In a period of 6.00 s, 6.00 × 10 23 oxygen molecules strike a wall with an area of 2.00 cm 2 . If the molecules move with a speed of 400.0 m/s and strike the wall head-on in elastic collisions, what is the pressure exerted on the wall? (The mass of one O2 molecule is 5.344 × 10 -26 kg.)
A. 28.1 kPa
B. 32.3 kPa
C. 21.4 kPa
D. 34.3 kPa
E. 47.6 kPa

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wagonbelleville wagonbelleville · Answer 1 · 2019-12-12T05:25:33+01:00

Answer:

option C

Explanation:

given,

Period of oxygen molecule = 6 s

number of molecule = 6 x 10²³

Area of wall = 2 cm²

speed of molecule = 400 m/s

mass of O₂ molecule = 5.344 × 10⁻²⁶ Kg

we know,

[tex]F = \dfrac{P}{A}[/tex]

Δp= F Δt

Δp change in momentum

Δp = N.m.Δv

where N is the number of oxygen molecules strike a wall

Δv = v - (-v)= 2 v

Pressure,

[tex]P = \dfrac{2mv.N}{\Delta t \times A}[/tex]

[tex]P = \dfrac{2\times 5.344 \times 10^{-26}\times 400\times 6 \times 10^{23}}{6 \times 2 \times 10^{-4}}[/tex]

P = 21376 Pa

P = 21.4 kpa

hence the correct answer is option C