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Chapter 19: Problem 45

In a period of \(6.00 \mathrm{~s}, 9.00 \cdot 10^{23}\) nitrogen molecules strike a section of a wall with an area of \(2.00 \mathrm{~cm}^{2}\). If the molecules move with a speed of \(400.0 \mathrm{~m} / \mathrm{s}\) and strike the wall head on in elastic collisions, what is the pressure exerted on the wall? (The mass of one \(\mathrm{N}_{2}\) molecule is \(4.68 \cdot 10^{-26} \mathrm{~kg}\).)

Short Answer

Expert verified
Answer: The pressure exerted on the wall is 2.81 x 10^4 Pa.

Step by step solution

01

Calculate the momentum change for one molecule after collision

After the elastic collision, the direction of the nitrogen molecule is reversed. The momentum change for one molecule can be calculated by subtracting the initial momentum from the final momentum. This can be expressed as: Momentum change \(= m(v_{final} - v_{initial})\) Since the molecule changes its direction, \(v_{final} = -v_{initial}\), and the momentum change becomes: Momentum change \(= 2mv_{initial}\) Substituting the given values, we have: Momentum change \(= 2(4.68\cdot 10^{-26}\mathrm{~kg})(400.0\mathrm{~m} / \mathrm{s}) = 3.744\cdot 10^{-23}\mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\)
02

Calculate the momentum change for all molecules

We are given that there are \(9.00\cdot 10^{23}\) nitrogen molecules striking the wall. To find the total momentum change for all molecules, multiply the momentum change for one molecule by the number of molecules: Total momentum change \(= \text{number of molecules} \times \text{momentum change per molecule}\) Total momentum change \(= (9.00\cdot 10^{23})(3.744\cdot 10^{-23}\mathrm{~kg} \cdot \mathrm{m} / \mathrm{s})= 3.37\cdot 10^1\mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\)
03

Use the impulse-momentum theorem to find the force

The impulse-momentum theorem states that the force times the time interval is equal to the momentum change: \(F\Delta t = \Delta p\) Where \(F\) is the force, \(\Delta t\) is the time interval, and \(\Delta p\) is the momentum change. We are given a time interval of \(6\mathrm{~s}\). Solving for the force, we have: \(F=(\Delta p)/(\Delta t) = (3.37\cdot 10^1\mathrm{~kg} \cdot \mathrm{m} / \mathrm{s})/(6\mathrm{~s}) = 5.62\mathrm{~N}\)
04

Calculate the pressure

Now, we can find the pressure exerted on the wall by dividing the force by the given area (\(2.00\mathrm{~cm}^2\)). First, convert the area from \(\mathrm{cm}^2\) to \(\mathrm{m}^2\): \(2.00\mathrm{~cm}^{2} = 2.00\cdot 10^{-4}\mathrm{~m}^{2}\) Now, calculate the pressure: Pressure \(= \frac{F}{A} = \frac{5.62\mathrm{~N}}{2.00\cdot 10^{-4}\mathrm{~m}^{2}} = 28100\mathrm{~Pa}\) So, the pressure exerted on the wall is \(2.81\cdot 10^4\mathrm{~Pa}\).

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