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Rasek [7]
3 years ago
8

If the volume of a sample of gas is reduced at constant temperature, the average velocity of the molecules _______, the average

force of an individual collision _________, and the average number of collisions with the wall, per unit area, per second_______.
Physics
1 answer:
VLD [36.1K]3 years ago
6 0

Answer:

Explanation:

If the volume of a sample of gas is reduced at constant temperature, the average velocity of the molecules increases, the average force of an individual collision increases, and the average number of collisions with the wall, per unit area, per second increases.

As volume is reduced, the gas molecules come closer together, which increases the number of collisions between them and their collisions with the container walls. Also, since the distance traveled by each molecule between successive collision decreases, the molecule velocity doesn't decrease much within collisions as a result of which, the average velocity is higher compared to when the gas is stored in a larger volume. Finally, due to constant collisions, the direction of molecule travel changes rapidly owing to which the acceleration of molecules increases.

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. Water is flowing at 12m/s in a horizontal pipe under a pressure of 600kpa radius 2cm. a. What is the speed of the water on the
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Explanation:

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Inlet Radius r₁ = 0.5 cm

Outlet Velocity, V₂ = not given (we are asked to find this)

Outlet Pressure,  P₂ = not given (we are asked to find this)

Outlet Radius, r₂ = 0.5 cm

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Inlet Area, A₁ = π (r₁)² = π(2)² = 4π cm²

Outlet Area, A₂ = π (r₂)² = π(0.5)² = 0.25π cm²

<u>Part A :</u>

Assuming that water is incompressible, we can reason that within the same given time, the amount of volume of water entering the inlet must equal the volume of water exiting the outlet. Hence by the continuity equation (i.e. conservation of mass)

Inlet Volume flow rate = Outlet Volume flow rate

(recall that Volume flow rate in a pipe is given by Velocity x Cross Section Area), Hence the equation becomes

V₁ x A₁ = V₂ x A₂  (substituting the values that we know from above)

12 x 4π = V₂ x 0.25π  (we don't have to change all to SI units because the conversion factors on the left will cancel out the conversion factors on the right).

V₂ = (12 x 4π) / (0.25π)

V₂ = 192 m/s  (Answer)

<u>Part B:</u>

For Part B, if we assume a closed ideal system (control volume method), we can simply apply the energy equation (i.e Bernoulli's equation)

P₁ + (1/2)ρV₁ + ρgh₁ = P₂ + (1/2)ρV₂ + ρgh₂

Because the pipe is horizontal, there is no difference between h₁ and h₂, hence we can neglet this term:

P₁ + (1/2)ρV₁ = P₂ + (1/2)ρV₂  (rearranging)

P₂ = P₁ + (1/2)ρV₁ - (1/2)ρV₂

= P₁  + (1/2)ρ (V₁-V₂)

Assuming that the density of water is approx, ρ = 1000 kg/m³

P₂ = 600,000  + (1/2)(1000) (12-192)

= 600,000  + ( -90,000)

= 510,000 Pa

= 510 kPa (Answer)

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