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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

(a) At what temperature (if any) does the numerical value of Celsius degrees equal the numerical value of Fahrenheit degrees? (b) At what temperature (if any) does the numerical value of kelvins equal the numerical value of Fahrenheit degrees?
The fuselage of an Airbus A340 has a circumference of \(17.72 \mathrm{m}\) on the ground. The circumference increases by \(26 \mathrm{cm}\) when it is in flight. Part of this increase is due to the pressure difference between the inside and outside of the plane and part is due to the increase in the temperature due to air drag while it is flying along at $950 \mathrm{km} / \mathrm{h} .$ Suppose we wanted to heat a full-size model of the airbus made of aluminum to cause the same increase in circumference without changing the pressure. What would be the increase in temperature needed?
The volume of air taken in by a warm-blooded vertebrate in the Andes mountains is \(210 \mathrm{L} /\) day at standard temperature and pressure (i.e., \(\left.0^{\circ} \mathrm{C} \text { and } 1 \text { atm }\right) .\) If the air in the lungs is at \(39^{\circ} \mathrm{C},\) under a pressure of $450 \mathrm{mm} \mathrm{Hg}$ and we assume that the vertebrate takes in an average volume of \(100 \mathrm{cm}^{3}\) per breath at the temperature and pressure of its lungs, how many breaths does this vertebrate take per day?
A square brass plate, \(8.00 \mathrm{cm}\) on a side, has a hole cut into its center of area \(4.90874 \mathrm{cm}^{2}\) (at $\left.20.0^{\circ} \mathrm{C}\right) .$ The hole in the plate is to slide over a cylindrical steel shaft of cross-sectional area \(4.91000 \mathrm{cm}^{2}\) (also at \(20.0^{\circ} \mathrm{C}\) ). To what temperature must the brass plate be heated so that it can just slide over the steel cylinder (which remains at \(20.0^{\circ} \mathrm{C}\) )? [Hint: The steel cylinder is not heated so it does not expand; only the brass plate is heated. \(]\)
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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