/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 68 A smoke particle has a mass of \... [FREE SOLUTION] | 91影视

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Chapter 13: Problem 68

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?

Short Answer

Expert verified
Question: Calculate the root-mean-square speed of a smoke particle with a mass of \(1.38 \times 10^{-17} \mathrm{kg}\) in a room at \(27^\circ\text{C}\). Answer: The root-mean-square speed of the smoke particle is approximately \(0.299\ \mathrm{m/s}\).

Step by step solution

01

Convert the given temperature to Kelvin

First, we need to convert the given temperature from Celsius to Kelvin. We can use the relation $$ T_{K} = T_{C} + 273.15 $$ So in our case, $$ T_K = 27^\circ\text{C} + 273.15 = 300.15\text K $$
02

Substitute the given values into the rms speed formula

Now that we have the temperature in Kelvin, we can substitute all the given values into the rms speed formula: $$ v_{rms} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \mathrm{J/K} \times 300.15 \mathrm{K}}{1.38 \times 10^{-17} \mathrm{kg}}} $$
03

Calculate the rms speed

Now, we'll simply calculate the rms speed using the values given: $$ v_{rms} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 300.15} {1.38 \times 10^{-17}}} = \sqrt{\frac{1.234 \times 10^{-20}}{1.38 \times 10^{-17}}} = \sqrt{0.08942} \arr 0.299\ \mathrm{m/s} $$ Thus, the rms speed of the smoke particle is approximately \(0.299\ \mathrm{m/s}\).

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Most popular questions from this chapter

Agnes Pockels \((1862-1935)\) was able to determine Avogadro's number using only a few household chemicals, in particular oleic acid, whose formula is \(\mathrm{C}_{18} \mathrm{H}_{34} \mathrm{O}_{2}\) (a) What is the molar mass of this acid? (b) The mass of one drop of oleic acid is \(2.3 \times 10^{-5} \mathrm{g}\) and the volume is $2.6 \times 10^{-5} \mathrm{cm}^{3} .$ How many moles of oleic acid are there in one drop? (c) Now all Pockels needed was to find the number of molecules of oleic acid. Luckily, when oleic acid is spread out on water, it lines up in a layer one molecule thick. If the base of the molecule of oleic acid is a square of side \(d\), the height of the molecule is known to be \(7 d .\) Pockels spread out one drop of oleic acid on some water, and measured the area to be \(70.0 \mathrm{cm}^{2}\) Using the volume and the area of oleic acid, what is \(d ?\) (d) If we assume that this film is one molecule thick, how many molecules of oleic acid are there in the drop? (e) What value does this give you for Avogadro's number?
At \(0.0^{\circ} \mathrm{C}\) and \(1.00 \mathrm{atm}, 1.00 \mathrm{mol}\) of a gas occupies a volume of \(0.0224 \mathrm{m}^{3} .\) (a) What is the number density? (b) Estimate the average distance between the molecules. (c) If the gas is nitrogen \(\left(\mathrm{N}_{2}\right),\) the principal component of air, what is the total mass and mass density?
Steel railroad tracks of length \(18.30 \mathrm{m}\) are laid at $10.0^{\circ} \mathrm{C} .$ How much space should be left between the track sections if they are to just touch when the temperature is \(50.0^{\circ} \mathrm{C} ?\)
A scuba diver has an air tank with a volume of \(0.010 \mathrm{m}^{3}\) The air in the tank is initially at a pressure of \(1.0 \times 10^{7} \mathrm{Pa}\) Assuming that the diver breathes \(0.500 \mathrm{L} / \mathrm{s}\) of air, find how long the tank will last at depths of (a) \(2.0 \mathrm{m}\) and (b) $20.0 \mathrm{m} .\( (Make the same assumptions as in Example \)13.6 .)$

Aluminum rivets used in airplane construction are made slightly too large for the rivet holes to be sure of a tight fit. The rivets are cooled with dry ice \(\left(-78.5^{\circ} \mathrm{C}\right)\) before they are driven into the holes. If the holes have a diameter of \(0.6350 \mathrm{cm}\) at \(20.5^{\circ} \mathrm{C},\) what should be the diameter of the rivets at \(20.5^{\circ} \mathrm{C}\) if they are to just fit when cooled to the temperature of dry ice?
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