# HEAT UNIT 1.1 KINETIC THEORY OF GASES Introduction Postulates of Kinetic Theory of Gases

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1 UNIT HEAT. KINETIC THEORY OF GASES.. Introduction Molecules have a diameter of the order of Å and the distance between them in a gas is 0 Å while the interaction distance in solids is very small. R. Clausius and J. C. Maxwell developed the mathematical concept of the random motion of the gas molecules in a gas which helped to explain not only the properties of gases but also that of free electrons in metals... ostulates of Kinetic Theory of Gases. A gas consists of a large number of minute particles known as molecules of the same size and mass.. The interaction between molecules is negligible unlike solids.. The molecules in a gas are in a state of random motion in all directions with different velocities ranging from zero to infinity. 4. The size of the molecules is negligible in comparison with the volume of a gas. 5. During the motion, molecules collide with each other which is perfectly elastic. 6. The impact of molecules on the walls of the container exerts a pressure. ressure of a gas is the average force per unit area of the molecules on the walls of the container during impact. Larger the number of impacts, greater will be the pressure. 7. Due to random motion of molecules in gas, the molecules have kinetic energy. The average kinetic energy of a gas molecule varies as absolute temperature. According to kinetic theory of gases, the molecules of a gas are in random motion. So, the average velocity of the molecules is zero. Hence we find the rms velocity of a molecule.

2 Engineering hysics ol. II If c, c, c,... c n are the velocities of the molecules of a gas at any instant, the mean square velocity of the molecules of the gas is, velocity of a molecule is c + c + c +... c. Hence root mean square n c rm... n c + c + c + c n.. ressure Exerted by a Gas Let us enclose a gas in a cubical box of dimension metre. Let m be the mass of the molecule and n be the total number of molecules in this cubical vessel of volume m. Consider a molecule D moving in the x-direction with a velocity v towards the wall B. It hits the wall B with a velocity v and rebounds with the same velocity v as the collision is perfectly elastic. Momentum of the molecule D before collision is mv and the momentum of the molecule after collision is equal to mv. Thus change in momentum of the molecule on the wall B due to a single collision of the molecule D is mv ( mv) mv. In metres the molecule undergo collision. For v metre it will undergo v collisions (in a sec). Thus force exerted by the single molecule on the wall B change in momentum v time mv mv. A B D Fig.. We shall now consider the combined effect of collisions of all molecules on the wall B. All the n molecules in the cubical vessel are not moving along the x-axis between the walls B and A. Since there are only three independent directions x, y and z, it is reasonable

3 Heat n to assume that, at any instant, there may be molecules moving along the x-axis between walls B and A. Therefore, force exerted by the molecules on the walls B mnv. Being the area of the surface B is sq. m., the pressure exerted on the wall B is mnv But all molecules are not moving with the same velocity. Hence it is reasonable to substitute c, the mean square velocity of the molecule, for v. If c, c, c,... c n are the velocities of the molecules at any instant, then Therefore,... n c c + c + c + c mnc Since m is the mass of the molecule and n is the number of molecules per unit volume of the gas, then mn ρ i.e., ρc Further discussion is possible by choosing unit volume or molar volume ( m ). Let us consider one mole of a gas which occupies a volume m. The pressure exerted by the gas is mnac m mn Ac RT, where R is universal gas constant. Thus mc m RT N R as Boltzmann constant k B. N A A B k T

4 4 Engineering hysics ol. II or mc k B T...() This is kinetic energy of a molecule. Hence, average kinetic energy of one mole of the gas is mc NA NAkBT RT...() It is now evident that K.E. of a molecule of the gas is k B T or K.E. µ T. Conclusion: Let us consider any amount of a gas. Let,, T and n be the pressure, volume, temperature and number of the molecules of the gas. K.E. of a molecule of a gas k B T Hence, K.E. of the gas nk BT. If M is the mass of a given gas with as the volume, then Mc Mc K. E. of the gas...() nk BT or nk BT...(4)..4 Boyle s Law and Charles Law (i) Kinetic energy of a gas is directly proportional to the absolute temperature. Hence from equations () and (), it is obvious that at constant temperature constant or µ. This is Boyle s law. (ii) Since K.E. of the gas is proportional to the absolute temperature, from equations () and (4), µ T. Conclusions (a) At constant pressure [from Eqns. and 4] µ T. This is Charles first law (b) At constant volume µ T. This is Charles second law.

5 Heat 5. SECIFIC HEAT.. Measurement of Heat and Unit of Heat Heat is a form of energy. Usually quantities of heat is measured by the effect they produce. Since heat is a form of energy it is measured in joule, in the same way as other forms of energy. Water was taken as the standard substance for defining heat units. Calorie is the unit of heat in C.G.S. unit. It is defined as the quantity of heat required to raise the temperature of one gram of water through C from 4.5 C to 5.5 C. Calorie 4. joule. Heat capacity of a body: It is the quantity of heat required to raise the temperature of the body by one degree kelvin, i.e., the unit of heat capacity of a body is joule/kelvin J/K. Specific heat capacity of a substance (C): It is the quantity of heat required to raise the temperature of the body by one degree kelvin. The unit is joule per kg per degree kelvin (J kg K ). The dimension is L T K. Specific heat of some substances Substance J kg K kj kg K J gm K Water Aluminium Copper Water equivalent: Water equivalent of a body is the mass of water having the same heat capacity as the given body. In S.I. unit, water equivalent (in kg) Heat capacity of the body. Specific heat of water Quantity of heat: If m is the mass of the body, C is the specific heat, then quantity of heat Q required to raise the temperature through θ is m C θ i.e., Q m C θ Method of mixtures: When a hot body is allowed to share its heat with a cold body, there is a flow of heat from the hot body to the cold body until both attain a common temperature. Then, if no heat is lost to or gained from the surroundings, then Heat lost by the hot body Heat gained by the cold body.

6 6 Engineering hysics ol. II.. Specific Heat of a Gas at Constant olume and at Constant ressure (i) C : The specific heat of a gas at constant volume is the amount of heat required to raise the temperature of unit mass of the gas through one degree kelvin keeping its volume constant. Thus the amount of heat required to raise the temperature of one mole of the gas through one degree kelvin at constant volume is called molar specific heat at constant volume (C ). (ii) C : The specific heat of a gas at constant pressure is the amount of heat required to raise the temperature of unit mass of the gas through one degree kelvin keeping its pressure constant. As discussed above, molar specific at constant pressure can be stated. The unit of specific heat is J kg K. Why C > C?: When a gas is heated at constant volume, the heat supplied is utilized only to increase the internal energy of the gas. But when it is heated at constant pressure, the heat supplied is used not only to increase the internal energy but also doing work during expansion. For the same temperature, the increase in internal energy is the same in both cases. Hence C > C... Mayer s Relations (C C R) Consider one mole of an ideal gas enclosed inside a non-conducting cylinder with a light smooth piston. d T Fig..

7 Heat 7 Let be its pressure, its volume and T its temperature. Keeping the volume constant, let the gas be heated so as to raise its temperature by one degree kelvin. The quantity of heat supplied, Q C This amount of heat is used only to raise, the internal energy of the gas. Now, let us imagine the gas be heated at constant pressure so as to raise the temperature by K. The quantity of heat supplied is given by Q C This amount of heat is partly used to raise the internal energy of the gas as the temperature raises by K and also used to do external work as the gas expands to keep the pressure constant. Therefore, C C + external work...() When the gas expands keeping pressure cosntant, let the piston be moved through a distance dx. Hence external work done by the gas, W F S A dx d where A is the area of cross-section of the piston and A dx d is the increase in volume of the gas. Thus, equation () can be written as C C + d...() For an ideal gas, RT After expansion, ( + d) R (T + ) i.e., + d RT + R RT d R i.e., C C + R or C C R This is the well known Mayer s relation. An Example: Let us consider air at constant pressure with C 966 J kg K. Density of air at ST.9 kg/m. Get the value of R using RT ressure (ST) N/m, T 7 K, ρ.9 kg/m. Gas constant for kg, M R as ρ T ρt.0 0 R.9 7 5

8 8 Engineering hysics ol. II R 87 J kg K...4 Isothermal and Adiabatic Changes Isothermal rocess: The process in which temperature is kept constant either by adding heat or taking heat away from a system is called isothermal process. Slow expansion of a gas enclosed in a perfectly conducting cylinder fitted with a perfectly conducting position. At constant temperature, suppose a certain mass of a perfect gas undergoes an isothermal expansion by d; then the total external work done is dw d. The total work done in an isothermal expansion from volume to under the same conditions is W ( ). If the pressure changes during the expansion, then the total work done is RT W d ;but RT or RT Thus, W d RT d. Hence in an isothermal change R and T are constants. Thus W RT log. e i.e., W RT.0log But Hence, W can also be written as or W RT.0 log. Adiabatic transformation: The process that takes place in a system under thermal isolation, such that heat is not transferred from the system to outside or from outside to a system is called an adiabatic process. It means that there is no exchange of heat between the system and surroundings. The expansion or compression of a gas enclosed in a perfectly non-conducting cylinder filled with a perfectly non-conducting frictionless piston. Since the system is thermally insulated no heat can enter or leave the system i.e., dq 0.

9 Heat 9 Hence du + d 0. Since perfect isolation is not practically possible, the physical changes should take place very rapidly so that heat produced during compression does not have enough time to leave the system and similarly during expansion heat does not have enough time to enter the system. When a certain mass of a perfect gas undergoes adiabatic changes, the total work done when the volume changes from and is W d For adiabatic change γ constant or k γ Thus, γ ( ) W k d k W k d γ γ γ γ γ γ γ Again, constant k Thus, W γ γ γ γ γ W γ γ [ ] ( ) If the temperature of the gas changes from T to T during the operation, then RT and RT R W T T γ ( ) Conclusion: The simple relation of isothermal change is k Differentiating this equation we get, d + d 0 d d d d

10 0 Engineering hysics ol. II Thus the slope of the isothermal equation is d d The adiabatic relation is γ constant. Differentiating this equation, γ γ d+ γ d 0 If, we get d + γ d 0 or d d γ Hence, the slope of adiabatic γ That is under same conditions of pressure and volume, the slope of adiabatic curve is γ-times that of isothermal formation. Mark Questions SHORT QUESTIONS. Give the dependence of root mean square velocity on absolute temperature.. Write down the relation connecting kinetic energy of a molecule, Boltzmann constant and absolute temperature.. State Boyle s Law..4 Give the unit of specific heat..5 What is Mayer s relation?.6 Dependence of K.E. of a molecule with temperature. Mark Questions.* State two important postulates of kinetic theory of gases.. Find the relation between and the kinetic energy of the gas.. The quantities required to measure the amount of heat in an experiment. Explain..4* Explain Charles Law..5 Discuss the principle of method of mixtures..6* Two kilogram of water is heated to raise the temperature from 0 C to 40 C. Compute the heat energy supplied in joule..7 Explain why C > C.

11 Heat ANSWERS TO STARRED QUESTIONS.* (i) The molecules in a gas are in a state of random motion in all directions with different velocities ranging from zero to infinity. (ii) Due to random motion of the molecules in a gas, the molecules have kinetic energy. The average kinetic energy of a gas molecule varies as absolute temperature..4* Since K.E. of the gas is proportional to the absolute temperature, µ T i.e., At constant pressure, µ T. This is Charles first law. Similarly at constant volume µ T. This is Charles second law..6* The general formula for heat lost or heat gained is ms θ. In this case m kg, s 400 joule kg K and θ (40 0) C. Thus, Q m s θ J Q joule. 0 Marks Questions REIEW QUESTIONS. (a) Bring out the important postulates of kinetic theory of gases. (b) Obtain the expression for the pressure exerted by a gas in a cubical box in terms of the density of the gas and root mean square velocity. (c) At ST, calculate the value of R using the equation RT.. (a) Discuss the pressure exerted by a gas and obtain an expression for the pressure using kinetic theory of gases. (b) Derive the expression for kinetic energy of a gas. (c) Deduce Boyle s law and Charles laws.. (a) Explain specific heat of a gas at constant volume and constant pressure (b) Obtain Mayer s relation. (c) Explain why C p > C?.4 (a) Explain isothermal expansion and obtain the expression for the work done during this process. (b) Show that for an adiabatic change in a perfect gas, γ constant. (c) A quantity of dry air at 7 C is compressed slowly to (/) of its original volume. What is the percentage change in pressure?

12 Engineering hysics ol. II ROBLEMS AND SOLUTIONS. The ratio of specific heat capacity of helium is.66. Calculate the specific heat capacity at constant volume and at constant pressure. Given: R 8.9 J/K/mol. Solution: C C R C C.66 or C.66C Thus.66C C R C (.66 ) R C R JK mol C.5 JK mol Answer.. Calculate the RMS velocity of a molecule of oxygen at 00 K. Given: 8. J/mol-K. Solution: The mean K.E., mc k B T or mn Ac N AkBT RT where N A is Avogadro s number and R is universal gas constant mn A M 0 A M Ac RT c RT M 0 A c m s Answer.. At what temperature, pressure remaining constant, the RMS velocity becomes double its value at T 7 K. Let C and C be the velocities corresponding to T and T.

13 Heat Solution: c c c c T T T T T or 4 T T K T 09 K Answer..4 Calculate the molecular energy of one gram of hydrogen gas at K. Given that the molecular weight of hydrogen is 0 kg and R 8. J/mol-K. Solution: 0 kg of hydrogen will have molecules. N 0 - A will have molecules. where N A is the Avogadro s number. The molecular energy is N A k T RT E 00.6 J Answer..5 Given C 4.65 J/mol-K for hydrogen. Find C for hydrogen. R kj/ mol-k. Solution: i.e., C C R C C R C 6.4 J mol-k Answer.

14 4 Engineering hysics ol. II.6 A quantity of air at 00 K and atmospheric pressure is suddenly compressed to half its original volume. Find the final pressure and temperature. Solution: (i) atmospheric pressure and During sudden compression, the process is adiabatic γ γ.4 ( ).66 atmosphere. (ii),, T 00 K and T? γ γ T T γ T [ ] [ ] T T 95.9 K Answer..7 Calculate the work done if one mole of an ideal gas expands isothermally at 7 C until its volume is doubled [R 8. J mol K ]. Express the answer in calorie. Solution: m mol, T 400 K;,. Work done RT ( ) log e i.e., W.0 0 J log0 or W calorie 4 W calorie Answer.

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