Kinetic Theory of Gases

KINETIC THEORY OF GASES The properties of the gases are entirely different from those of solid and liquid. In case of gases, thermal expansion is very large as compared to solids and liquids. To state the conditions of a gas, its volume, pressure and temperature must be specified. Intermolecular force Solid > liquid > real gas > ideal gas (zero) Potential energy Solid < liquid < real gas < ideal gas (zero) At a given temperature for solid, liquid and gas: (i) Internal kinetic energy : Same for all (ii) Internal potential Energy : Maximum for deal gas (PE = 0) and minimum for solids (PE = –ve) (iii) Internal Energy : Maximum for Ideal gas and Minimum for solid Notation : N.T.P. (Normal temperature and pressure) Temperature 0°C = 273.15 K Pressure 1 atm = 1.01325 × 105 M/m2 = 1.01325 × 105 pascal Volume 22.4 litre S.T.P. (Standard Temperature and Pressure 0.01° C = 273.16K 1 atm 22.4 litre 1 lit = 10–3 m3 = 103 cm3 = 103 cc = 103 ml water = 1000 kg/m3 = 1 g/cc, ice = 900 kg/m3 = 0.9 g/cc Ideal Gas Concept • Volume of gas molecules is negligible as compared to volume container. So volume of gas = volume of container (Except 0 K) • No intermoleculer forces act between gas molecules. Properties of Ideal Gas • A gas which follows all gas laws and gas equation at every possible temperature and pressure is known as ideal or perfect gas. • Ideal gas molecules can do only translational motion, so their kinetic energy is only translational kinetic energy • Ideal gas can not be liquified because IMF is zero • Potential energy of ideal gas is zero so internal energy of ideal gas is perfectly translational K.E. of gas. It is directly proportional to absolute temperature. So, internal energy depends only and only on its temperature. Etrans  T For a substance U = UKE + UPE UKE : depends only on T, UPE: depends upon intermolecular force (Always negative) • Specific heat of ideal gas is constant quantity and it does not change with temperature • All real gases behaves as an ideal gas at high temperature and low pressure and low density. • Gas molecules have point mass and negligible volume and velocity is very high (107 cm/s). That's why there is no effect of gravity on them. Equation of state for Ideal gas  mN  R PV = RT Þ PV =   RT    N T = NkT  mN 0   N0  where  = number of moles of gas k = Boltzamann constant P RT kT     Mw m Gas Laws • Boyle’s Law According to it for a given mass of an ideal gas at constant temperature, the volume of 1 a gas is inversely proportional to its pressure, i.e., V  P if m and T = Constant  P1V1 = P2V2 • Charle’ Law According to it for a given mass of an ideal gas at constant pressure, volume of a gas is directly proportional to its absolute temperature, i.e. V µ T V T if m and P = Constant  1  1 V2 T2 • Gay-Lussac’s Law According to it, for a given mass of an ideal gas at constant volume, pressure of a gas is directly proportional to its absolute temperature, i.e., P  T P T if m and V = constant Þ 1  1 P2 T2 Avogadro’s Law According to it, at same temperature and pressure, equal volume of all gases contain equal number of molecules, i.e., N1 = N2 if P, V and T are same. Dalton’s Partial Pressure Mixture Law : According to it, the pressure exerted by mixture of non-reactive gases is equal to the sum of partial pressure of each component gases present in the mixture, ie., P=P1+ P2+…. Illustrations Illustration 39. By increasing temperature of gas by 5° C its pressure increases by 0.5% from its initial value at constant volume then what is initial temperature of gas ? Solution: T P 5  100  At constant volume T µ P  × 100 = × 100 = 0.5  T = = 1000K P 0.5 T Illustration 40. Calculate the value of universal gas constant at STP. Solution: PV Universal gas constant is given by R = T One mole of all gases at S.T.P. occupy volume V = 22.4 litre = 22.4 × 10–3 m3 P = 760 mm of Hg = 760 × 10–3 × 13.6 × 103 × 9.80 N m2, T = 273.16 K 760 103 13.6 103  9.80  22.4 10 3 R= = 8.314 L mol–1 K–1 273.16 Illustration 41. A closed container of volume 0.02 m3 contains a mixture of neon and argon gases at a temperature of 27°C and pressure of 1 × 105 N/m2. The total mass of the mixture is 28 g. If the gram molecular weights of neon and argon are 20 and 40 respectively, find the mass the individual gases in the container, assuming them to be ideal. Given : R = 8.314 J/mol-K. Solution: Let m gram be the mass of neon. Then, the mass of argon is (28 – m)g. m 28  m 28  m   Total number of moles of the mixture, = ….(i) 20 40 40 PV 1105  0.02  Now, = = 0.8 ….(ii) RT 8.314  300 28  m By (i) and (ii), = 0.8  28 + m = 32  m = 4 gram 40 or mass of argon = (28 – 4)g = 24 g Illustration 42. Calculate the temperature of the Sun if density is 1.4 g cm–3, pressure is 1.4 × 109 atmosphere and average molecular weight of gases in the Sun in 2 g/mole. [Given R = 8.4 J mol–1K–1] Solution: PV M V M PV = RT  T = ….(i) But  = and  = = Mw Mw V R From equation (i) Illustration 43. T= PVM w PM w 1.4 109 1.01105  2 103 = 2.4 × 107 K   VR R 1.4 1000  8.4 At the top of a mountain a thermometer reads 7°C and barometer reads 70 cm of Hg. At the bottom of the mountain they read 27°C and 76 cm of Hg respectively. Compare the density of the air at the top with that at the bottom. Solution:  P R  M M M  and    By gas equation PV = RT  Q   Mw T M w  Mw V   P   P  Now as Mw and R are same for top and bottom       T  T  T  B T PT TB 70 300 75      So = 0.9868 B PB TT 76 280 76 Illustration 44. A sample of oxygen of volume of 500 cc at a pressure of 2 atm is compressed to a volume of 400 cc. What pressure is needed to do this if the temperature is kept constant ? Solution: V  500  Temperature is constant, so P1 V1 = P2V2 P2 = P1 1  2  = 2.5 atm V2  400  Illustration 45. A vessel of volume 8.0 × 10–3 m3 contains an ideal gas at 300 K and 200 k Pa. The gas is allowed to leak till the pressure falls to 125 kPa. Calculate the amount of the gas leaked assuming that the temperature remains constant. Solution: As the gas leaks out the volume and the temperature of the remaining gas do not change. The PV number of moles of the gas in the vessel in given by  = . RT PV The number of moles in the vessel before the leakage is = 1 and that after the leakage is RT P2 V 2 = RT (200  125) 103  8.0 103 The amount leaked is 1 – 2 = = 0.24 mole 8.3  300 Illustration 46. 1500 ml of a gas at a room temperature of 23°C is inhaled by a person whose body temperature is 37°C, if the pressure and mass stay constant, what will be the volume of the gas in the lungs of the person ? Solution: T1 = 273 + 23 = 296 K; T2 = 273 + 37 = 310 K. Pressure and amount of the gas are kept constant, V1 V2 T 310  So  V2 = V1 × 2 = 1500 × = 1570.95 ml T1 T2 T1 296 Illustration 47. A sample of O2 is at a pressure of 1 atm when the volume is 100 ml and its temperature is 27°C. What will be the temperature of the gas if the pressure becomes 2 atm and volume remains 100 ml. Solution: T1 = 273 + 27 = 300 K P P P 2 For constant volume 1  2  T2 = T1 × 2 = 300 × = 600 K = 600 – 273 = 327°C T1 T2 P1 1 1. BEGINNER’S BOX - 5 A vessel of volume 8.3 × 10–3 m3 contains an ideal gas at temperature 27°C and pressure 200 kPa. The gas is allowed to leak till the pressure falls to 100 kPa and temperature increases to 327°C. What is the amount of gas in moles will be leaked out ? 2. Two closed vessels of equal volume contain air at 105 kPa, 300 K and are connected through a narrow tube. If one of the vessels is now maintained at 300 K and other at 400 K, What will be the pressure in the vessels ? 3. The volume of a gas is 1 litre at the pressure 1.2 × 107 Nm–2 and temperature 400 K. Calculate the number of molecules in the gas. 4. A vessel contains two non-reactive gases : neon (monatomic) and oxygen (diatomic) The ratio of their partial pressures is 3 : 2. Estimate the ratio of (a) number of molecules and (b) mass density of neon and oxygen in the vessel. Atomic mass of Ne = 20.2 u. molecular mass of O2 = 32.0 u. The kinetic theory of gases • Rudolph Claussius (1822–88) and James Clark Maxwell (1831–75) developed the kinetic theory of gases in order to explain gas laws in terms of the motion of the gas molecules. The theory is based on following assumptions regarding the motion of molecules and the nature of the gases. Basic postulates of Kinetic theory of gases. • Every gas consists of extremely small particles known as molecules. The molecules of a given gas are all identical but are different than those of another gas. • The molecules of a gas are identical, spherical, rigid and perfectly elastic point masses. • The size is negligible in comparison to inter molecular distance (10– 9 m) Assumptions regarding motion : • Molecules of a gas keep on moving randomly in all possible direction with all possible velocities. • The speed of gas molecules lie between zero and infinity (very high speed). • The number of molecules moving with most probable speed is maximum. Assumptions regarding collision: • The gas molecules keep colliding among themselves as well as with the walls of containing vessel. These collision are perfectly elastic. (ie., the total energy before collision = total energy after the collision.) Assumptions regarding force: • No attractive or repulsive force acts between gas molecules. • Gravitational attraction among the molecules is ineffective due to extremely small masses and very high speed of molecules. Assumptions regarding pressure: • Molecules constantly collide with the walls of container due to which their momentum changes. This change in momentum is transferred to the walls of the container. Consequently pressure is exerted by gas molecules on the walls of container. Assumptions regarding density: • The density of gas is constant at all points of the container. Properties/ Assumptions of Ideal gas • The molecules of a gas are in a state of continuous random motion. They move with all possible velocities in all possible directions. They obey Newton's law of motion. r • Mean momentum = 0; Mean velocity < v > = 0; < v2 >  0 (Non zero); < v3 > = < v5 > = 0 • The average distance travelled by a molecule between two successive collisions is called as mean free path (m) of the molecule. • The time during which a collision takes place is negligible as compared to time taken by the molecule to cover the mean free path. At NTP ratio of time of collision to time of motion to cover mean free path is 10–8 : 1. • When a gas is taken into a vessel it is uniformly distributed in entire volume of vessel such that its mass density, moleculer density, motion of molecules etc. all are identical for all direction, therefore root mean velocity. vx2  vy2  vz2 are equal. Pressure exerted by the gas in all direction Px = Py = Pz = P  equal • All those assumptions can be justified, if number of gas molecules are taken very large i.e., 1023 molecules/cm3. Different speeds of gas molecules • Average velocity Because molecules are in random motion in all possible direction with all possible velocity. Therefore, the average velocity of the gas molecules in container is zero. r r r v1  v 2  ........v n r < v >= =0 N 3P 3RT 3kT kT    1.73 RMS speed of molecules vrms =  Mw m m Average speed or Mean speed of molecules : r By maxwell’s velocity distribution law vM or <| v |>  vmean r r r 8P 8RT 8kT kT | v |  | v 2 | ........ | v n | r   <| v |>  vmean = 1 = = 1.59 n  M w m m Most probable speed of molecules (vmp) At a given temperature the speed to which maximum number of molecules belongs is called as most probable speed (vmp) 2P 2RT 2kT kT   vmp = = 1.41  Mw m m P RT kT    Mw m At any temperature vrms > vMean > vMP > vsound (always) v rms vrms 3 3 For a gas at any temperature (T) ,   vsound  vMP 2 A temperature is not possible at which above order can be changed vrms  vMean  vMP  vsound (always) Velocity of sound in gas medium (vs) vsound = • • • Maxwell’s law of distribution of velocities Expression for Pressure of An Ideal gas Consider an ideal gas enclosed in a cubical vessel of length . Suppose there are 'N' molecules in a gas which are moving r r r with velocities v1 , v 2 .......v N . r If we consider any single, molecule then its instantaneous. v r can be expressed as v  v x ˆi  v y ˆj  v z kˆ Due to random motion of the molecule r vx = vy = vz ; | v | = v x 3  v y 3  v z 3 = v 2x  v 2y  v 2z Suppose a molecule of mass m is moving with a velocity vx towards the face ABCD, It strikes the face of the cubical vessel and returns back to strike the opposite face. Change in momentum of the molecule per collision p = –mvx – mvx = –2 mvx Momentum transferred to the wall of the vessel per molecule per collision p = 2 mvx The distance travelled by the molecule in going to face ABCD and coming back is 2. 2l So, the time between two successive collision is t = vx v v v x molecule velocity ; fc = rms or fc = m  m m 2l mean free path Hence momentum transferred to the wall per second by the molecule is equal to force applied on the wall v x mv 2x mv 2   Therefore, force F = (2 mvx) [As V = 3 Vx] 2l l 3l F 1 mv 2 1 mv 2 Pressure exerted by gas molecule P = P= [ A ×  = V]  A 3 l A 3 V 1 mv 2 N 1 mN  v 2 1 mv 2 1 mN 2 Pressure exerted by gas P =  =  =   v rms 3 V 3 V N 3 V N 3 V 3RT 3PV 3RT 1M 2 1 v 2rms    vrms = ,P= v rms  v 2rms M M w 3V 3 Mw Number of collision per sec per molecule is fc = • Average number of molecules for each wall = No. of molecules along each axis = ● ● (Nx = Ny = Nz) v 2rms v v v = • Root mean square velocity along any axis for gas molecule is 3 v (vrms)= (vrms)y = (vrms)z = rms 3 All gas laws and gas equation can be obtained by expression of pressure of gas (except Joule’s law) Degree of freedom (f) The number of independent ways in which a molecule or an atom can exhibit motion or have energy is called its degrees of freedom. The number of independent coordinates required to specify the dynamic state of a system is called its degrees of freedom. The degrees of freedom are of three types : (a) Translational Degree of freedom : There are maximum three degree of freedom corresponding to translational motion. (b) Rotational Degree of freedom : The number of degrees of freedom in the case depends on the structure of the molecule. (c) Vibrational Degree of freedom : It is exhibited at high temperature. 2 x ● N 3 N . 6 2 y 2 z Degree of freedom for different gases according to atomicity of gas at low temperature At high temperature a diatomic molecule has 7 degrees of freedom. (3 translational, 2 rotational and 2 vibrational) Maxwell’s law of equipartition of energy The total kinetic energy of a gas molecules is equally distributed among its all degree of 1 freedom and the energy associated with each degree of freedom at absolute temperature T is kT 2 For on molecule of gas 1 Energy related with each degree of freedom = kT 2 v2 f 1 3 Energy related with all degree of freedom = kT Q vx2  vy2  vz2  rms  mv 2rms  kT 2 2 2 3 2 1 v 3 kT 1  kT So energy related with one degree of freedom = m rms  2 3 2 3 2 Different K.E. of gas (Internal Energy) • Translational kinetic energy (ET) ET = 1 3 Mv 2rms  PV 2 2 1 Mv 2rms 2 [Note : Total internal energy of ideal gas is total kinetic energy] Kinetic energy of volume V is • Energy per unit volume or energy density (EV) 1 M 1 2 Total energy E  ; EV =   v 2rms  v rms EV = 2V 2 Volume V 3 1 3  P =  v 2rms   EV = P 2 2 2  • Molar K.E. or Mean Molar K.E. (E) 1 2 for N0 molecules or Mw (gram) M w v rms 2 3 3 E = RT  N0kT 2 2 E= • Molecular kinetic energy or mean molecular K.E. (E) E 3 RT 3 1 2 E   kT E = M w v rms , N0 2 N0 2 2 • Kinetic energy of 1 g mass (Em) = 3 RT 3 N 0  KT 3 KT   2 M w 2 N0m 2 m SPECIFIC HEAT CAPACITY Monoatomic Gases The molecule of a monoatomic gas has only three translational degrees of freedom. Thus, the average energy of a molecule at temperature T is (3/2)kT. Total internal energy of a mole of such a gas is 3 3 U = kT × NA = RT 2 2 The molar specific heat at constant volume, Cv, is dU 3  R Cv (monoatomic gas) = dT 2 Diatomic Gases As explained earlier, a diatomic molecule treated as a rigid rotator like a dumbbell has 5 degrees of freedom : 3 translational and 2 rotational. Using the law of equipartition of energy, the total internal energy of a mole of such a gas is 5 5 U = kT × NA = RT 2 2 The molar specific heats are then given by 5 Cv (rigid diatomic) = R 2 MEAN FREE PATH A molecule in its path undergoes a number of collisions so the path traversed by it is not a straight line but somewhat zigzag. Between two successive collisions a molecule travels in a straight line with uniform velocity. As motion is random thus the distance travelled by molecule between two succesive collisions is not always equal. The average distance travelled by a molecule between two succesive collisions is called as mean free path (m) of the molecule. 1 m = 2d 2 n N Here n = = No. of molecules per unit volume. V d = diameter of a molecule GOLDEN KEY POINTS ● ● ● ● ● At a given temperature graph drawn between molecular velocity and number of molecules is known as velocity distribution curve. The velocities of molecules of a gas are in between zero and infinity (0 – ) With the increase in the temperature, the most probable velocity and maximum molecule velocity both increases. The number of molecules within certain velocity range is constant although the velocity of molecule changes continuously at particular temperature. The area enclosed between the (N – v) curve and the velocity axis represents the total number of molecules. On the basis of velocity distribution Maxwell established the law of equipartition of energy for gases at any temperature. • Except 0 K, at any temperature T, E > Em > E • At a common temperature, for all ideal gas E and E are same while Em is different and depends upon nature of gas (Mw or m) ● For thermal equilibrium of gases, temperature of each gas is same and this temperature called as temperature of mixture (Tm) which can be find out on basis of conservation of energy (All gases are of same atomicity).  NT N1T1  N 2 T2  .......  N nTn  Tm = N N1  N 2 .......N n ● 1 mole gas : Mean translational kinetic energy = ● 3 f RT ; Total kinetic energy = RT 2 2 3 f 1 molecule of gases : Mean translational kinetic energy = kT ; Total kinetic energy = kT; 2 2 ƒ  Degree of freedom Specific heats of all substances approach zero as T  0. This is related to the fact that degrees of freedom get frozen and ineffective at low temperatures . Illustrations Illustration 48. The velocities of ten particles in ms–1 are 0, 2, 3, 4, 4, 4, 5, 5, 6, 9. Calculate (i) average speed and (ii) rms speed (iii) most probable speed. Solution: 0  2  3 4  4  4 5 5 6 9 (i) average speed, vav = 10 42 = = 4.2 ms–1 10 1/2  (0) 2  (2) 2  (3) 2  (4) 2  (4) 2  (5) 2  (5) 2  (6) 2 (9) 2  vrms =   10   (ii) rms speed, (iii)  228  =  = 4.77 ms–1   10  most probable speed vmp = 4 m/s 1/2 Illustration 49. At what temperature root mean square velocity of hydrogen becomes double of its value at S.T.P., pressure remaining constant ? Solution: Let v1 be the r.m.s. velocity at S.T.P. and v2 be the r.m.s. velocity at unknown temperature T2. v12 T1   v 22 T2 2 or v  T2 = T1  2  = 273 × (2)2 = 273 × 4 = 1092 K = (1092 – 273) = 819°C  v1  Illustration 50. Calculate rms velocity of oxygen molecule at 27°C Solution: Temperature, T = 27° C  273 + 27 = 300 K. Molecular weight of oxygen = 32 × 10–3 kg and R = 8.31 J mol–1 K–1 3RT 3  8.31  300  rms velocity is vrms = = 483.5 ms–1 3 M 32 10 Illustration 51. Calculate the kinetic energy of a gram molecule of argon at 127°C. Solution: Temperature, T = 127°C = 273 + 127 = 400 K, R = 8.31 J/mol K 3 3 K.E. per gram molecule of argon = R T = × 8.31 × 400 = 4986 J 2 2 Illustration 52. The mass of a hydrogen molecule is 3.32 × 10–27 kg. If 1023 molecules are colliding per second on a stationary wall of area 2 cm2 at an angle of 45° to the normal to the wall and reflected elastically with a speed 103 m/s. Find the pressure exerted on the wall. (in N/m2) Solution: As the impact is elastic r r  | p1 | | p 2 | = p = mv = 3.32 × 10–24 kg m/s r r The change in momentum along the normal p = | p1  p 2 | = 2p cos45° = If f is the collision frequency then force applied on the wall F =  1. F  Pressure P = A 2pf  A 2p p = p × f = t 2pf 2  2.32 10 24 10 23 = 2.347 × 103 N/m2 4 2  10 BEGINNER’S BOX - 6 Given : Avogadro number N = 6.02 × 1023 molecules/mole and Boltzmann’s constant k = 1.38 × 10–23 J/(molecule-K). Calculate.: (a) (b) (c) The average kinetic energy of translation of an oxygen molecule at 27°C The total kinetic energy of an oxygen molecule at 27°C The total kinetic energy of 1 mole of oxygen gas at 27°C. 2. The temperature of gas is –73°C To what temperature should it be heated so that (a) average kinetic energy of 1 he molecule is doubled ? (b) the root mean square velocity of the molecules is doubled? 3. For a gas (a) (b) (c) R = 0.4. For this gas calculate the following CP atomicity and degree of freedom value of Cv and  mean gram – molecular kinetic energy at 300 K temperature 4. If three molecules are having speeds v1, v2 and v3 respectively, then what will be their average speed and root mean square speed ? 5. Four molecules of a gas are having speeds of 1, 4, 8 and 16 ms–1. Find the root mean square velocity of the gas molecules ? 6. A flask contains argon and chlorine in the ratio of 2 : 1 by mass. The temperature of the mixture is 27°C Obtain the ratio of (a) Average translational kinetic energy per molecule, and (b) root mean square speed (vrms) of the molecules of the two gases. Atomic mass of argon = 39.9 u : Molecular mass of chlorine= 70.9 u.