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Chapter 13: Problem 70

What are the rms speeds of helium atoms, and nitrogen, hydrogen, and oxygen molecules at \(25^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
Answer: The approximate rms speeds at 25掳C are: - Helium atoms: 1367 m/s - Nitrogen molecules: 515 m/s - Hydrogen molecules: 1929 m/s - Oxygen molecules: 482 m/s

Step by step solution

01

Temperature conversion

To convert from Celsius to Kelvin, add 273.15: \(T=25^{\circ} \mathrm{C} + 273.15 = 298.15\ \mathrm{K}\) Now, we will find the molar mass of each gas particle and convert it to the mass of a single particle in kilograms: For helium atom: \(M_{He} = 4.0026\ \mathrm{g/mol}\) or \(4.0026 \times 10^{-3}\ \mathrm{kg/mol}\) For nitrogen molecule: \(M_{N_2} = 28.0134\ \mathrm{g/mol}\) or \(28.0134 \times 10^{-3}\ \mathrm{kg/mol}\) For hydrogen molecule: \(M_{H_2} = 2.01588\ \mathrm{g/mol}\) or \(2.01588 \times 10^{-3}\ \mathrm{kg/mol}\) For oxygen molecule: \(M_{O_2} = 31.9988\ \mathrm{g/mol}\) or \(31.9988 \times 10^{-3}\ \mathrm{kg/mol}\) The Avogadro constant denoted as \(N_A\) is \(6.022 \times 10^{23}\ \mathrm{mol^{-1}}\). Now, we will calculate the mass of a single particle for each type of gas using the formula: \(m = \frac{M}{N_A}\) For helium atom: \(m_{He} = \frac{4.0026 \times 10^{-3}}{6.022 \times 10^{23}}\ \mathrm{kg}\) For nitrogen molecule: \(m_{N_2} = \frac{28.0134 \times 10^{-3}}{6.022 \times 10^{23}}\ \mathrm{kg}\) For hydrogen molecule: \(m_{H_2} = \frac{2.01588 \times 10^{-3}}{6.022 \times 10^{23}}\ \mathrm{kg}\) For oxygen molecule: \(m_{O_2} = \frac{31.9988 \times 10^{-3}}{6.022 \times 10^{23}}\ \mathrm{kg}\) Now, we will calculate the rms speed for each particle using the equation: \(v_{rms} = \sqrt{\frac{3kT}{m}}\) For helium atom:
02

rms speed of helium

\(v_{rms_{He}} = \sqrt{\frac{3(1.38 \times 10^{-23})(298.15)}{4.0026 \times 10^{-3}/(6.022 \times 10^{23})}} \approx 1367\ \mathrm{m/s}\) For nitrogen molecule:
03

rms speed of nitrogen

\(v_{rms_{N_2}} = \sqrt{\frac{3(1.38 \times 10^{-23})(298.15)}{28.0134 \times 10^{-3}/(6.022 \times 10^{23})}} \approx 515\ \mathrm{m/s}\) For hydrogen molecule:
04

rms speed of hydrogen

\(v_{rms_{H_2}} = \sqrt{\frac{3(1.38 \times 10^{-23})(298.15)}{2.01588 \times 10^{-3}/(6.022 \times 10^{23})}} \approx 1929\ \mathrm{m/s}\) For oxygen molecule:
05

rms speed of oxygen

\(v_{rms_{O_2}} = \sqrt{\frac{3(1.38 \times 10^{-23})(298.15)}{31.9988 \times 10^{-3}/(6.022 \times 10^{23})}} \approx 482\ \mathrm{m/s}\) The rms speeds at \(25^{\circ} \mathrm{C}\) are approximately: Helium atoms: 1367 m/s Nitrogen molecules: 515 m/s Hydrogen molecules: 1929 m/s Oxygen molecules: 482 m/s

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Most popular questions from this chapter

A temperature change \(\Delta T\) causes a volume change \(\Delta V\) but has no effect on the mass of an object. (a) Show that the change in density $\Delta \rho\( is given by \)\Delta \rho=-\beta \rho \Delta T$. (b) Find the fractional change in density \((\Delta \rho / \rho)\) of a brass sphere when the temperature changes from \(32^{\circ} \mathrm{C}\) to \(-10.0^{\circ} \mathrm{C}\)
Air at room temperature and atmospheric pressure has a mass density of $1.2 \mathrm{kg} / \mathrm{m}^{3} .$ The average molecular mass of air is 29.0 u. How many molecules are in \(1.0 \mathrm{cm}^{3}\) of air?
A smoke particle has a mass of \(1.38 \times 10^{-17} \mathrm{kg}\) and it is randomly moving about in thermal equilibrium with room temperature air at \(27^{\circ} \mathrm{C} .\) What is the rms speed of the particle?
A \(10.0-\mathrm{L}\) vessel contains \(12 \mathrm{g}\) of \(\mathrm{N}_{2}\) gas at \(20^{\circ} \mathrm{C}\) (a) Estimate the nearest-neighbor distance. (b) Is the gas dilute? [Hint: Compare the nearest-neighbor distance to the diameter of an \(\left.\mathrm{N}_{2} \text { molecule, about } 0.3 \mathrm{nm} .\right]\)
A flat square of side \(s_{0}\) at temperature \(T_{0}\) expands by \(\Delta s\) in both length and width when the temperature increases by \(\Delta T\). The original area is \(s_{0}^{2}=A_{0}\) and the final area is $\left(s_{0}+\Delta s\right)^{2}=A .\( Show that if \)\Delta s \ll s_{0}$$$\frac{\Delta A}{A_{0}}=2 \alpha \Delta T$$(Although we derive this relation for a square plate, it applies to a flat area of any shape.)
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