MHT CET · Physics · Laws of Motion
\(10^{23}\) molecules of a gas, each having a mass of \(3 \times 10^{-27} \mathrm{~kg}\) strike per second per sq.cm of a rigid wall at an angle of \(60^{\circ}\) with the normal and rebound with a velocity of \(500 \mathrm{~m} / \mathrm{s}\). The pressure exerted by the gas molecules on the wall is
- A \(2000 \mathrm{~N} / \mathrm{m}^2\)
- B \(500 \mathrm{~N} / \mathrm{m}^2\)
- C \(1000 \mathrm{~N} / \mathrm{m}^2\)
- D \(1500 \mathrm{~N} / \mathrm{m}^2\)
Answer & Solution
Correct Answer
(D) \(1500 \mathrm{~N} / \mathrm{m}^2\)
Step-by-step Solution
Detailed explanation
According to the Newton's second law of motion:
\(\text {Pressure exerted on the wall}\) \(=\frac{\text { Rate of change of mometum } \times \text { no. of molecules }}{\text { Area }}\)
\(\Rightarrow P=\frac{\left(2 m V \cos 60^{\circ}\right) \times N}{A}=\) \(\frac{2 \times 3 \times 10^{-27} \times 5 \times 10^{-2} \times \cos 60^{\circ} \times 10^{23}}{10^{-4}}\)
\(\therefore P=1500 \mathrm{~N} / \mathrm{m}^2\)
\(\text {Pressure exerted on the wall}\) \(=\frac{\text { Rate of change of mometum } \times \text { no. of molecules }}{\text { Area }}\)
\(\Rightarrow P=\frac{\left(2 m V \cos 60^{\circ}\right) \times N}{A}=\) \(\frac{2 \times 3 \times 10^{-27} \times 5 \times 10^{-2} \times \cos 60^{\circ} \times 10^{23}}{10^{-4}}\)
\(\therefore P=1500 \mathrm{~N} / \mathrm{m}^2\)
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