound does not change, so we can write it for the two points
P = I₁ A₁ = I₂ A₂
I₁ r₁² = I₂ r₂²
we substitute the ratio of intensities
I₁ r₁² = (10³ I₁ ) r₂²
r₁² = 10³ r₂²
r₂ = r₁ / √10³
we calculate
r₂ =
r₂ = 0.316 m