The important relations you should remember in kinetic theory of gases are the following:

(1) Pressure exerted by a gas,

(2) R.M.S. speed of molecule,

Remember all the three expressions for r.m.s.speed. The second one gives the r.m.s.speed in terms of molar mass M and the universal gas constant R. The third one gives the r.m.s. speed in terms of molecular mass ‘m’ and Boltzman’s constant ‘k’.

(3) Average translational kinetic energy of any type of gas molecule is

(4) If the molecule has ‘n’ degrees of freedom, the average kinetic energy per molecule is (n/2 )kT.

The following points are worth noting in the present context:

(i) A mono atomic gas molecule has 3 degrees of freedom and has translational kinetic energy only [equal to

(ii) A diatomic molecule has 5 degrees of freedom (three translational and two rotational). In this case, the total average kinetic energy per molecule is

(iii) Tri-atomic and polyatomic molecules have 6 degrees of freedom (three translational and three rotational). The total average kinetic energy per molecule is ( 6/2 )kT =

The K.E. per mole in all the above cases is N times (N is the Avogadro number). Since Nk=R, the average K.E. per mole is

Note that the molar heat capacity at constant volume (C

The molar heat capacity of a gas at constant pressure (C

It will be useful to remember the relation connecting the ratio of specific heats ‘γ’ and the number of degrees of freedom ‘f’:

Let us now consider the following question:

The following question appeared in the Kerala Engineering Entrance Test paper of 2002:

(1) Pressure exerted by a gas,

**P = (1/3) nmc**where ‘n’ is the number of molecules per unit volume, ‘m’ is the molecular mass, ‘c’ is the r.m.s. speed of the molecule ‘ρ’ is the density of the gas, ‘k’ is Boltzman’s constant and T is the absolute temperature.^{2}= (1/3)ρc^{2}= nkT(2) R.M.S. speed of molecule,

**c = √(3P/ρ) = √(3RT/M) = √(3kT/m).**Remember all the three expressions for r.m.s.speed. The second one gives the r.m.s.speed in terms of molar mass M and the universal gas constant R. The third one gives the r.m.s. speed in terms of molecular mass ‘m’ and Boltzman’s constant ‘k’.

(3) Average translational kinetic energy of any type of gas molecule is

**(3/2)kT**since translational motion along three directions only are possible in our three dimensional space.(4) If the molecule has ‘n’ degrees of freedom, the average kinetic energy per molecule is (n/2 )kT.

The following points are worth noting in the present context:

(i) A mono atomic gas molecule has 3 degrees of freedom and has translational kinetic energy only [equal to

**(3/2)kT**].(ii) A diatomic molecule has 5 degrees of freedom (three translational and two rotational). In this case, the total average kinetic energy per molecule is

**(5/2 )kT**.(iii) Tri-atomic and polyatomic molecules have 6 degrees of freedom (three translational and three rotational). The total average kinetic energy per molecule is ( 6/2 )kT =

**3kT**.The K.E. per mole in all the above cases is N times (N is the Avogadro number). Since Nk=R, the average K.E. per mole is

**(3/2 )RT**for mono atomic,**(5/2 )RT**for diatomic and**3RT**for triatomic and polyatomic gas molecules.Note that the molar heat capacity at constant volume (C

_{V}) is obtained by putting T=1(corresponding to a temperature rise of 1K) in the above expressions. The values are therefore**(3/2 )R**for mono atomic gas,**(5/2)R**for diatomic gas and**3R**for triatomic and polyatomic gases.The molar heat capacity of a gas at constant pressure (C

_{P}) is given by**C**. Therefore, the values are_{P}= C_{V}+ R**(5/2)R**for mono atomic gas,**(7/2)R**for diatomic gas and**4R**for triatomic and polyatomic gases.It will be useful to remember the relation connecting the ratio of specific heats ‘γ’ and the number of degrees of freedom ‘f’:

**γ = 1+ (2/f)**

Note: In the above discussion, the vibrational modes of the molecules have been ignored. Even though the above values are in agreement with the values obtained from experiment in the case of several gases, there are discrepancies in the case of certain diatomic gases and several polyatomic gases. The vibrational modes also are therefore to be taken into account in more rigorous treatment.Let us now consider the following question:

**The average translational kinetic energy of a helium gas molecule (molar mass 4) at a particular temperature is 0.05electron volt. The average translational kinetic energy of an oxygen molecule (molar mass 32) at the same temperature will be**

(a) 0.4eV (b) 0.08eV (c) 0.2eV (d) 0.05eV (e) 0.1eV

Don’t be concerned about the type of the gas and the molar mass. The translational kinetic energy of all types of molecules is the same at a given temperature. The correct option therefore is (d).(a) 0.4eV (b) 0.08eV (c) 0.2eV (d) 0.05eV (e) 0.1eV

The following question appeared in the Kerala Engineering Entrance Test paper of 2002:

**An electron tube was sealed off during manufacture at a pressure of 1.2×10**

(a) 2×10^{-7}mm of mercury at 27˚C. Its volume is 100cm^{3}. The number of molecules that remain in the tube is(a) 2×10

^{16}(b) 3×10^{15}(c) 3.86×10^{11}(d) 5×10^{ 11}(e) 2.5×10^{12}You may do this problem by using one of the following relations:

(1) P = nkT (2) PV = rT where ‘r’ is the gas constant for the mass of gas in the electron tube.

If you use the first relation, you should remember the value of Boltzman’s constant ‘k’, which is 1.38×10

n = P/(kT) = hdg/(kT)

The number of molecules in the electron tube is n×V = hdgV/(kT)

= (1.2×10

If you use the second relation you will write PV = n’RT where n’ is the number of moles of gas in the tube. Therefore, n’ = PV/(RT).

The number of molecules in the tube = n’N = PVN/(RT) = hdgVN/(RT).

Note that this is the same relation as we obtained in the first method since k = R/N.

If you are not very quick in arithmetical manipulations, don’t attempt questions like this during your initial trial.

Consider now the following M.C.Q.:

(1) P = nkT (2) PV = rT where ‘r’ is the gas constant for the mass of gas in the electron tube.

If you use the first relation, you should remember the value of Boltzman’s constant ‘k’, which is 1.38×10

^{-23}. Since k = R/N you can substitute the values of the universal gas constant R (= 8.31 J/mol/K ) and the Avogadro number N (= 6.02×10^{23}), which you should definitely remember, to get ‘k’. The number of molecules per unit volume therefore is,n = P/(kT) = hdg/(kT)

The number of molecules in the electron tube is n×V = hdgV/(kT)

= (1.2×10

^{-10}×13600×9.8×100×10^{-6}) / (1.38×10^{-23}×300) = 3.86×10^{11}[Option (c)].If you use the second relation you will write PV = n’RT where n’ is the number of moles of gas in the tube. Therefore, n’ = PV/(RT).

The number of molecules in the tube = n’N = PVN/(RT) = hdgVN/(RT).

Note that this is the same relation as we obtained in the first method since k = R/N.

If you are not very quick in arithmetical manipulations, don’t attempt questions like this during your initial trial.

Consider now the following M.C.Q.:

**1.2 mole of helium gas (mono atomic), 0.4 mole of nitrogen (diatomic) and 0.4 mole of oxygen (diatomic) are contained in a vessel of volume 10 litre at a temperature of 27˚C. The pressure of this mixture of gases is (in Nm**

(a) 5×10

Some of you may have certain doubts regarding this simple question. Your doubts are added when you see the distractions ‘mono atomic’ and ‘diatomic’. But your doubts will be cleared the moment you remember the expression for pressure in the form, P = nkT. Here ‘n’ is the number of molecules per unit volume, ‘k’ is the Boltzman constant and T is the absolute temperature. The type of the gas does not come into the picture and you require the number density ‘n’ only for substituting in the expression for P. Since the mixture contains 2 moles (1.2+0.4+0.4), the total number of molecules in the mixture is 2N where N is Avogadro number. The number of molecules per unit volume (n) therefore is 2N/(10×10^{-2})(a) 5×10

^{5}(b) 10^{6}(c) 2.5×10^{5}(d) 2.5×10^{6}(e) 5×10^{6}^{-3}) = N/(5×10^{-3}). Therefore, P = NkT/(5×10^{-3}). But, k = R/N so that P = RT/(5×10^{-3}) = 8.3×300/(5×10^{-3}) = 5×10^{5}Nm^{-2}.Let us consider another question:

**Oxygen and hydrogen gases are at the same temperature T. The kinetic energy of an oxygen molecule will be**

(a) 32 times the kinetic energy of a hydrogen molecule (b) 16 times the kinetic energy of a hydrogen molecule (c) twice the kinetic energy of a hydrogen molecule (d) 4 times the kinetic energy of a hydrogen molecule (e) the same as that of the hydrogen molecule.

The correct option is (e). The kinetic energy (total) of a gas molecule is (n/2)kT where ‘n’ is the number of degrees of freedom. Oxygen and hydrogen are diatomic and they have the same degrees of freedom (n=5) and hence the same kinetic energy at a given temperature.

(a) 32 times the kinetic energy of a hydrogen molecule (b) 16 times the kinetic energy of a hydrogen molecule (c) twice the kinetic energy of a hydrogen molecule (d) 4 times the kinetic energy of a hydrogen molecule (e) the same as that of the hydrogen molecule.

Certain simple questions may confuse you if you are in a hurry and you may give wrong answers! Here is one such question:

**To decrease the volume of an ideal gas by 10% at constant temperature, the pressure should be increased by**

(a) 5% (b) 8.91% (c) 10% (d) 11.1% (e) 12.25%

(a) 5% (b) 8.91% (c) 10% (d) 11.1% (e) 12.25%

Don’t pick out option (c) in a hurry. We have PV = P’×0.9V so that P’ = P/0.9 = 1.111P. The increase in pressure is 11.1% [option (d)].

Let us discuss one more question:

**When a gas contained in a closed vessel is heated through 1˚C, the pressure of the gas increases by 0.2%. the final temperature of the gas is**

(a) 200K (b) 361K (c) 500K (d) 501K (e) 601K

We have, P/T = 1.002P/(T+1) since the volume is constant. Solving this we obtain T = 500K. The final temperature is T+1 = 501K.

(a) 200K (b) 361K (c) 500K (d) 501K (e) 601K

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