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The kinetic theory of gases states that the kinetic energy of a gas is directly proportional...

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The kinetic theory of gases states that the kinetic energy of a gas is directly proportional to the temperature of the gas. A relationship between the microscopic properties of the gas molecules and the macroscopic properties of the gas can be derived using the following assumptions: The gas is composed of pointlike particles separated by comparatively large distances. The gas molecules are in continual random motion with collisions being perfectly elastic. The gas molecules exert no long-range forces on each other. One of the most important microscopic properties of gas molecules is velocity. There are several different ways to describe statistically the average velocity of a molecule in a gas. The most obvious measure is the average velocity Vavg. However, since the molecules in a gas are moving in random directions, the average velocity is approximately zero. Another measure of velocity is (V^2)avg, the average squared velocity. Since the square of velocity is always positive, this measure does not average to zero over the entire gas. A third measure is the root-mean-square (rms) speed,Vrms , equal to the square root of (V^2)avg. The rms speed is a good approximation of the the typical speed of the molecules in a gas. This histogram (Figure 1) shows a theoretical distribution of speeds of molecules in a sample of nitrogen () gas. In this problem, you'll use the histogram to compute properties of the gas.


Part A
What is the average speedv_avgof the molecules in the gas?
Express your answer numerically to three significant digits.
Part B

Because the kinetic energy of a single molecule is related to its velocity squared, the best measure of the kinetic energy of the entire gas is obtained by computing the mean squared velocity,(v^2)_{\rm avg}, or its square rootv_rms. The quantityv_rmsis more common than(v^2)_{\rm avg}because it has the dimensions of velocity instead of the less-familiar velocity-squared.

What is the rms speedv_rmsof the molecules in the nitrogen gas?

Express your answer numerically to three significant digits.
Part C

What is the temperatureTof the sample of\rm N_2gas described in the histogram?

Express your answer in degrees Celsius to three significant figues.
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Answer #1
Concepts and reason

The concepts used to solve this problem are average and root mean square speed.

First find the average velocity of the each point like particles from the graph.

Then find the rms speed of the molecule in the nitrogen gas by relating the kinetic energy of the single molecule to its velocity squared.

Finally find the temperature of the sample gas in the histogram by using the root mean square speed expression.

Fundamentals

Average is a middle number of some data.

Kinetic theory narrates a gas as a large number of submicroscopic particles.

The molecules are random, rapid motion and have same mass.

The speed of the particle in a gas is measured as a root mean square speed.

The square root of average velocity squared of the molecules in a gas is defined as root mean square speed.

The expression for root mean square speed is,

3RT
гms
т

Here, R
is the gas constant, T
is the temperature, т
is the mass of the molecule, and V.
rms
is the root mean square.

(A)

Percentage
of molecules
16
12
4
0
Speed range (m/s)
0 200
200 400
009 00
600 800
800 1000
1000 1200

From the diagram,

The average speed of the molecule is,

100(6)+[300 (34)]+[so0 (36)]+[700(17)] + [ 900( 5)]+100(2) m/s
V
avg
100
=474m/s

(B)

Kinetic energy of the single molecule is related to its velocity square.

So from the graph,

([10(6)]+[300 (34)]+[s00 (36)]
700 (17)]+[900 (5)]+[10 (2)]/
m/s
rms
100
= /269200 m/s
=518.8m/s
519m/s

(C)

The expression for root mean square speed is,

3RT
гms
т

Substitute 519m/s
for V.
rms
, |8.314J/mol K
for R
, and 0.028kg/mol
for т
.

519m/s=3(8.314 J/mol - K ) 7
0.028kg/mol
|(519m/s) (890.785 J/kg K)T

Rearrange the above expression for T
.

T(519m/s)
(890.785J/kg K)
=302.386 K

To change the temperature from Kelvin to degree subtract 273.15 Celsius

T 302.386 K-273.15C|
29.2°C

Ans: Part A

The average speed of molecule is 474 m/s
.

Part B

The average speed of the molecules in nitrogen gas is 519m/s
.

Part C

The temperature of the sample gas in the histogram is 29.2°C
.

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