The mass of hydrogen molecule is 3.32 ×10-27 kg. If 1023 hydrogen molecules strike per second at 2 cm2 area of a rigid wall at an angle of 45° from the normal and rebound back with a speed of 1000 ms-1, then the pressure exerted on the wall is

F = rate of change of momentum =ΔpΔt=(nm)(2vcosθ)Pressure, p=FA=(nm)(2vcosθ)A =1023×3.32×10-27×2×1000×cos45°2×10-4 =2.34 × 103 Nm-2 or Pa

The mass of hydrogen molecule is 3.32 ×10-27 kg. If 1023 hydrogen molecules strike per second at 2 cm2 area of a rigid wall at an angle of 45° from the normal and rebound back with a speed of 1000 ms-1, then the pressure exerted on the wall is

F = rate of change of momentum =ΔpΔt=(nm)(2vcosθ)Pressure, p=FA=(nm)(2vcosθ)A =1023×3.32×10-27×2×1000×cos45°2×10-4 =2.34 × 103 Nm-2 or Pa

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The mass of hydrogen molecule is $3.32 \times 10^{- 27} kg .$ If 10²³ hydrogen molecules strike per second at 2 cm² area of a rigid wall at an angle of 45° from the normal and rebound back with a speed of $1000 {ms}^{- 1}$ , then the pressure exerted on the wall is

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0.23 $\times$ 10⁶ Pa

23.4 $\times$ 10³ Pa

0.23 $\times$ 10³ Pa

2.34 $\times$ 10³ Pa

answer is A.

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Detailed Solution

F = rate of change of momentum $= \frac{Δ p}{Δ t} = (n m) (2 v \cos θ)$

Pressure, $p = \frac{F}{A} = \frac{(n m) (2 v \cos θ)}{A}$

$= \frac{10^{23} \times 3.32 \times 10^{- 27} \times 2 \times 1000 \times \cos 45 °}{2 \times 10^{- 4}}$

$= 2.34 \times 10^{3} {Nm}^{- 2} or Pa$

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The mass of hydrogen molecule is 3.32 ×10-27 kg. If 1023 hydrogen molecules strike per second at 2 cm2 area of a rigid wall at an angle of 45° from the normal and rebound back with a speed of 1000 ms-1, then the pressure exerted on the wall is

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The mass of hydrogen molecule is $3.32 \times 10^{- 27} kg .$ If 10²³ hydrogen molecules strike per second at 2 cm² area of a rigid wall at an angle of 45° from the normal and rebound back with a speed of $1000 {ms}^{- 1}$ , then the pressure exerted on the wall is