Contents
Some of the most important Physics Topics include energy, motion, and force.
What is the Relationship Between the Boltzmann Constant and the Avogadro Constant?
The principle of equipartition of energy in kinetic theory came essentially from the concept of degrees of freedom. In Section 1.5.1, we have seen that the pressure of an ideal gas is
p = \(\frac{1}{3}\)(px + py + pz)
The number of degrees of freedom of an ideal gas molecule = 3. Clearly, it led to the factor of \(\frac{1}{3}\).
So, we get p = \(\frac{1}{3}\)ρc2 or, c2 = \(\frac{3 p}{\rho}\) (c = rms speed of the gas molecules).
Again, the total kinetic energy of the molecules in 1 mol of an ideal gas is \(\frac{1}{3}\)pV = \(\frac{3}{2}\)RT (Section 1.6). So, it may be said that the molecular kinetic energy of 1 moI of an ideal gas for each degree of freedom is \(\frac{1}{2}\)RT. Scientist Ludwig Boltzmann analysed the motion of a single molecule in an ideal gas and theoretically established the principle of equipartition of energy
Statement of the principle of equipartition of energy: The average molecular kinetic energy of any substance is equally shared among the degrees of freedom; the average kinetic energy of a single molecule associated with each degree of freedom is \(\frac{1}{2}\)kT (T = absolute temperature, k = Boltzmann constant = 1.38 × 10-23 J ᐧ K-1).
The number of degrees of freedom of an ideal gas molecule = 3. So, from the equipartition principle, the average kinetic energy of a molecule = 3 × \(\frac{1}{2}\)kT = \(\frac{3}{2}\)kT. The molecule has no potential energy So, the average total energy of a molecule is
e = \(\frac{3}{2}\)kT
Again, the total energy of 1 mol of an ideal gas is E = \(\frac{3}{2}\)RT.
So, the number of molecules in 1 mol of a gas = \(\frac{E}{e}\) = \(\frac{R}{k}\).
Clearly, this is the Avogadro number NA, i.e.,
NA = \(\frac{R}{k}\) or, R = NAk
This is the relation between the universal gas constant R, the Avogadro number NA and the Boltzmann constant k. Then the total energy of 1 mol of an ideal gas is
E = \(\frac{3}{2}\)NAkT
For any amount of ideal gas containing N molecules, E = \(\frac{3}{2}\)NkT. This relation is widely used particularly in chemistry.
It is to be noted that the equipartition principle is applicable to all substances, not only to gases. For any substance, solid, liquid or gas, the average molecular kinetic energy associated with each degree of freedom is \(\frac{1}{2}\)kT. For ideal gases the molecular potential energy is zero; so it is easy to calculate the total energy But for real gases, liquids or solids, the potential energy calculations are not easy. However, for these substances, the kinetic energy strictly follows the principle of equipartition.
Specific heat of a gas: The first law of thermo-dynamics is written as
dQ = dE + dW [E is taken for internal energy]
At constant volume, dV = 0; so dW = pdV = 0. Then, dQ = dE.
So, the heat absorbed or released at constant volume for a temperature change dT of 1 mol of a gas is dQ = CvdT. Here, Cv = molar specific heat at constant volume.
Then, Cv = \(\frac{d Q}{d T}\) = \(\frac{d E}{d T}\)
In case of monatomic gas: The kinetic theory assumes that the ideal gas molecules are monatomic. Actually gases like helium, neon, argon are monatomic. For 1 mol of such a gas, E = \(\frac{3}{2}\)RT
So, Cv = \(\frac{d E}{d T}\) = \(\frac{3}{2}\)R
The molar specific heat at constant pressure is Cp.
∴ Cp – Cv = R or Cp = Cv + R = \(\frac{3}{2}\)R + R = \(\frac{5}{2}\)R
The ratio between the two specific heats is
γ = \(\frac{C_p}{C_v}\) = \(\frac{5}{3}\) = 1.67
This value tallies with the experimentally determined values of γ in case of helium, neon, etc.
In case of diatomic gas: The molecules of gases like oxygen, nitrogen, hydrogen, etc., are diatomic. The number of degrees of freedom of a diatomic molecule = 5; so for 1 mol of such a gas, E = \(\frac{5}{2}\)RT.
Then, Cv = \(\frac{d E}{d T}\) = \(\frac{5}{2}\)R; Cp = Cv + R = \(\frac{5}{2}\)R + R = \(\frac{7}{2}\)R;
∴ γ = \(\frac{C_p}{C_v}\) = \(\frac{7}{5}\) = 1.4
This value of γ is also supported by experiments.
In general cases: If f is the number of degrees of freedom of an ideal gas molecule then, the energy of 1 mol of such a gas E = \(\frac{f}{2}\)RT
Cv = \(\frac{f}{2}\)R; Cp = \(\frac{f}{2}\)R + R = \(\left(\frac{f+2}{2}\right)\)R
So, γ = \(\frac{C_p}{C_v}\) = \(\frac{f+2}{f}\) = 1 + \(\frac{2}{f}\).
Specific heats of helium and hydrogen gases:
We know that, R ≈ 2 cal ᐧ mol-1 ᐧ °C-1
For helium gas,
Cv = \(\frac{3}{2}\)R = \(\frac{3}{2}\) × 2 = 3 cal ᐧ mol-1 ᐧ °C-1
Cp = \(\frac{5}{2}\)R = \(\frac{5}{2}\) × 2 = 5 cal ᐧ mol-1 ᐧ °C-1
The molecular weight of helium = 4. So the specific heats are,
The molecular weight of hydrogen = 2. So, the specific heats of hydrogen gas are,
cv = \(\frac{C_v}{2}\) = \(\frac{5}{2}\) = 2.5 cal ᐧ g-1 ᐧ °C-1
and cp = \(\frac{C_p}{2}\) = \(\frac{7}{2}\) = 3.5 cal ᐧ g-1 ᐧ °C-1
It may be noted that the value of cp for helium gas and the values of both cv and cp for hydrogen gas are greater than the specific heat of water (1 cal ᐧ g-1 ᐧ °C-1).
Cv, Cp and γ of a gas mixture:
Let n1 mol of a gas (molar specific heat at constant volume and constant pressure be \(C_{\nu_1}\) and \(C_{p_1}\) respectively) is mixed with n2 mol of a gas (molar specific heat at constant volume and constant pressure be \(C_{v_2}\) and \(C_{p_2}\) respectively) such that they do not react chemically.
Therefore, thermal capacity of (n1 + n2) mol gas of the mixture at constant volume is n1\(C_{v_1}\) + n2\(C_{v_2}\).
Hence, at constant volume effective molar specific heat of the mixture is, Cv = \(\frac{n_1 C_{v_1}+n_2 C_{v_2}}{n_1+n_2}\)
Similarly, at constant pressure, the effective molar specific heat of the mixture is, Cp = \(\frac{n_1 C_{p_1}+n_2 C_{p_2}}{n_1+n_2}\)
∴ Ratio of the two molar specific heats of the mixture,
γ = \(\frac{C_p}{C_v}\) = \(\frac{n_1 C_{p_1}+n_2 C_{p_2}}{n_1 C_{v_1}+n_2 C_{v_2}}\) …… (1)
Now, if f1 and f2 be the degrees of freedom of the molecules of two gases respectively then,
\(C_{v_1}\) = \(\frac{f_1}{2}\)R; \(C_{p_1}\) = \(C_{v_1}\) + R = (\(\frac{f_1}{2}\) + 1)R
and \(C_{v_2}\) = \(\frac{f_2}{2}\)R; \(C_{p_2}\) = \(C_{v_2}\) + R = (\(\frac{f_2}{2}\) + 1)R
Putting these values in equation (1), we can find γ of the gas mixture.
Numerical Examples
Example 1.
The rms speed of the molecules of an ideal gas at STP is 0.5 km ᐧ s-1. Find the density of the gas. What will be the density at 21°C if pressure remains the same? Given, atmospheric pressure = 105 N ᐧ m-2.
Solution:
rms speed, c = \(\sqrt{\frac{3 p}{\rho}}\); so ρ = \(\frac{3 p}{c^2}\)
Here, p = 105 N ᐧ m-2,
c = 0.5 km ᐧ s-1 = 0.5 × 1000 m ᐧ s-1 = 500 m ᐧ s-1
∴ ρ = \(\frac{3 \times 10^5}{(500)^2}\) = 1.2 kg ᐧ m-3
At constant pressure, c ∝ \(\frac{1}{\sqrt{\rho}}\); also c ∝ \(\sqrt{T}\). So, ρ ∝ \(\frac{1}{T}\)
Then, \(\frac{\rho_0}{\rho_{21}}\) = \(\frac{T_{21}}{T_0}\)
or, ρ21 = ρ0\(\frac{T_0}{T_{21}}\) [T0 = 0°C = 273K, T21 = 21°c = (21 + 273)K = 294K]
= 1.2 × \(\frac{273}{294}\) = 1.11 kg ᐧ m-3.
Example 2.
Find out the energy of 1 mol of a gas and its average molecular kinetic energy at 27°C. Given, R = 8.3 × 107 erg ᐧ mol-1 ᐧ K-1 and NA = 6.02 × 1023 mol-1.
Solution:
Energy of 1 mol of the gas is
E = \(\frac{3}{2}\)RT = \(\frac{3}{2}\) × (8.3 × 107) × 300 [T = 27°C = 300K]
= 3.735 × 1010 erg
Average molecular kinetic energy is
e = \(\frac{3}{2}\)kT = \(\frac{3}{2} \frac{R}{N_A} T\) = \(\frac{3}{2}\) × \(\frac{\left(8.3 \times 10^7\right) \times 300}{6.02 \times 10^{23}}\)
= 6.2 × 10-14 erg.
Example 3.
The average kinetic energy of a molecule in gas at STP is 5.6 × 10-14erg. Find out the number of molecules per volume of the gas. Given, density of mercury = 13.6 g ᐧ cm-3.
Solution:
Pressure of the gas, p
= \(\frac{2}{3}\) × energy of gas molecules per unit volume
= \(\frac{2}{3}\) × n × average kinetic energy of 1 molecule
where n = number of molecules per unit volume of the gas.
∴ n \(=\frac{3 p}{2 \times \text { average kinetic energy of } 1 \text { molecule }}\)
= \(\frac{3 \times(76 \times 13.6 \times 980)}{2 \times\left(5.6 \times 10^{-14}\right)}\) = 2.71 × 1019.
Example 4.
Find out the temperature at which the rms speed of nitrogen molecules will be equal to the escape velocity from the earth’s gravity. Given, mass of a nitrogen atom = 23.24 × 10-24g ; average radius of the earth = 6390 km; g = 980 cm ᐧ s-2; Boltzmann constant = 1.37 × 10-16erg ᐧ °C-1.
Solution:
rms speed of a molecule = \(\sqrt{\frac{3 R T}{M}}\);
escape velocity = \(\sqrt{2 g R_1}\), where R1 = radius of the earth.
According to the question,
\(\sqrt{\frac{3 R T}{M}}\) = \(\sqrt{2 g R_1}\)
or, T = \(\frac{2}{3} \frac{g M R_1}{R}\)
Now, R = Nk and M = mN
where m = mass of a nitrogen molecule
= 2 × (23.24 × 10-24) g ;
N = number of nitrogen molecules.
So, T = \(\frac{2}{3} \cdot \frac{g m N R_1}{N k}\) = \(\frac{2}{3} \cdot \frac{g m R_1}{k}\)
= \(\frac{2}{3}\) × \(\frac{980 \times\left(2 \times 23.24 \times 10^{-24}\right) \times\left(6390 \times 10^5\right)}{1.37 \times 10^{-16}}\)
= 1.42 × 105K.
Example 5.
Find out the temperature at which the average kinetic energy of a gas molecule will be equal to the energy gained by an electron on acceleration across a potential difference of 1 V. Given, Boltzmann constant = 1.38 × 10-23 J ᐧ K-1; charge of an electron 1.6 × 10-19 C.
Solution:
Energy gained by the electron
= 1 eV = (1.6 × 10-19 C) × 1 V = 1.6 × 10-19 J.
Average kinetic energy of a gas molecule
= \(\frac{3}{2}\)kT = \(\frac{3}{2}\) × (1.38 × 10-23) × T J.
∴ 1.6 × 10-19 = \(\frac{3}{2}\) × (1.38 × 10-23) × T
or, T = \(\frac{2}{3}\) × \(\frac{1.6 \times 10^{-19}}{1.38 \times 10^{-23}}\) = 7729K = 7456°C.
Example 6.
Find out the molecular kinetic energy of 1 mol of oxygen gas at STP. Given, molecular weight of oxygen = 32; density of oxygen at STP = 1.43 g ᐧ L-1; density of mercury = 13.6 g ᐧ cm-3.
Solution:
Volume of 1 mol or 32 g oxygen
= \(\frac{32}{1.43}\)L = \(\frac{32 \times 10^3}{1.43}\)cm3;
At STP, pressure p = 76 × 13.6 × 981 dyn ᐧ cm-2;
temperature, T = 0°C = 273 K.
Molecular kinetic energy of 1 mol oxygen gas at STP (diatomic gas) is
E = \(\frac{5}{2}\)RT = \(\frac{5}{2} \frac{p V_T}{T}\) = \(\frac{5}{2} p V\)
= \(\frac{5}{2}\) × (76 × 13.6 × 981) × \(\frac{32 \times 10^3}{1.43}\)
= 5.67 × 1010 erg.
Example 7.
At what temperature will the rms speed of a hydrogen molecule be equal to that of an oxygen molecule at 47°C? [AIEEE ‘02]
Solution:
Molecular weights of oxygen and hydrogen, respectively, are M1 = 32 and M2 = 2.
rms speed, c = \(\sqrt{\frac{3 R T}{M}}\)
∴ \(\frac{c_1}{c_2}\) = \(\sqrt{\frac{T_1}{T_2} \cdot \frac{M_2}{M_1}}\)
Here, c1 = c2 and T1 = 47°C = 320 K
∴ T2 = T1 ᐧ \(\frac{M_2}{M_1}\) = 320 × \(\frac{2}{32}\)
= 20 K = (20 – 273)°C = -253°C
Example 8.
0.76 g of a mixture of hydrogen and oxygen gases has a volume of 2 L, temperature of 300 K and pressure of 105 N ᐧ m-2. Find out the individual masses of hydrogen and oxygen In the mixture.
Solution:
Let the number of moles of hydrogen and oxygen be n1 and n2 respectively.
Then, the pressure of the mixture,
p = \(\frac{n_1 R T}{V}\) + \(\frac{n_2 R T}{V}\) = (n1 + n2)\(\frac{R T}{V}\)
or, n1 + n2 = \(\frac{p V}{R T}\) = \(\frac{10^5 \times\left(2 \times 10^{-3}\right)}{8.3 \times 300}\) = 0.08 …….. (1)
Now, mass of hydrogen gas = 2n1 and mass of oxygen gas = 32n2.
So, 2n1 + 32n2 = 0.76
or, n1 + 16n2 = 0.38 ……. (2)
(2) – (1) gives,
15n2 = 0.3 or, = 0.02
Then from (1), n1 = 0.08 – 0.02 = 0.06.
∴ Mass of hydrogen = 2 × 0.06 = 0.12 g and mass of oxygen = 32 × 0.02 = 0.64 g.
Example 9.
Find out the number of molecules in a gas of volume 20 cm3 at a pressure of 76 cm of mercury, and at 27°C. Given, average molecular kinetic energy at 27°C = 2 × 10-14erg.
Solution:
Pressure of the gas,
ρ = \(\frac{1}{3} \rho c^2\) = \(\frac{1}{3} \frac{m N}{V} c^2\)
[m = mass of a molecule, N = number of molecules]
or, N = \(\frac{3 p V}{m c^2}\) = \(\frac{3 p V}{2 \times \frac{1}{2} m c^2}\)
[average molecular kinetic energy = \(\frac{1}{2} m c^2\) = 2 × 10-14 erg]
= \(\frac{3 \times(76 \times 13.6 \times 980) \times 20}{2 \times\left(2 \times 10^{-14}\right)}\)
= 1.52 × 1021
Example 10.
Find the temperature at which the average kinetic energy of a gas molecule will be equal to the energy of a photon in 6000A radiation. Given, Holtzmann constant, k = 1.38 × 10-23 J ᐧ K-1; Planck’s constant, h = 6.625 × 10-34 J ᐧ s.
Solution:
Let the required temperature = T
Average molecular kinetic energy = \(\frac{3}{2}\)kT;
Energy of a photon = hv = \(\frac{h c}{\lambda}\)
[Here, λ = 6000A = 6000 × 10-1o m]
According to the question, \(\frac{3}{2}\)kT = \(\frac{h c}{\lambda}\)
∴ T = \(\frac{2}{3} \frac{h c}{\lambda k}\) = \(\frac{2}{3}\) × \(\frac{\left(6.625 \times 10^{-34}\right) \times\left(3 \times 10^8\right)}{\left(6000 \times 10^{-10}\right) \times\left(1.38 \times 10^{-23}\right)}\)
= 1.6 × 104 K.