CAPITULO 3: DIAGNÓSTICO, ANÁLISIS Y DISCUSIÓN DE RESULTADOS
4. PROPUESTA DE MEJORA
4.5. Breve descripción del curso
4.5.2. Desarrollo de los contenidos
Although all the gas laws could be derived from Eq. 3.8. we shall derive the ideal gas equation only and proceed to calculate different molecular parameters.
Eq. 3.8 can be rewritten as.
pV = 2/3 N (l/2mu ') .... 3.S)
From the Kinetic theory of gases (postulate 6) it is known that the absolute temperature of a gas sample is directly proportional to the mean kmet1c energy of the molecules, i.e.
T oc 1/2 mu'
Or 1/2 mu·l = K T where K is constant.
Substituting this in Eq. 3.9. we obtain.
pV
=
213 N K T.. 3.10
3 II
This can be \ritten as
pV = NkT 3.12
\\here k. kno\\n :1s Boltzmann const:1nt is equal to 2/3K. The value ofk is I JX x 10 '3 J K 1 Eq 3 12 1s the idc:1l gas equation for N molecules For a gas having nmoles. the number of molecules N is grven by.
N c nN.
' 3 l.i
\\/hen: "l,,. ts Avogrado·s constant and Jt IS equal to the number of molecules (or species} in one mole ,1f a substance. It rs equal to 6 022 x I O'' mol ' Hence. the equation torn moles of the gas can be
\\ rittcn by rsing Eqs 312 and 3. 13
PV
=
n, '
k,
T = n RT 3 14Where R i·s equal to N . k.
"
Let us now calcul:1tc some parameters of the g'ls molecules bv the combined usc of Eqs. 3.. J 10.
J 12 :1nd l 14
.1 4 I Calculation of Average Kinetic Energy
A\cragc kinetic energy per molecule can be calculated from Eq. 3.10_ knm\·mg that K J/2k
A\·Cragc kinetic energy per molecule
=
l 11 mu: 3/2 k T ... J 15 Sm1ilarl- average kinetic energy per mole (N) ( 1/2 mu ')IN.) (3/2 k T) l/2N.kT
3/2 RT (\\here R rs N, k) 3 16 The energy calculated using this l'\.prcssi0n ts also called the translational energy: this energy is due to the mot1on of the molecules in space
Thus. \\C can"" that the tr,msbtronal cnergv (K.E) of an ideal gas is independent of the tvpc and pressure of the gas. What then do vou think it depends on from the equation vvc derived for the kmetic energy') It depends on the absolute tcnw .:raturc.
Example I
Let us lllustratc the usc of Eqs 3 IS 3.!1d 3 J fi in doing some calculations.
Calculation of the a\"l::ragc translational energy per molecule
Calcubte the average translational kinetic cnerg.v values of nitrogen at 300 K Using Eq 3 I :i. the translational encrgv of nitrogen per molecule at 300 K 3;2 kT
2. Calculation of the average translatiOnal cnery of nitrogen per mole
Similarly using Eq. 3.16. the translational energy of nitrogen per mole at 300 K 3/2xX.3141mol 1 K 1 x300K
3.74 x 103 1 mol'
312 RT
3.4.2
Calculation of Number Density and Concentration
Number density. (n ,.) is defined as the number of molecules of a gas in unit volume. We can obtain number density by re-:uranging the equation.
pV NkT
as
Number of density of a gas ( nJ = NIV = p/k T ... 3 17 We can also do this for the concentration of a gas \\hich is defined as the number of moles of a gas in unit volume. What do vou obtam when vou re-arrange equation?
pY
=
n N,kT = nRTConcentration of a gas (c) = n/Y p/RT ... ··· ... 3.18 Example 2
Calculate the number density and concentration of nitrogen molecules at 298 2 K and 1.013 x I 0' Pa.
Using Eqs. 3.17 and 3.1X.
Number of density (n) of nitrogen molecules at 29X.2 K = p/k T I 013 x I 0' Pa
1.3Sx 10 1'1 K-'x29X.2 K 2.462 X 10"5 111 '
Concentration (c) of nitrogen at 29X.2 K = p/RT 1.013xlll'Pa 8.3141 mol·' K' x 2'! .2 K 40. X6 mol m ·'
You should note that the number dcns1ty or the concentration of a gas is directly proportional to the gas pressure and mvcrsdy proportional to 1ts t mpcrature.
3 4.3 Calculation of Mean Square Speed and Root Mean Square Speed
In section 3.2.2. \\·e have dctincd mean square speed (u'). The square root of its value is called root mean square speed and 1s represented as u,,, For one mole of the gas. combimng Eqs. 3.8 and 3.14.
\\C can wnte
PV
=
RT = Mmu· (n = I and Nm=
Mm 31.e. mean square speed (
u')
= 3RT/MmMolar mass)
... ... . . . ... 3.19
Root mean square speed (--iu') = u,,, = --!3RT/Mm ... 3.20
Example 3
Let us calculate uof methane molecules at 515 K using Eq 3.20
Um• = V3RT/Mm = V3 x 8.314 J mol- 1 K-1 x 515 K/0.0 16 kg mol 8.96 x 10' m s- 1
Using the above example, answer the following exercises:
Exercise 3
Calculate the root mean square speed of hydrogen molecules at 500 K (Molar mass of hydrogen= 0.002 kg).
Exercise 4
At what temperature will hydrogen molecules have the same root mean square speed that oxygen molecules have at l27°C?
4.0 Conclusion
We can conclude by saying that the kinetic theory is applicable to any gas. The theory also accounts for the properties of gases as expressed by the various gas laws we studied in Module 2, Unit I. It is also postulated that since the molecules are in constant random motion and they always get farther apart they fill up any container. In the process. the molecules bombard the walls of the container explaining the existence of pressure by gaseous molecules. However. absolute zero is the temperature at which the molecules can be said to have no translational motion.
5.0 Summary
In this Unit. we have learnt that:
• Kinetic theory of gases refers to the idea that gas molecules are in continuous motion.
• The absolute temperature of an ideal gas is proportional to the average translation to the average translational kinetic energy of the molecules present in it.
• The Number density of a gas is the number of molecules of a gas in unit volume.
• The concentration of a gas is the number of moles of a gas in unit volume.
• At any given instant, the molecules in a sample of a gas have a range of velocities.
• One application of Graham's law is the gaseous diffusion method of separating isotopes.
6.0 References and Other Resources
Goldberg, David and Dillard, Clyde (1974). College Chemistry 3rd Revised Edition New York Macmillan Publishing Co. Inc.
IGNOU ( 1997). States of Matter: Physical Chemistry CHE-04 New Delhi.
7.0 Tutor-Marked Assignment
(2) 1.000 x
to-'
m' of argon at a certain pressure and temperature took 151 s to effuse through a porous barrier. How long will it take for the same volume of oxygen to effuse under identical conditions''(Hint: The time taken by a gas to effuse varies inversely as its rate of effussion.)
(3) A mixture of 2.00 x JO-' kg ofand 2.00 x JO·' kg of He exerts a pressure of 1.50 x I 0' Pa What arc the partial pressures of H, and He''
(4) Calculate the ratio of mean square speed of oxygen to nitrogen at 300 K.
(5) Calculate the number density and concentration of oxygen at 1.013 x 105 Pa and 300 K.