Assume patmos=1.00atm. what is the gas pressure pgas? express your answer in pascals to three significant figures.

Physics

2 answers:

hodyreva [135]3 years ago

8 0

Answer: Well, let's start by finding the pressure due to the "extra" height of the mercury. p = 1.36e4 kg/mÂł Â· (0.105m - 0.05m) Â· 9.8m/sÂ˛ = 7330 N/mÂ˛ = 7330 Pa The pressure at B is clearly p_b = p_atmos = p_gas + 7330 Pa The pressure at A is p_a = p_gas = p_atmos - 7330 Pa c) 1 atm = 101 325 Pa Then p_gas = 101325 Pa - 7330 Pa = 93 995 Pa

ANTONII [103]3 years ago

3 0

The gas pressure is about 9.40 × 10⁴ Pa

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<h3>Further explanation</h3>

Let's recall Hydrostatic Pressure formula as follows:

$\boxed{ P = \rho g h}$

where:

P = hydrosatic pressure ( Pa )

ρ = density of fluid ( kg/m³ )

g = gravitational acceleration ( m/s² )

h = height of a column of liquid ( m )

Let us now tackle the problem!

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Complete Question:

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?

Given:

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

Asked:

gas pressure = p_gas = ?

Solution:

Because the height difference of mercury column is 5.5 cm , then the pressure difference between the left arm and the right arm will be 5.5 cmHg.

$\Delta p = p_o - p_{gas}$