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vivado [14]
2 years ago
5

Can pockets of vacuum persist in an ideal gas? Assume that a room is filled with air at 20∘C and that somehow a small spherical

region of radius 1 cm within the room becomes devoid of air molecules. Estimate how long it will take for air to refill this region of vacuum. Assume the atomic mass of air is 29 u.
Physics
1 answer:
Ilya [14]2 years ago
3 0

Answer:

time taken is 20 μs

Explanation:

given data

temperature = 20°C  = 293 K

radius = 1 cm

atomic mass of air = 29 u

to find out

how long it will take for air to refill

solution

we find here rms velocity of air particle that is

\frac{1}{2}mv^2 = \frac{3}{2}RT

here m is mass and t is temperature and v is speed and R is ideal gas constant i.e. 8.3145 (kg·m²/s²)/K·mol

v = \sqrt{\frac{3RT}{M} }  ............................1

v = \sqrt{\frac{3(8.314)293}{29*10{-3}kg} }

v = 501.99 m/s

so now for cover 1 cm

time taken by air

time take = \frac{r}{v}

time taken = \frac{1*10^{-2}m}{501.99}

time taken = 19.92 × 10^{6} s = 20μs

so time taken is 20 μs

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2 years ago
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When a surface is submerged in a fluid, the resultant pressure force on the body acts in what manner? Assume no shear forces are
Alex17521 [72]

Answer:

Perpendicular to the surface of the body.

Explanation:

When a surface is submerged in a fluid the resultant pressure force on the body acts perpendicular to the surface of the body. This is because fluids cannot withstand nor exert sideways forces. One could obtain this analytically considering that the origin of this force comes from the movement of the fluid molecules.

5 0
2 years ago
Calculate the pressure, in atmospheres, exerted by each of the following:
gregori [183]

Answer:

a) 14.2 atm

b) 4.46 atm

c) 1.06 atm

Explanation:

For an ideal gas,

PV = nRT

P = pressure of the gas

V = volume occupied by the gas

n = number of moles of the gas

R = molar gas constant = 0.08206 L.atm/mol.K

T = temperature of the gas in Kelvin

a) For HF,

P =?, V = 2.5L, n = 1.35 moles, T = 320K

P = 1.35 × 0.08206 × 320/2.5

P = 14.2 atm

b) For NO₂

P =?, V = 4.75L, n = 0.86 moles, T = 300K

P = 0.86 × 0.08206 × 300/4.75

P = 4.46 atm

c) For CO₂

P =?, V = 5.5 × 10⁴ mL = 55L, n = 2.15 moles, T = 57°C = 330K

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2 years ago
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3 0
2 years ago
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23. While sliding a couch across a floor, Andrea and Jennifer exert forces F → A and F → J on the couch. Andrea’s force is due n
PSYCHO15rus [73]

Answer:

a)  (95.4 i^ + 282.6 j^) N , b) 298.27 N  71.3º and c)   F' = 298.27 N   θ = 251.4º

Explanation:

a) Let's use trigonometry to break down Jennifer's strength

      sin θ = Fjy / Fj

      cos θ = Fjx / Fj

Analyze the angle is 32º east of the north measuring from the positive side of the x-axis would be

          T = 90 -32 = 58º

         Fjy = Fj sin 58

         Fjx = FJ cos 58

         Fjx = 180 cos 58 = 95.4 N

         Fjy = 180 sin 58 = 152.6 N

Andrea's force is

         Fa = 130.0 j ^

We perform the summary of force on each axis

X axis

       Fx = Fjx

       Fx = 95.4 N

Axis y

       Fy = Fjy + Fa

       Fy = 152.6 + 130

       Fy = 282.6 N

       F = (95.4 i ^ + 282.6 j ^) N

b) Let's use the Pythagorean theorem and trigonometry

       F² = Fx² + Fy²

       F = √ (95.4² + 282.6²)

       F = √ (88963)

       F = 298.27 N

       tan θ = Fy / Fx

       θ = tan-1 (282.6 / 95.4)

       θ = tan-1 (2,962)

       θ = 71.3º

c) To avoid the movement they must apply a force of equal magnitude, but opposite direction

       F' = 298.27 N

       θ' = 180 + 71.3

       θ = 251.4º

4 0
2 years ago
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