# 2.2 Pressure, temperature, and rms speed  (Page 4/18)

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${v}_{\text{rms}}=\sqrt{\frac{3\phantom{\rule{0.2em}{0ex}}RT}{M}}.$

We digress for a moment to answer a question that may have occurred to you: When we apply the model to atoms instead of theoretical point particles, does rotational kinetic energy change our results? To answer this question, we have to appeal to quantum mechanics. In quantum mechanics, rotational kinetic energy cannot take on just any value; it’s limited to a discrete set of values, and the smallest value is inversely proportional to the rotational inertia. The rotational inertia of an atom is tiny because almost all of its mass is in the nucleus, which typically has a radius less than ${10}^{-14}\phantom{\rule{0.2em}{0ex}}\text{m}$ . Thus the minimum rotational energy of an atom is much more than $\frac{1}{2}\phantom{\rule{0.2em}{0ex}}{k}_{\text{B}}T$ for any attainable temperature, and the energy available is not enough to make an atom rotate. We will return to this point when discussing diatomic and polyatomic gases in the next section.

## Calculating kinetic energy and speed of a gas molecule

(a) What is the average kinetic energy of a gas molecule at $20.0\phantom{\rule{0.2em}{0ex}}\text{ºC}$ (room temperature)? (b) Find the rms speed of a nitrogen molecule $\left({\text{N}}_{2}\right)$ at this temperature.

## Strategy

(a) The known in the equation for the average kinetic energy is the temperature:

$\stackrel{\text{–}}{K}=\frac{1}{2}\phantom{\rule{0.2em}{0ex}}m\stackrel{\text{–}}{{v}^{2}}=\frac{3}{2}{k}_{\text{B}}T.$

Before substituting values into this equation, we must convert the given temperature into kelvin: $T=\left(20.0+273\right)\phantom{\rule{0.2em}{0ex}}\text{K}=293\phantom{\rule{0.2em}{0ex}}\text{K}\text{.}$ We can find the rms speed of a nitrogen molecule by using the equation

${v}_{\text{rms}}=\sqrt{\stackrel{\text{–}}{{v}^{2}}}=\sqrt{\frac{3{k}_{\text{B}}T}{m}},$

but we must first find the mass of a nitrogen molecule. Obtaining the molar mass of nitrogen ${\text{N}}_{2}$ from the periodic table, we find

$m=\frac{M}{{N}_{A}}=\frac{2\phantom{\rule{0.2em}{0ex}}\left(14.0067\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}\text{kg/mol}\right)}{6.02\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{23}\phantom{\rule{0.2em}{0ex}}{\text{mol}}^{\text{-1}}}=4.65\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-26}\phantom{\rule{0.2em}{0ex}}\text{kg}.$

## Solution

1. The temperature alone is sufficient for us to find the average translational kinetic energy. Substituting the temperature into the translational kinetic energy equation gives
$\stackrel{\text{–}}{K}=\frac{3}{2}\phantom{\rule{0.2em}{0ex}}{k}_{\text{B}}T=\frac{3}{2}\left(1.38\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-23}\phantom{\rule{0.2em}{0ex}}\text{J/K}\right)\left(293\phantom{\rule{0.2em}{0ex}}\text{K}\right)=6.07\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-21}\phantom{\rule{0.2em}{0ex}}\text{J}.$
2. Substituting this mass and the value for ${k}_{\text{B}}$ into the equation for ${v}_{\text{rms}}$ yields
${v}_{\text{rms}}=\sqrt{\frac{3{k}_{\text{B}}T}{m}}=\sqrt{\frac{3\left(1.38\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-23}\phantom{\rule{0.2em}{0ex}}\text{J/K}\right)\left(293\phantom{\rule{0.2em}{0ex}}\text{K}\right)}{4.65\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-26}\phantom{\rule{0.2em}{0ex}}\text{kg}}}=511\phantom{\rule{0.2em}{0ex}}\text{m/s}\text{.}$

## Significance

Note that the average kinetic energy of the molecule is independent of the type of molecule. The average translational kinetic energy depends only on absolute temperature. The kinetic energy is very small compared to macroscopic energies, so that we do not feel when an air molecule is hitting our skin. On the other hand, it is much greater than the typical difference in gravitational potential energy when a molecule moves from, say, the top to the bottom of a room, so our neglect of gravitation is justified in typical real-world situations. The rms speed of the nitrogen molecule is surprisingly large. These large molecular velocities do not yield macroscopic movement of air, since the molecules move in all directions with equal likelihood. The mean free path (the distance a molecule moves on average between collisions, discussed a bit later in this section) of molecules in air is very small, so the molecules move rapidly but do not get very far in a second. The high value for rms speed is reflected in the speed of sound, which is about 340 m/s at room temperature. The higher the rms speed of air molecules, the faster sound vibrations can be transferred through the air. The speed of sound increases with temperature and is greater in gases with small molecular masses, such as helium (see [link] ).

#### Questions & Answers

What is differential form of Gauss's law?
Rohit Reply
help me out on this question the permittivity of diamond is 1.46*10^-10.( a)what is the dielectric of diamond (b) what its susceptibility
OLUWA Reply
a body is projected vertically upward of 30kmp/h how long will it take to reach a point 0.5km bellow e point of projection
Abu Reply
i have to say. who cares. lol. why know that t all
Jeff
is this just a chat app about the openstax book?
Lord Reply
kya ye b.sc ka hai agar haa to konsa part
MPL Reply
what is charge quantization
Mayowa Reply
it means that the total charge of a body will always be the integral multiples of basic unit charge ( e ) q = ne n : no of electrons or protons e : basic unit charge 1e = 1.602×10^-19
Riya
is the time quantized ? how ?
Mehmet
What do you meanby the statement,"Is the time quantized"
Mayowa
Can you give an explanation.
Mayowa
there are some comment on the time -quantized..
Mehmet
time is integer of the planck time, discrete..
Mehmet
planck time is travel in planck lenght of light..
Mehmet
it's says that charges does not occur in continuous form rather they are integral multiple of the elementary charge of an electron.
Tamoghna
it is just like bohr's theory. Which was angular momentum of electron is intral multiple of h/2π
Aditya
determine absolute zero
OFERE Reply
The properties of a system during a reversible constant pressure non-flow process at P= 1.6bar, changes from constant volume of 0.3m³/kg at 20°C to a volume of 0.55m³/kg at 260°C. its constant pressure process is 3.205KJ/kg°C Determine: 1. Heat added, Work done, Change in Internal Energy and Change in Enthalpy
Opeyemi Reply
U can easily calculate work done by 2.303log(v2/v1)
Abhishek
Amount of heat added through q=ncv^delta t
Abhishek
Change in internal energy through q=Q-w
Abhishek
please how do dey get 5/9 in the conversion of Celsius and Fahrenheit
Gwam Reply
what is copper loss
timileyin Reply
this is the energy dissipated(usually in the form of heat energy) in conductors such as wires and coils due to the flow of current against the resistance of the material used in winding the coil.
Henry
it is the work done in moving a charge to a point from infinity against electric field
Ashok Reply
what is the weight of the earth in space
peterpaul Reply
As w=mg where m is mass and g is gravitational force... Now if we consider the earth is in gravitational pull of sun we have to use the value of "g" of sun, so we can find the weight of eaeth in sun with reference to sun...
Prince
g is not gravitacional forcé, is acceleration of gravity of earth and is assumed constante. the "sun g" can not be constant and you should use Newton gravity forcé. by the way its not the "weight" the physical quantity that matters, is the mass
Jorge
Yeah got it... Earth and moon have specific value of g... But in case of sun ☀ it is just a huge sphere of gas...
Prince
Thats why it can't have a constant value of g ....
Prince
not true. you must know Newton gravity Law . even a cloud of gas it has mass thats al matters. and the distsnce from the center of mass of the cloud and the center of the mass of the earth
Jorge
please why is the first law of thermodynamics greater than the second
Ifeoma Reply
every law is important, but first law is conservation of energy, this state is the basic in physics, in this case first law is more important than other laws..
Mehmet
First Law describes o energy is changed from one form to another but not destroyed, but that second Law talk about entropy of a system increasing gradually
Mayowa
first law describes not destroyer energy to changed the form, but second law describes the fluid drection that is entropy. in this case first law is more basic accorging to me...
Mehmet
define electric image.obtain expression for electric intensity at any point on earthed conducting infinite plane due to a point charge Q placed at a distance D from it.
Mateshwar Reply
explain the lack of symmetry in the field of the parallel capacitor
Phoebe Reply
pls. explain the lack of symmetry in the field of the parallel capacitor
Phoebe

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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