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Chapter 22: Problem 21

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.

Short Answer

Expert verified
The root-mean-square speed of the smoke particles is approximately \(325.46 m/s\).

Step by step solution

01

Convert the Mass from Grams to Kilograms

The mass of smoke particles is given as \(5.2 \times 10^{-14}\) grams. To convert this to kilograms, multiply by \(1 \times 10^{-3} Kg/g\): \(5.2 \times 10^{-14} g \times 1 \times 10^{-3} Kg/g = 5.2 \times 10^{-17} Kg\)
02

Convert the Temperature from Celsius to Kelvin

The temperature is given as 14°C. To convert this to Kelvin, add 273.15: \(14°C + 273.15 = 287.15 K\)
03

Substitute Values Into the Speed Formula

Now, we can substitute the known values into the root-mean-square speed formula: \(V_{rms} = \sqrt{\frac{3(1.38 \times 10^{-23} J/K)(287.15 K)}{5.2 \times 10^{-17} Kg}}\)
04

Solve the Equation

Carry out the necessary calculations to solve for \(V_{rms}\). Be careful with the order of operations: multiplication and division first, then the square root. This yields a root-mean-square speed of approximately \(325.46 m/s\).

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Most popular questions from this chapter

A cylindrical container of length \(56.0 \mathrm{~cm}\) and diameter \(12.5 \mathrm{~cm}\) holds \(0.350\) moles of nitrogen gas at a pressure of \(2.05\) atm. Find the rms speed of the nitrogen molecules.

Consider a sample of argon gas at \(35.0^{\circ} \mathrm{C}\) and \(1.22 \mathrm{~atm}\) pressure. Suppose that the radius of a (spherical) argon atom is \(0.710 \times 10^{-10} \mathrm{~m} .\) Calculate the fraction of the container volume actually occupied by atoms.

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 \(v_{\mathrm{p}}=\sqrt{3 \mathrm{kT} / \mathrm{m}}\).

At \(44.0^{\circ} \mathrm{C}\) and \(1.23 \times 10^{-2}\) atm the density of a gas is \(1.32 \times 10^{-5} \mathrm{~g} / \mathrm{cm}^{3} .(a)\) Find \(v_{\mathrm{rms}}\) for the gas molecules. (b) Using the ideal gas law, find the number of moles per unit volume (molar density) of the gas. ( \(c\) ) By combining the results of (a) and \((b)\), find the molar mass of the gas and identify it.

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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