Solutions for Exercises of Kinetic Theory of Gases and Radiation

1. Choose the correct option.

i) In an ideal gas, the molecules possess

(A) only kinetic energy

(B) both kinetic energy and potential energy

(D) neither kinetic energy nor potential energy

ii) The mean free path X of molecules is given by

$latex \left(A\right)\;\sqrt{\frac2{\pi nd^2}}$

$latex \left(B\right)\;\frac2{\pi nd^2}$

$latex \left(C\right)\;\frac2{\sqrt2\pi nd^2}$

$latex \left(D\right)\;\frac2{\sqrt{2\pi nd}}$,

where n is the number of molecules per unit volume and d is the diameter of the molecules.

iii) If pressure of an ideal gas is decreased by 10% isothermally, then its volume will

(A) decrease by 9%

(B) increase by 9%

(D) increase by 11.11%

iv) If a = 72 and r = 0.24, then the value of t_r is

(A) 0.02

(B) 0.04

(D) 0.2

v) The ratio of emissive power of perfectly blackbody at 1327 °C and 527 °C is

(A) 4:1

(B) 16 : 1

(D) 8 : 1

2. Answer in brief.

i) What will happen to the mean square speed of the molecules of a gas if the temperature of the gas increases?

ii) On what factors do the degrees of freedom depend?

iii) Write ideal gas equation for a mass of 7 g of nitrogen gas.

iv) If the density of oxygen is 1.44 kg/m³ at a pressure of 10⁵N/m², find the root mean square velocity of oxygen molecules.

v) Define athermanous substances and diathermanous substances.

3. When a gas is heated its temperature Explain this phenomenon based on kinetic theory of gases.

4. Explain, on the basis of kinetic theory, how the pressure of gas changes if its volume is reduced at constant temperature.

5. Mention the conditions under which a real gas obeys ideal gas equation.

6. State the law of equipartition of energy and hence calculate molar specific heat of mono- and di-atomic gases at constant volume and constant pressure.

7. What is a perfect blackbody ? How can it be realized in practice?

8. State (i) Stefan-Boltmann law and (ii) Wein’ s displacement law.

9. Explain spectral distribution of blackbody radiation.

10. State and prove Kirchoff s law of heat radiation.

11. Calculate the ratio of mean square speeds of molecules of a gas at 30 K and 120 K.[Ans: 1:4]

12. Two vessels A and B are filled with same gas where volume, temperature and pressure in vessel A is twice the volume, temperature and pressure in vessel B. Calculate the ratio of number of molecules of gas in vessel A to that in vessel B.

$latex P_A=2P_B,\; V_A=\;2V_B,\; T_A\;=\;2T_B,\;\frac{n_A}{n_B}=?$

We know that,

PV = nRT

$latex P_AV_A=n_ART_A\;and\;P_BV_P=\;n_BRT_B$

$latex \therefore\frac{P_AV_A}{P_BV_B}=\frac{n_AT_A}{n_BT_B}$

$latex \therefore\frac{n_A}{n_B}=\frac{P_A}{P_B}\times\frac{V_A}{V_B}\times\frac{T_B}{T_A}$

$latex \therefore\frac{n_A}{n_B}=\frac{2P_B}{P_B}\times\frac{2V_B}{V_B}\times\frac{T_B}{2T_B}$

$latex \therefore\frac{n_A}{n_B}=\;2\times2\times\frac12$

$latex \therefore\frac{n_A}{n_B}=\;2$

$latex \therefore\frac{n_A}{n_B}=\frac21$

13. A gas in a cylinder is at pressure P. If the masses of all the molecules are made one third of their original value and their speeds are doubled, then find the resultant pressure.

$latex M_2=\frac13M_1,\;c_2=2c_1,\;P_2=?

We know that,

$latex P=\frac{Mc^2}{3V}(c=rms\;speed)$

$latex P\propto Mc^2\;for\;cons\tan t\;volume$

$latex \therefore\frac{P_2}{P_1}=\frac{M_2}{M_1}\times\frac{c_2^2}{c_1^2}$

$latex \therefore\frac{P_2}{P_1}\;=\;\frac{{}_3^1M_1}{M_1}\times\left(\frac{2c_1}{c_1}\right)^2$

$latex \therefore\frac{P_2}{P_1}\;=\;\frac13\times\frac41$

$latex \therefore\frac{P_2}{P_1}\;=\;\frac43$

$latex \therefore P_2=\frac43P_1$

14. Show that rms velocity of an oxygen molecule is $latex \sqrt2$ times that of a sulfur dioxide molecule at S.T.P.

15. At what temperature will oxygen molecules have same rms speed as helium molecules at S.T.P.? (Molecular masses of oxygen and helium are 32 and 4 respectively)

$latex M_{0_2}=32,\;{M_H}_e=4,\;T_{H_e}=STP=273K,$

$latex T_{0_2}=?,\;C_{RMS(O_2)}=C_{RMS(H_e)}$

We know that, $latex C_{RMS}=\sqrt{\frac{3RT}M}$

$latex \therefore C_{RMS(O_2)}=\sqrt{\frac{3RT_{(O_2)}}{M_{0_2}}}$

$latex \therefore C_{RMS(He)}=\sqrt{\frac{3RT_{He}}{M_{He}}}$

$latex \therefore\sqrt{\frac{3RT_{O_2}}{M_{O_2}}}=\sqrt{\frac{3RT_{He}}{M_{He}}}$

$latex \;\;\therefore\sqrt{\frac{T_{O_2}}{32}}\;=\;\sqrt{\frac{273}4}$

$latex \therefore\;\frac{T_{O_2}}{32}\;=\;\frac{273}4$

$latex \therefore\;T_{O_2}\;=\;\frac{273}4\;\times\;32$

$latex \therefore T_0=2184$

16. Compare the rms speed of hydrogen molecules at 127 °C with rms speed of oxygen molecules at 27 °C given that molecular masses of hydrogen and oxygen are 2 and 32 respectively.

$latex M_{O_2}=32,\;M_{H_2}=?$

$latex T_{H_2}=127^\circ C=127+273=300K,$

$latex \frac{C_{RMS(H_2)}}{C_{RMS(O_2)}}=?$

We know that,$latex C_{RMS}=\sqrt{\frac{3R}M}$

$latex \therefore\frac{C_{RMS(H_2)}}{C_{RMS(0_2)}}=\frac{\sqrt{\displaystyle T_{H_2}/M_{H_2}}}{\sqrt{T_{0_2}/M_{0_2}}}$

$latex \therefore\frac{C_{RMS(H_2)}}{C_{RMS(0_2)}}\;=\;\sqrt{\frac{T_{H_2}}{M_{H_2}}\times\frac{M_{0_2}}{T_{0_2}}}$

17. Find kinetic energy of 5 litre of a gas at S.T.P. given standard pressure is 1.013 x 10⁵ N/m².

V = 5 liters = 5 × 10^-3m³,

P = 1.013 × 10⁵N/m²,

K.E. = ?

We know that,

Average K.E.of gas = 3PV/2

$latex \therefore K.E\;of\;gas\;=\frac{3\times1.013\times10^5\times5\times10^{-3}}2$

$latex \therefore K.E\;of\;gas\;=\frac{15.195\times10^2}2$

$latex \therefore K.E\;of\;gas\;=7.5975\times10^2J$

$latex \therefore K.E\;of\;gas\;=0.7597kJ$

18. Calculate the average molecular kinetic energy (i) per kmol (ii) per kg (iii) per molecule of oxygen at 127 °C, given that molecular weight of oxygen is 32, R is 8.31 J mo1^-1 K^-1 and Avogadro’s number N_A is 6.02 x 10²³ molecules mol^-1.

T = 127°C = 127 + 273 = 400K,

M_O2 = 32gm = 32 × 10^-3kg,

R = 8.31Jmol^-1K^-1, N_A = 6.02 × 10²³

i) K.E per kmol of gas = ?

K.E.of one (per)mole =3/2 RT

∴ K.E.of one (per) mole $latex =\frac32\times8.31\times400$

= 4986 J

∴ K.E.of per kmol,

$latex =4986J\;\times\;1000$

= 4.986× 10⁶J

ii) K.E. per kg of gas = ?

$latex K.E\;per\;unit\;mass=\frac32\frac{RT}M$

∴ K.E.per kg of gas,

$latex =\frac32\times\frac{8.31\times400}{32\times10^{-3}}$

$latex =\frac{4986}{32\times10^{-3}}$

∴ K.E.per kg of gas,

$latex =155.81\times10^3J$

iii) K.E. per molecule of gas = ?

$latex K.e\;per\;molecule=\frac32\frac{RT}{N_A}$

∴ K.E.per molecule,

$latex =\frac{4986}{6.02\times10^{23}}$

$latex =828.23\times10^{-23}$

$latex =8.28\times10^{-21}J$

19. Calculate the energy radiated in one minute by a blackbody of surface area 100 cm² when it is maintained at 227 °C.

$latex T\;=\;227^\circ C\;=\;227\;+\;273$

$latex \;T\;=\;500K,\;t\;=\;60\;s,$

$latex A\;=\;100cm^2\;=\;100\;\times\;10^{-4}m^2$

$latex A\;=\;10^{-2}m^2,$

$latex \sigma=5.67\times10^{-8}J/m^2sK^4,\;Q=?$

We know that, Energy radiated by the body per unit

time per unit area = σT⁴

∴ Energy Radiated = AtσT⁴

∴ Energy Radiated $latex =10^{-2}\times60\times5.67\times10^{-8}\times{(500)}^4$

$latex =340.2\times625\times10^8\times10^{-10}$

$latex =2126.25\;J$

20. Energy is emitted from a hole in an electric furnace at the rate of 20 W, when the temperature of the furnace is 727 °C. What is the area of the hole? (Take Stefan’s constant a to be 5.7 x 10-8 J _s-1 _m-2 K^-4)

P = 20 W, T = 727°C = 727+ 273 = 1000K,

A = ? σ = 5.67 × 10^-8Jm^-2s^-1K^-4

We know that, Energy radiated by the body per unit time per unit area = σT⁴

$latex =\;\frac Q{At}\;=\;\sigma T^4$

$latex \therefore\;\frac PA\;=\;\sigma T^4$

$latex \therefore\;P\;=\;A\sigma T^4$

$latex A\;=\;\frac P{\sigma T^4}$

$latex A\;=\;\frac{20}{5.67\times10^{-8}\times{(1000)}^4}$

$latex A\;=\;\frac2{5.67\times10^{-4}}$

$latex A=3.52\times10^{-4}m^2$

21. The emissive power of a sphere of area 0.02 m² is 0.5 kcal s^-1 m^-2. What is the amount of heat radiated by the spherical surface in 20 second?

A = 0.02m², E = 0.5kcal s^-1m^-2, t = 20 s, Q = ?

$latex E\;=\;\frac Q{At}$

$latex \therefore\;Q\;=\;EAt$

$latex \therefore\;Q\;=\;0.5\times0.02\times20$

$latex \therefore\;Q\;=\;0.2kcal$

22. Compare the rates of emission of heat by a blackbody maintained at 727 °C and at 227 °C, if the blackbodies are surrounded by an enclosure (black) at 27 °C. What would be the ratio of their rates of loss of heat?

T₁ = 727°C = 727 + 273 = 1000K,

T₂= 227°C = 227 + 273 = 500K,

T₀ = 27°C = 227 + 273 = 300K , P₁/P₂=?

Rate of energy supplied = ?

We know that, Energy lost by the black body per unit time per unit area = σ(T₄ − T₀⁴ )

$latex \therefore\;\frac{P_1}{P_2}\;=\;\frac{\sigma(T_1^4-T_0^4)}{\sigma(T_2^4-T_0^4)}$

$latex \frac{1000^4-300^4}{500^4-30^4}=\frac{{(10\times10^2)}^4-{(3\times10^2)}^4}{{(5\times10^2)}^4-{(3\times10^2)}^4}$

$latex \frac{P_1}{P_2}\;=\;\frac{\lbrack1000-81\rbrack\times10^8}{\lbrack625-81\rbrack\times10^8}$

$latex \frac{P_1}{P_2}\;=\;\frac{9919}{544}$

$latex \frac{P_1}{P_2}\;=\;18.23$

23. Earth’s mean temperature can be assumed to be 280 K. How will the curve of blackbody radiation look like for this temperature? Find out λ_max. In which part of the electromagnetic spectrum, does this value lie?

$latex \lambda_{max}\propto\frac1T\;\;\;\;\;\;\;\therefore\lambda_{max}=\frac bT$

$latex \therefore b=2.898\times10^{-3}mK$

$latex \therefore\;\lambda_{max}\;=\;\frac{2.898\times10^{-3}}{280}$

$latex \therefore\;\lambda_{max}\;=\;1.035\times10^{-5}m(Microwave)$

24. A small-blackened solid copper sphere of radius 2.5 cm is placed in an evacuated chamber. The temperature of the chamber is maintained at 100 °C. At what rate energy must be supplied to the copper sphere to maintain its temperature at 110 °C? (Take Stefan’s constant a to be 5.76 x 10-8 J s^-1 m^-2 K^-4 and treat the sphere as blackbody.)

r = 2.5cm = 2.5 × 10^-2m,

T = 100°C = 100 + 273 = 373K,

T = 110°C = 110 + 273 = 383K,

σ = 5.67 × 10^-8JmsK,

Rate of energy supplied = ?

We know that,

Energy lost by the black body per unit time per unit area,

$latex =\sigma(T^4-T_0^4)$

$latex \therefore\;\frac Q{At}\;=\;\sigma(T^4-T_0^4)$

$latex \therefore\;\frac PA\;=\;\sigma(T^4-T_0^4)$

$latex \therefore\;P\;=\;A\sigma(T^4-T_0^4)$

$latex \therefore\;P\;=\;4\pi r^2\sigma(T^4-T_0^4)$

Hence rate of energy supplied is given as,

$latex \therefore P=4\pi r^2\sigma(T^4-T_0^4)$

$latex =4\times3.14\times{(2.5\times10^{-2})}^2\times5.67\times10^{-8}\times\lbrack383^4-373^4\rbrack$

=0.962W (using log)

25. Find the temperature of a blackbody if its spectrum has a peak at (a) λ_max = 700 nm (visible), (b) λ_max= 3 cm (microwave region) and (c) λ_max₌ 3 m (FM radio waves) (Take Wien’s constant b = 2.897 x 10^-3 m K).

$latex \lambda_{max}\;\propto\;\frac1T$

$latex \therefore\;\lambda_{max}\;=\;\frac bT$

$latex \therefore\;\lambda_{max}\;=\;\frac{2.897\times10^{-3}}T$

$latex a)\;\lambda_{max}=700nM=700\times10^{-9}m$

$latex \therefore\;T\;=\;\frac{2.897\times10^{-3}}{\lambda_{max}}$

$latex \therefore\;T\;=\;\frac{2.897\times10^{-3}}{700\times10^{-9}}$

$latex T=4.138\times10^3=4138\;K$

$latex b)\;\lambda_{max}=3cm=3\times10^{-2}m$

$latex \therefore\;T\;=\;\frac{2.897\times10^{-3}}{\lambda_{max}}$

$latex \therefore\;T\;=\;\frac{2.897\times10^{-3}}{3\times10^{-2}}$

$latex T=0.9656\times10^{-1}=0.0966\;K$

$latex b)\;\lambda_{max}=3m=3m$

$latex \therefore\;T\;=\;\frac{2.897\times10^{-3}}{\lambda_{max}}$

$latex \therefore\;T\;=\;\frac{2.897\times10^{-3}}3$

$latex \therefore\;T\;=\;0.9656\;\times\;10^{-3}$

$latex \therefore\;T\;=\;0.966\;\times\;10^{-3}K$