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Chapter 22: Problem 14

The speeds of a group of ten molecules are \(2.0,3.0,4.0, \ldots\), \(11 \mathrm{~km} / \mathrm{s}\). ( \(a\) ) Find the average speed of the group. (b) Calculate the root-mean-square speed of the group.

Short Answer

Expert verified
(a) The average speed of the group is 6.0 km/s (b) The root-mean-square speed of the group is approximately 7.10 km/s.

Step by step solution

01

Calculate Average Speed

The average speed can be calculated by adding up all the molecule speeds, and then dividing by the count of molecules. Here are the speeds: \(2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0, 10.0, 11.0\) km/s, totaling ten molecules. Summing these up equals 60 km/s. Thus, the average speed equals \(\frac{60 \text{ km/s}}{10} = 6.0 \text{ km/s}\).
02

Calculate Square of Each Speed

In order to calculate the root-mean-square speed, we first need to find the square of each molecule's speed. The squared speeds are: \(4.0, 9.0, 16.0, 25.0, 36.0, 49.0, 64.0, 81.0, 100.0, 121.0\) (km/s)\(^2\).
03

Calculate Root-Mean-Square Speed

The root-mean-square speed is found by taking the square root of the average of these squared speeds. Add up the squared speeds to get 505 (km/s)\(^2\), and divide by the number of molecules (10) to find the average, which equals 50.5 (km/s)\(^2\). Taking the square root of this gives approximately 7.10 km/s.

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Most popular questions from this chapter

Calculate the root-mean-square speed of smoke particles of mass \(5.2 \times 10^{-14} \mathrm{~g}\) in air at \(14^{\circ} \mathrm{C}\) and \(1.07\) atm pressure.

(a) Find the number of molecules in \(1.00 \mathrm{~m}^{3}\) of air at \(20.0^{\circ} \mathrm{C}\) and at a pressure of \(1.00 \mathrm{~atm} .(b)\) What is the mass of this volume of air? Assume that \(75 \%\) of the molecules are nitrogen \(\left(\mathrm{N}_{2}\right)\) and \(25 \%\) are oxygen \(\left(\mathrm{O}_{2}\right)\).

(a) Consider \(1.00 \mathrm{~mol}\) of an ideal gas at \(285 \mathrm{~K}\) and \(1.00 \mathrm{~atm}\) pressure. Imagine that the molecules are for the most part evenly spaced at the centers of identical cubes. Using Avogadro's constant and taking the diameter of a molecule to be \(3.00 \times 10^{-8} \mathrm{~cm}\), find the length of an edge of such a cube and calculate the ratio of this length to the diameter of a molecule. The edge length is an estimate of the distance between molecules in the gas. (b) Now consider a mole of water having a volume of \(18 \mathrm{~cm}^{3}\). Again imagine the molecules to be evenly spaced at the centers of identical cubes and repeat the calculation in \((a)\).

Show that, for atoms of mass \(m\) emerging as a beam from a small opening in an oven of temperature \(T\), the most probable speed is \(v_{\mathrm{p}}=\sqrt{3 \mathrm{kT} / \mathrm{m}}\).

The mass of the \(\mathrm{H}_{2}\) molecule is \(3.3 \times 10^{-24} \mathrm{~g}\). If \(1.6 \times 10^{23}\) hydrogen molecules per second strike \(2.0 \mathrm{~cm}^{2}\) of wall at an angle of \(55^{\circ}\) with the normal when moving with a speed of \(1.0 \times 10^{5} \mathrm{~cm} / \mathrm{s}\), what pressure do they exert on the wall?
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