Kinetic Theory of Gases Notes

THERMAL PROPERTIES OF MATTER 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) Internal energy, internal kinetic energy, internal potential energy At a given temperature for solid, liquid and gas: (i) Internal kinetic energy (ii) Internal potential Energy : Maximum for ideal gas (PE = 0) and Minimum for solids (PE = –ve) (iii) Internal Energy : Maximum for Ideal gas and Minimum for solid Same for all At a given temperature for rared and compressed gas : (i) Internal kinetic energy (ii) Internal potential energy  (PE)Rared > (PE)compressed (iii) Internal Energy (U)Rared > (U)compressed   N.T.P. Same (Normal temperature) Temperature 0° C = 273.15 K Pressure 1 atm = 1.01325 × 105 N/m2 S.T.P. (Standard Temperature and Pressure) 0.01° C = 273.16K 1 atm = 1.01325 × 10 pascal 5 Volume 22.4 litre 22.4 litre IDEAL GAS CONCEPT (i) A gas which follows all gas laws and gas equation at every possible temperature and pressure is known as ideal or perfect gas. (ii) Volume of gas molecules is negligible as compared to volume of container so volume of gas = volume of container (Except 0K) (iii) No intermolecular force act between gas molecules. (iv) 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 For a substance U = UKE + UPE (v) UKE : depends only on T, UPE : depends upon intermolecular forces (Always negative) Specific heat of ideal gas is constant quantity and it does not change with temperature (vi) CV ideal gas CV real gas All real gases behaves as ideal gas at high temperature and low pressure. temperature temperature (vii) Volume expansion coefficient () and pressure expansion coefficient () is same for a ideal gas and value of each is 1 per °C 273 (viii)  = = 1 per °C 273 Gas molecule 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 PV = µRT where  = number of moles of gas  PV   mN  R  M RT =   RT =   N T = NkT Mw  m N0   N0  Ex. 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 ? Sol.  At constant volume T  P  Ex. Calculate the value of universal gas constant at STP. Sol. Universal gas constant is given by R  PV T 5  1 00 T P  10 0   1 0 0  0.5  T   10 00K 0.5 T P 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 m–2 R = T = 273 K 7 6 0  10 3  13 .6  1 0 3  9 .8 0  22 .4  1 0 3 = 8·31 J mol–1 K–1 273 Ex. 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 Nm2. 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 masses of the individual gases in the container, assuming them to be ideal. Given : R = 8.314 J/mol/K. Sol. Let m gram be the mass of neon. Then, the mass of argon is (28 – m)g. Total number of moles of the mixture, Now, By (i) and (ii),  m 28  m 28  m   20 40 40 ...(i)  PV 1  1 0 5  0 .0 2   0 .8 RT 8 .3 1 4  3 0 0 ...(ii) 28  m  0 .8  28 + m = 32 m = 4 gram or mass of argon = (28 – 4)g = 24 g 40 Ex. 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] Sol. PV = RT T  PV R From equation (i) T  ...(i) But  M M and   V Mw   V Mw P V Mw P Mw 1.4  10 9  1 .01  10 5  2  10 3  = = 2.4 × 107 K 1.4  1000  8.4 VR R THERMAL PROPERTIES OF MATTER Ex. 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. Sol. By gas equation PV   P R  M M M  and   RT  Q    T M M V Mw w  w   P T 7 0 30 0 75 P  P    Now as MW and R are same for top and bottom      So T  T  B  = 0.9868 7 6 28 0 7 6 B PB TT  T  T  T  B Ex. During an experiment an ideal gas is found to obey an additional law VP2 = constant. The gas is initially at temperature T and volume V. What will be the temperature of the gas when it expands to a volume 2V. Sol. By gas equation PV = RT and VP2 = constant on eliminating P  A    V  RT   V V R T A  V1 T   1  V2  T2  V 2V  T  T' = T'  2 T GAS LAWS 1. Boyle's Law According to it for a given mass of an ideal gas at constant temperature, the volume of a gas is inversely proportional to its pressure, i.e., V  1 if m and T = Constant P m=constant T=constant PV P PV V V P Ex. A sample of oxygen with 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 ? Sol. Temperature is constant, so P1 V1 = P2V2 Ex. An air bubble doubles in radius on rising from bottom of a lake to its surface. If the atmosphere pressuer is equal to that due to a column of 10 m of water, then what will be the depth of the lake.  P2  P1 V1  500  2  = 2.5 atm V2  400  (Assuming that surface tension is negligible) ? Sol. Given that constant temperature, we use P1V1 = P2V2 P2 = (10) dg (for water column) V1  P1 = (10+h) dg (where h=depth of lake) 4 3  4 3  4 3 2r  8  r = 8V Thus for P V = P V ,  r , V2  1 2 2 1 1  3  3 3 We have 10 dg (8V1) = (10 + h) dg V1  80 = 10 + h  h = 70 m Ex. 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. Sol. As the gas leaks out, the volume and the temperature of the remaining gas do not change. The number of moles of the gas in the vessel in given by n = PV . RT The number of moles in the vessel before the leakage is n 1  The amount leaked is n1 – n2 = 2. P1 V PV and that after the leakage is n 2  2 . RT RT (P1  P2 )V (2 00  1 2 5 )  1 0 3  8 .0  1 0 3  = 0.24 mole RT 8 .3  3 0 0 Charle's 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  T if m and P = Constant V T(k) V T V T T V Ex. 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 ? Sol. T1 = 273 + 37 = 310 K; T2 = 273 + 23 = 296 K. Pressure and amount of the gas are kept constant, So 3. V1 V2  T1 T2  V2  V1  293 T2 = 1500  = 1417.74 ml T1 310 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 if m and V = constant P T(k) P T P T P T Ex. 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. Sol. T1 = 273 + 27 = 300 K For constant volume 2 P P1 P2   T2  T1  2 = 3 0 0  = 600 K = 600 – 273 = 327°C 1 P1 T1 T2 THERMAL PROPERTIES OF MATTER 4. Avogadro's Law According to it, at same temperature and pressure of equal volumes of all gases contain equal number of molecules, i.e., N1 = N 2 if P,V and T are same. Dalton's Partial Pressure Mixture Law : According to it, the pressure exerted by a gasesous mixture is equal to the sum of partial pressure of each component gases present in the mixture, ie., P = P1 + P2 + .... Ex. Sol. The mass percentage in composition of dry air at sea level contains approximately 75.5% of N 2. If the total atmospheric pressure is 1 atm then what will be the partial pressure of nitrogen ? Mnitrogen   29  0 .7 5 5  The mole fraction of nitrogen 1   = 0.78  28  Molecular weight  The partial pressure of nitrogen P1  1 RT 1 RT  1     P = (0.78) × 1 = 0.78 atm    V  V 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 as regards to the motion of molecules and the nature of the gases. Basic Postulates of Kinetic Theory of Gases (a) Every gas consists of extremely small particles known as molecules. The molecules of a given gas are all identical but are different than those another gas. (b) The molecules of a gas are identical, spherical, rigid and perfectly elastic point masses. (c) The size is negligible in comparision to inter molecular distance (10–9 m) Assumptions regarding motion : (a) Molecules of a gas keep on moving randomly in all possible direction with all possible velocities. (b) The speed of gas molecules lie between zero and infinity (very high speed). (c) 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 collisions.) Assumptions regarding force: (a) No attractive or repulsive force acts between gas molecules. (b) 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 (i) 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.  (ii) Mean momentum = 0; Mean velocity = 0.< v > = 0; < v2 >  0 (Non zero); < v3 > = < v5 > = 0 (iii) The average distance travelled by a molecule between two successive collisions is called as mean free path (m) of the molecule. (iv) The time during which a collision takes place is negligible as compared to time taken by the molecule to cover the mean free path so NTP ratio of time of collision to free time of motion 10–8 : 1. (v) When a gas taken into a vessel it is uniformly distributed in entire volume of vessel such that its density, moleculer density, motion of molecules etc. all are identical for all direction, therefore root mean velocity v2x  v2y  vz2 equal Pressure exerted by the gas in all direction Px = Py = Pz = P equal (vi) All those assumptions can be justified, if number of gas molecules are taken very large i.e., 1023 molecules/cm3. 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 r r r moving with velocities v1 , v2 ........vN . r r If we consider any single molecule than its instantaneous velocity v can be expressed as v  v xˆi  v yˆj  v z kˆ Due to random motion of the molecule vx = vy = vz v  vx 3  vy 3  vz 3 = v2x  v2y  v2z 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. So, the time between two successive collision is t  Number of collision per sec per molecule is fc  2l vx vx v molecule velocity = , fc  rms or fc  v m 2l m mean free path m Hence momentum transferred in the wall per second by the molecule is = force on the wall force F = (2 mvx) v x mv2x mv2  = 2l l 3l THERMAL PROPERTIES OF MATTER Pressure exerted by gas molecule P  Pressure exerted by gas P = v2rms  1 mv2 F 1 mv2 =  P= 3 V A 3 l A 1 mv2 = V 3 v 1 mv2 N 1 mN  = V N 3 V 3 3 PV 3 RT  v  M µMw  rms (i) Average number of molecules for each wall = (ii) v 2x  v 2y  vz2  Q A  l  V 2 N = 1 mN 2 vrms 3 V 3 RT 1M 2 1 vrms  v2rms ,P= Mw 3 V 3 N N . No. of molecules along each axis = (Nx = Ny = Nz) 6 3 2 v rms vrms Root mean square velocity along any axis for gas molecule is (vrms)= (vrms)y= (vrms)z= 3 3 All gas laws and gas equation can be obtained by expression of pressure of gas (except Joule’s law) Ex. 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 10 3 m/s. Find the pressure exerted on the wall will be (in N/m2) Sol. As the impact is elastic r r  p 1  p 2  p  mv = 3.32 × 10–24 kg m/s A P1 45° 45° P2 r r The change in momentum along the normal p  p 2  p 1  2 p cos 45   2p If f is the collision frequency then force applied on the wall F   Pressure P  F 2pf  = A A p  p  f  2 pf t O B 2  3.32  10 24  10 23  2.347  10 3 N / m 2 2  10 4 DEGREE OF FREEDOM (f) (i) The number of independent ways in which a molecule or an atom can exhibit motion or have energy is called it's degrees of freedom. (ii) The number of independent coordinates required to specify the dynamical state of a system is called it's degrees of freedom. f=1 For example (a) Block has one degree of freedom, because it is confined to move in a straight line and has only one transistional degree of freedom. f=1 y (b) The projectile has two degrees of freedom becomes it is confined to move f=2 in a plane and so it has two translational degrees of freedom. x (c) The sphere has two degrees of freedom one rotational and another translational. Similarly a particle free to move in space will have three translational degrees of freedom. Note : In pure rolling sphere has one degree of freedom as KE = 1 7 K2 mv2 (1+ 2 ) = mv2 2 10 R f=2 THE DEGREES OF FREEDOM ARE OF THREE TYPES : (a) Translational Degree of freedom : Maximum three degree of freedom are there corresponding to translational motion. (b) Rotational Degree of freedom : The number of degrees of freedom in this case depends on the structure of the molecule. (c) Vibrational Degree of freedom : It is exhibited at high temperatures. Degree of freedom for different gases according to atomicity of gas at low temperature Ato micity of g as Translational R otational To tal 3 0 3 Monoatomic Ex. Ar, Ne, Id eal gas etc y z Diatomic 3 2 5 Triato mic (linear) Ex. CO2, C2H 2 3 2 5 Triatomic (Non–linear) o r Polyatomic Ex. H2O, NH3, CH4 3 3 6 Ex. O 2, Cl2, N2 etc. x O=C=O At high temperatures a diatomic molecule has 7 degrees of freedom. (3 translational, 2 rotational and 2 vibrational) Ex. Sol. Calculate the total number of degrees of freedom possessed by the molecules in one cm3 of H2 gas at NTP. 22400 cm3 of every gas constains 6.02 × 1023 molecules. 6 .02  10 23 = 0.26875 × 1020 2 24 00  Number of molecules in 1 cm3 of H2 gas =  Number of degrees of freedom of a H2 gas molecule = 5 Total number of degrees of freedom of 0.26875 × 1020 × 5 = 1.34375 × 1020. MAXWELL'S LAW OF EQUIPARTITION OF ENERGY The total kinetic energy of a gas molecules is equally distributed among its all degree of freedom and the energy associated with each degree of freedom at absolute temperature T is 1 kT 2 For one molecule of gas Energy related with each degree of freedom = Energy related with all degree of freedom = 1 kT 2 f 1 3 v2 kT v 2x  v 2y  vz2  rms  mv2rms  kT 2 2 2 3 So energy related with one degree of freedom = 1 v2rms 3 kT 1 m   kT 2 3 2 3 2 THERMAL PROPERTIES OF MATTER Ex. A cubical box of side 1 meter contains helium gas (atomic weight 4) at a pressure of 100 N/m2. During an observation time of 1 second, an atom travelling with the root–mean–square speed parallel to one of the edges of the cube, was found to make 500 hits with a particular wall, without any collision with other atoms. Take R = k = 1.38 × 10–23 J/K. (a) Evaluate the temperature of the gas. (b) Evaluate the average kinetic energy per atom. (c) Evaluate the total mass of helium gas in the box. Sol. 25 J/mol–K and 3 Volume of the box = 1m3, Pressure of the gas = 100 N/m2. Let T be the temperature of the gas (a) Time between two consecutive collisions with one wall = This time should be equal to 2 l v rms 2I , where l is the side of the cube. vrms  1 0 00 2 M 10 6  3  1 0 3  3 RT   1 6 0K = 1000  T  M 3R  25  3   3  1   vrms = 1000 m/s  50 0 (b) Average kinetic energy per atom = (c) From PV = nRT = 1 sec 500 3 3 kT = [(1.38 × 10–23) = 160] J = 3.312 ×–21 J 2 2 m PVM RT, Mass of helium gas in the box m = M RT 100  1   4  10 3  Substituting the values, m =  25     1 60  3 = 3.0 × 10–4 kg DIFFERENT K.E. OF GAS (INTERNAL ENERGY) (i) Translatory kinetic energy (ET) ET = Kinetic energy of volume V is = (ii) Total energy E  ; Volume V EV  1 2 1 2 2 1 2  M 2  V  vrms  2  vrms P  3  2 v rms      Molar K.E. or Mean Molar K.E. (E) E (iv) 1 Mv2rms Note : Total internal energy of ideal gas is kinetic 2 Energy per unit volume or energy density (EV) EV  (iii) 1 3 Mv2rms = PV 2 2 1 3 3 Mw v2rms for N0 molecules or Mw (gram) E  RT  N 0 kT 2 2 2 Molecular kinetic energy or mean molecular K.E. ( E ) E E 3 RT 3 1   kT Mw v2rms , E  2 N0 2 N0 2  E V  3 P 2 ETOOS KEY POINTS (i) Except 0 K, at any temperature T , E > Em > E (ii) 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) (iii) For thermal equilibrium of gases, temperature of each gas is same and this temperature called as temperature of mixture (T m ) which can be find out on basis of conservation of energy (All gases are of same atomicity). Tm   NT  N T N 1 1  N 2 T2  .........  N n Tn N1  N 2 ......N n (iv) 1 mole gas : Mean kinetic energy = (v) 3 RT 2 1 molecule of gases : Mean kinetic energy = Total kinetic energy = ; f RT 2 f 3 kT ; Total kinetic energy = kT 2 2  Degree of freedom Ex. Two ideal gases at temperature T1 and T2 are mixed. There is no loss of energy. If the masses of molecules of the two gases are m1 and m2 and number of their molecules are n1 and n2 respectively. Find the temperature of the mixture. Sol. Total energy of molecules of first gas = 3 3 n 1 kT1 , Total energy of molecules of second gas = n 2 kT2 2 2 Let temperature of mixture be T then total energy of molecules of mixture =  3 k (n 1  n 2 )T 2 n T  n 2 T2 3 3 (n 1  n 2 )kT  k(n 1 T1  n 2 T2 )  T  1 1 (n 1  n 2 ) 2 2 Ex. The first excited state of hydrogen atom is 10.2 eV above its ground state. What temperature is needed to excite hydrogen atoms to first excited level. Sol. K.E. of the hydrogen atom 3 kT = 10.2 eV = 1 0 .2  1 .6  10 1 9 J 2   19  T  2  1 0 .2  1.6  1203  7 .8 8  1 0 4 K 3 1 .3 8  1 0 EXPLANATION OF GAS LAWS FROM KINETIC THEORY 1. Boyle's Law According to this law, the product of the pressure and the volume of a given mass of gas at constant temperature is constant. From the kinetic theory of gases, the pressure of a given mass of an ideal gas is given by P  1 mN 2 vrms . mN is the mass of the gas 3 V P T=constant V THERMAL PROPERTIES OF MATTER which is constant. If the temperature remains constant, the mean– PV square–velocity of the molecules. ( v2rms ) also remains constant. T=constant Thus, from the above equation, we have PV = constant. This is Boyle's law. V  2. 1 (T = constant) or P P or V PiVi = PfVf Charle's Law V According to this law, the volume of a given mass of gas at constant pressure is directly proportional to the absolute temperature of the gas. From kinetic theory, we have V V 2 N3  N kT  kT 3 P  2  P  Q  P=constant 2 N 1  mv2rms   3 P  2 T(in K) 1 3  mv2rms  kT  2 2  V T P=constant If the pressure P is constant, then for a given mass of the gas, we have V  T This is Charle's Law. V or T 3. Gay Lussac's law of Pressure law P For a given mass of a gas the pressure of a gas at constant volume (called Isochoric process) is directly proportional to its absolute temperature, P  or 1 mN 2 vrms   v2rms T  3 V P = constant T or P  T (V = constant) T(in K) Pi Pf  Ti Vf P–T graph in an isochoric process is a straight line passing through origin or P T P v/s P or T graph is a straight line parallel to P or T axis. T P or T 4. Avogadro's Law Equal volumes of 'all gases' under the same conditions of temperature and pressure contain equal number of molecules. At same pressure equal volumes V of different gases contain respectively N1 and N2 molecules of masses m1 and m2. PV  1 m 1 N 1 v 2rms1 3 and PV  1 m 2 N 2 v 2rms2 3  m 1 N 1 v2rms1  m 2 N 2 v2rms2 Now, if the gases are at the same temperature, their average kinetic energies of translation per molecule are 1 1 2 2 equal. That is m 1 vrms1  m 2 vrms2  N1 = N2. 2 2 5. Dalton's Law of Partial Pressures The total pressure exerted by a mixture of non–reacting gases occupying a vessel is equal to the sum of the individual pressures which each gas would exert if it alone occupied the whole vessel. Let as consider a mixture of gases occupying a volume V. Suppose the first gas contains N1 molecules, each of mass m1 having mean–square–speed v 2rms1 , the second gas contains N2 molecules each of mass m2 and mean–square–speed v 2rms2 , and so on. Let P1, P2,........ respectively the partial pressures of the gases. Each gas fills the whole volume V. According to kinetic theory, we have P1 V  Additing, we get 1 m 1 N 1 v 2rm s1 , 3 P2 V  (P1  P2  .......)V  1 m 2 N 2 v2rms2 , and so on. 3 1 (m 1 N1 v2rms1  m 2 N 2 v2rms2  ......) 3 Now, the whole mixture is at the same temperature.  ......(i) 1 1 1 m 1 v2rms1  m 2 v2rms2 ......  mv2rms 2 2 2 Substituting this result in eqn. (i) we have (P1  P2  ......)V  1 (N1  N 2  ......)mv2rms 3 The mixture has a total number of molecules (N1 + N2 + .......). Hence the pressure P exerted by the mixture is given by PV  That 1 (N1  N 2  ......) mv2rms 3 P = P1 + P2 + ........ This is Dalton's law of partial pressures. DIFFERENT SPEEDS OF GAS MOLECULES Average Velocity Because molecules are in random motion in all possible direction in all possible velocity. Therefore, the average    v1  v2 ............ v N  velocity of the gas in molecules in container is zero. < v > = =0 N RMS Speed of Molecules v = rms 3P =  3RT = Mw 3kT kT = 1.73 m m Mean Speed of Molecules  By maxwell’s velocity distribution law vM or < | v| > = v mean r r r | v | | v2 | ..........| v n | r 8P < v> =v = 1 = = mean N  8 RT = M w 8 kT kT = 1.59 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) vmp = 2P =  2RT Mw = 2kT = 1.41 m kT m THERMAL PROPERTIES OF MATTER MAXWELL'S LAW OF DISTRIBUTION OF VELOCITIES ETOOS KEY POINTS (i) At any given temperature graph drawn in between molecular velocity and number of molecules is known as velocity distribution curve.  (ii) The velocities of molecules of a gas are in between zero and infinity (0 –  (iii) With the increase in the temperature, the most probable velocity and maximum molecule velocity both increases. (iv) The number of molecules within certain velocity range is constant although the velocity of molecule changes continuously at particular temperature. (v) The area enclosed between the (N – v) curve and the velocity axis presents the total number of molecules.    ) On the basis of velocity distribution Maxwell established gives the law of equipartition of energy for gases of an temperature. Velocity of Sound in Gas Medium (V S) (i) At any temperature vsound = P =  RT = Mw vrms > vMean > vMP > vsound vrms (ii) For a gas at any temperature (T) v = sound v rms 3 , v =  MP kT m (always) 3 2 (iii) A temperature is not possible at which above order can be changed vrms  vMean  vMP  vsound Ex. The velocities of ten particles in ms–1 are 0, 2, 3, 4, 4, 4, 5, 5, 6, 9. Calculate (i) average speed and Sol. (always ) (i) (ii) rms speed average speed, vav= (iii) most probable speed. 0 2 3  4  4  4 5 5 6 9 42 = = 4·2 ms–1 10 10  (0 )2  (2 )2  (3 )2  (4 )2  (4 )2  (4 )2  (5 )2  (5 )2  (6 )2  (9 )2  =  10   (ii) rms speed, vrms (iii) most probable speed vmp = 4 m/s 12  228  =    10  1/ 2 = 4.77 ms–1 184 Ex. At what temperature, will the root mean square velocity of hydrogen be double of its value at S.T.P., pressure remaining constant ? Sol. Let v1 be the r.m.s. velocity at S.T.P. and v2 be the r.m.s. velocity at unknown temperature T2.  Ex. Sol. v12 v22  T1 T2 2 or  v2  T2 = T1   = 273 × (2)2 = 273 × 4 = 1092 K = (1092 – 273) = 819°C  v1  Calculate rms velocity of oxygen molecule at 27°C Temperature, T = 27° C  273 + 27 = 300 K, Molecular weight of oxygen = 32 × 10–3 kg rms velocity is vrms = 3RT = M and 3  8  31  3 00 32  1 0 3 R = 8·31 J mol–1 K–1 = 483·5 ms–1 Ex. Calculate the kinetic energy of a gram molecule of argon at 127°C. Sol. Temperature, T = 127°C = 273 + 127 = 400 K, R = 8.31 J/mol K K.E. per gram molecule of argon = 3 3 RT= × 8.31 × 400 = 4986 J 2 2