{"id":252,"date":"2023-03-02T13:26:14","date_gmt":"2023-03-02T13:26:14","guid":{"rendered":"https:\/\/www.vartmaaninstitutesirsa.com\/?page_id=252"},"modified":"2023-03-17T15:20:17","modified_gmt":"2023-03-17T15:20:17","slug":"unit-viii-thermodynamics","status":"publish","type":"page","link":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/unit-viii-thermodynamics\/","title":{"rendered":"Unit VIII: Thermodynamics"},"content":{"rendered":"<p><strong>Chapter\u201312: Thermodynamics<\/strong><\/p>\n<p>Thermal equilibrium and definition of temperature (zeroth law of thermodynamics), heat, work and internal energy. First law of thermodynamics, isothermal and adiabatic processes. Second law of thermodynamics: reversible and irreversible processes, Heat engine and refrigerator.<\/p>\n<p><strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <u>Unit-8 Thermodynamics<\/u><\/strong><\/p>\n<p><strong><u>1 Thermodynamics: <\/u><\/strong><\/p>\n<p>The name thermodynamics is consist of two words thermo and dynamics. Here thermo means heat and dynamics means motion, hence the branch of physics which deals with the study of motion of heat is called thermodynamics. In thermodynamics we study the macroscopic properties of a body such as temperature, pressure, volume, internal energy, entropy and enthalpy etc.<\/p>\n<p><strong><u>2 Thermodynamic System:<\/u><\/strong><\/p>\n<p>A collection of large number of particles having same value of pressure, volume and temperature is called a thermo dynamical system.<\/p>\n<p><strong><u>Thermodynamical system may be of three types.<\/u><\/strong><\/p>\n<ul>\n<li><strong>Open System:-<\/strong><\/li>\n<\/ul>\n<p>A system which can exchange both matter and energy w.r.t surrounding is called open system. E.g.:- an open cup of tea.<\/p>\n<ul>\n<li><strong>Closed System:- <\/strong><\/li>\n<\/ul>\n<p>A system which can exchange heat only but no matter w.r.t surrounding is called closed system. e.g.:- a covered cup of tea.<\/p>\n<ul>\n<li><strong>Isolated System:- <\/strong><\/li>\n<\/ul>\n<p>A system which can exchange both matter and energy w.r.t surrounding is called isolated system. Tea in a thermostat or thermos<\/p>\n<p><strong><u>3 Surrounding:<\/u><\/strong><\/p>\n<p>Except system everything\u2019s which have a direct effect on the system is, called surrounding.<\/p>\n<p><strong><u>4 Thermodynamic Variables: <\/u><\/strong><\/p>\n<p>The quantities like pressure, volume and temperature which help us to study the behavior of a system are called thermo dynamical variables.<\/p>\n<p><strong><u>5 Equation of State<em>:<\/em><\/u><\/strong><\/p>\n<p>The mathematical relation between the pressure, volume and temperature of a thermo dynamical system is called the equation of state. E.g.: the eq. of state for \u00a0moles of an ideal gas is<\/p>\n<p>Here \u00a0is real Gas constant.<\/p>\n<ul>\n<li><strong>The thermo dynamical variable are of two types:-<\/strong><\/li>\n<\/ul>\n<p><strong><em>I <\/em><\/strong><strong>. Intensive thermo dynamical variable:<\/strong><\/p>\n<p>The variables which are independent on size of system e.g.: temperature pressure and specific heat capacity.<\/p>\n<ul>\n<li><strong> Extensive thermo dynamical variable:<\/strong><\/li>\n<\/ul>\n<p>The variable which depends on size of system is called extensive variables. e.g.: volume, energy, entropy, heat capacity and enthalpy.<\/p>\n<p><strong><u>7 Thermo Dynamical Equilibrium:<\/u><\/strong><\/p>\n<p>If two systems are at same temperature then they are said to be in thermal equilibrium and if the variables of a system like pressure, volume and temperature etc. do not changes w.r.t time then system is said to be in thermo dynamical equilibrium.<\/p>\n<ul>\n<li>An isolated system is always in thermo dynamical equilibrium.<\/li>\n<\/ul>\n<p><strong><u>8\u00a0 Zeroth Law of Thermodynamics:<\/u><\/strong><\/p>\n<table width=\"100%\">\n<tbody>\n<tr>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table width=\"100%\">\n<tbody>\n<tr>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table width=\"100%\">\n<tbody>\n<tr>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table width=\"100%\">\n<tbody>\n<tr>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table width=\"100%\">\n<tbody>\n<tr>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><em>According to 0<sup>th<\/sup> law of thermodynamics if two systems A and B are separately in thermal equilibrium with a third system C, than A and B are also in thermal equilibrium with each other.<\/em><\/p>\n<p>Here A and B are separated by a adiabatic wall from where no heat change may occurs and C is in contact with A and B by diathermal wall. All the three are in thermal equilibrium.<\/p>\n<ul>\n<li>The Zeroth law was formulated by R.H fowler in 1931 after the first and second law of thermodynamic was stated. But this law more basic then first then first and second law as this lead to the concept of the temp. So this law as called Zeroth law of thermo dynamics.<\/li>\n<\/ul>\n<p><strong><u>9\u00a0 Heat work and internal Energy:<\/u><\/strong><\/p>\n<p><strong>Internal Energy:-<\/strong><\/p>\n<p>The internal energy of a system is defined as the sum of molecular kinetic energy and molecular potential energy.<strong><em>\u00a0\u00a0\u00a0 <\/em><\/strong>i.e.<\/p>\n<ul>\n<li>Internal kinetic energy of a gas is a function of temperature.<\/li>\n<li>Internal energy of a ideal gas is purely kinetic in nature as ideal gas have no intermolecular force of attraction. So no potential energy.<\/li>\n<li>Internal energy of a system is thermo dynamical state variable as it does not depend on the path along which the state have been brought about.<\/li>\n<\/ul>\n<table width=\"698\">\n<tbody>\n<tr>\n<td width=\"334\"><strong><em>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/em><\/strong><strong>Heat<\/strong><\/td>\n<td width=\"363\"><strong><em>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/em><\/strong><strong>Work<\/strong><\/td>\n<\/tr>\n<tr>\n<td width=\"334\">1.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1. Heat is a form of energy which is transferred from one body to another body due to temperature difference.<\/td>\n<td width=\"363\">1.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1 Work is the mode of energy transfer which do not involve temperature difference between the two bodies.<\/td>\n<\/tr>\n<tr>\n<td width=\"334\">2.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2 If heat is supplied to a system then molecules of the system moves faster.<\/td>\n<td width=\"363\">2.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2 When work is done on a system then motions of the molecules of system becomes well organized.<\/td>\n<\/tr>\n<tr>\n<td width=\"334\">3.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3 Heat absorbed by a system is taken as \u00a0and heat given out by the system is taken as<\/td>\n<td width=\"363\">3.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3 Work done by the system is taken as \u00a0 and work done on the system is taken as<\/td>\n<\/tr>\n<tr>\n<td width=\"334\">4.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 4 Heat is not a state variable.<\/td>\n<td width=\"363\">4.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 4 Work is also not a state variable.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ul>\n<li>Increase in internal energy is taken while decrease in internal energy is taken<\/li>\n<\/ul>\n<p><strong>10IndicatorDiagram:<\/strong><\/p>\n<p>A graphical representation of the state of a system with the help of two thermo dynamical variable is called an indicator diagram and a graph between <em>P <\/em>and <em>V<\/em> of a system is called <em>P-V<\/em> diagram.<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>The area under the \u00a0diagram is numerically equal to work done by the thermo dynamical process.<\/p>\n<p>Suppose two points \u00a0on \u00a0diagram very close to each other having same pressure \u00a0and draw \u00a0on the volume axis. Here<\/p>\n<p>Now work done by the system against pressure is<\/p>\n<p>&nbsp;<\/p>\n<p>Now total done by the system to change its state from A to B is given by<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Example Calculate the work done by a gas as it is taken from the state a to b, b to c and c to a as shown in figure <\/strong><\/p>\n<p>Solution: The work done by the gas in the process \u00a0to \u00a0is the area of .<\/p>\n<p>This is<\/p>\n<p>In the process b to c the volume remains constant and the work done is zero.<\/p>\n<p>In the process \u00a0to \u00a0the gas is compressed. The volume is decreased and the work done by the gas is negative. The magnitude is equal to the area of .<\/p>\n<p>This area is<\/p>\n<p>Thus, the work done in the process\u00a0 \u00a0to \u00a0is \u221240 J.<\/p>\n<p><strong><u>11 First law of thermodynamics:<\/u><\/strong><\/p>\n<p>The first laws of thermodynamic explain law of conservation of energy. According to this law, the amount of heat given to a system will be used to do work and to increases the internal energy of the system.<\/p>\n<p>Suppose \u0394Q is the amount of heat given to the system, \u0394W is the amount of work done by the system and \u00a0is the increase in internal energy of the system.<\/p>\n<p>Then according to 1<sup>st<\/sup> law<\/p>\n<p>But<\/p>\n<p>So<\/p>\n<ul>\n<li>If heat is given to a system then taken and heat released by the system is taken<\/li>\n<li>Work done on the system is taken and work done by the system is taken<\/li>\n<li>Increase in internal energy of the system is taken and decrease in internal energy of the system is taken<\/li>\n<\/ul>\n<p><strong>Question<\/strong> A<strong> gas is contained in a vessel fitted with a movable piston. The container is placed on a hot stove. A total of100 cal of heat is given to the gas and the gas does 40 J of work in the expansion resulting from heating.<\/strong><\/p>\n<p><strong>Calculate the increase in internal energy in the process.<\/strong><\/p>\n<p>Solution: Heat given to the gas is \u2206Q = 100 cal = 418 J.<\/p>\n<p>Work done by the gas is \u2206W = 40 J.<\/p>\n<p>The increase in internal energy is\u00a0\u00a0\u00a0 \u2206U = \u2206Q \u2212 \u2206W = 418 J \u2212 40 J = 378 J.<\/p>\n<p><strong><u>12 Application of First Law of thermodynamics:<\/u><\/strong><\/p>\n<ul>\n<li><strong><u>Isothermal Process<em>:-<\/em><\/u><\/strong><\/li>\n<\/ul>\n<p>In an isothermal process the temperature of the system remains constant so change in internal energy of the system is zero i.e.<\/p>\n<p>So from first law of thermo dynamics\u00a0\u00a0\u00a0 \u00a0\u00a0 \u00a0\u00a0But\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u27f9<\/p>\n<p><em>Hence in isothermal process, heat supplied to the system will be used to do work.<\/em><\/p>\n<ul>\n<li><strong><u>Adiabatic Process<\/u><\/strong><strong>:- <\/strong>In an adiabatic process, heat change of the system is zero <strong><em>\u00a0\u00a0<\/em><\/strong>e.<\/li>\n<\/ul>\n<p>So first law of thermo dynamics becomes\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0 Or<\/p>\n<p><em>Hence in adiabatic process, the work done by the system will decrease the internal energy of the system<\/em>.<\/p>\n<ul>\n<li><strong><u>Isochoric Process:-<\/u><\/strong><\/li>\n<\/ul>\n<p>In a isochoric process, volume of the system remains constant i.e.<\/p>\n<p>So first law of thermo dynamic will become<\/p>\n<p>Hence in isochoric process, heat supplied to the system will increase the internal energy of the system.<\/p>\n<ul>\n<li><strong><u>Isobaric Process (Boiling Process):-<\/u><\/strong><\/li>\n<\/ul>\n<p>In isobaric process, the pressure of the system remains constant. It happens when a liquid boils and converts into vapor during boiling process will whole of the liquid converts into vapor, its pressure remains constant. So heat supplied to the system is used to increase the internal energy of the liquid molecules and to work during expression of the liquid. I.e. heat given to system to convert it into<\/p>\n<p>Where \u00a0is the mass of liquid and \u00a0is latent heat of vaporization<\/p>\n<p>Or<\/p>\n<ul>\n<li><strong><u>Cyclic Process:- <\/u><\/strong>In a cyclic process, initial and final stage of the system remains same. So<\/li>\n<\/ul>\n<p>Then first law of thermo dynamics becomes<\/p>\n<p>Hence heat given to the system will converts into work in cyclic process.<\/p>\n<ul>\n<li><strong><u>Melting Process:- <\/u><\/strong><\/li>\n<\/ul>\n<p>When a solid is melted at its melting point then change in volume \u00a0is negligible small so work done by the solid is<strong><em>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/em><\/strong><\/p>\n<p>If \u00a0is the latent heat of fusion and \u00a0is the mass of the solid then \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0or<\/p>\n<p>&nbsp;<\/p>\n<ol start=\"9\">\n<li><strong> 9. An electric heater supplies heat to a system at a rate of 100 W. If the system performs work at a rate of 75 joules per second, at what rate is the internal energy increasing?<\/strong><\/li>\n<\/ol>\n<p>Ans.\u00a0 Given:, \u00a0,<\/p>\n<p>From first law of thermodynamics,\u00a0\u00a0 \u00a0or<\/p>\n<p>Or<\/p>\n<p><strong><u>13 Isothermal Process:-<\/u><\/strong><\/p>\n<p><em>A thermo dynamical process in which temperature of the system remains constant however pressure and volume may change is called <strong>isothermal process.<\/strong><\/em><\/p>\n<p><strong><em>Essential Conditions for an isothermal process:-<\/em><\/strong><\/p>\n<ul>\n<li>The wall of the container must be perfectly conducting to exchange heat from system to surrounding.<\/li>\n<li>The process of compression or expansion should be very slow, so as to provide sufficient time for the exchange of heat.<\/li>\n<\/ul>\n<p><strong><u>Equation of Isothermal Process:-<\/u><\/strong><\/p>\n<p>According to Boyle&#8217;s law at constant temperature, the relation between <em>P<\/em> and <em>V<\/em> may be given<\/p>\n<p>as\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 or<\/p>\n<p>The above eq. is called equation of isothermal process.<\/p>\n<p><strong><u>Work Done in an Isothermal Process:<\/u><\/strong><\/p>\n<p>Suppose \u00a0moles of an ideal gas is filled in a cylinder having conducting walls and frictionless piston and let <em>P<\/em> is the pressure of the gas,<\/p>\n<p>Then work done by the gas to move piston through small Volume dV is<\/p>\n<p>Now again suppose that gas expend from initial stage \u00a0to final stage \u00a0then total amount of work done will be<\/p>\n<p>For isothermal process<\/p>\n<p>Using in eq. (i) we get\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0 \u00a0=<\/p>\n<p>&nbsp;<\/p>\n<p>Or<\/p>\n<p>Also\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0 So<\/p>\n<p>Here\u00a0\u00a0\u00a0 eq. (ii) and (iii) represents work done for an isothermal expansion.<\/p>\n<p>&nbsp;<\/p>\n<p><strong><u>14 Adiabatic Process:<\/u><\/strong><\/p>\n<p><em>A thermo dynamic process in which heat remains constant however temperature, pressure and volume of the system may change is called <strong>adiabatic process<\/strong><\/em>.<\/p>\n<p><strong><u>Essential Conditions for Adiabatic Process:<\/u><\/strong><\/p>\n<ul>\n<li>In an adiabatic process the wall of the container must be perfectly insulator so that no heat can exchange from system to surrounding.<\/li>\n<li>The process of compression or expansion should be sudden so that heat does not have time to change.<\/li>\n<li><strong>Adiabatic Relation between P and V:<\/strong><\/li>\n<\/ul>\n<p>As according to first law of thermodynamics<\/p>\n<p>And for one mole of gas<\/p>\n<p>And for adiabatic process \u00a0 so from (1) we get<\/p>\n<p>Also according to ideal gases eq<\/p>\n<p>Differentiating both sides\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u27f9<\/p>\n<p>Using in eq. (ii) we get<\/p>\n<p>\u27f9<\/p>\n<p>\u27f9<\/p>\n<p>\u27f9<\/p>\n<p>\u27f9<\/p>\n<p>Dividing by \u00a0both sides we get<\/p>\n<p>or<\/p>\n<p>Now integrating both sides, we get<\/p>\n<p>Where <em>c<\/em> is constant of integration<\/p>\n<p>or<\/p>\n<p>or<\/p>\n<p>or<\/p>\n<p>This eq. is called equation of adiabatic process\/change.<\/p>\n<ul>\n<li><strong><u>Adiabatic Relation between P and T:<\/u><\/strong><\/li>\n<\/ul>\n<p>For one mole of gas \u00a0\u00a0\u00a0\u00a0\u00a0 so<\/p>\n<p>So adiabatic eq. \u00a0becomes<\/p>\n<p>&nbsp;<\/p>\n<p>i.e.<\/p>\n<p>This is a adiabatic relation between pressure and temperature of a ideal gas.<\/p>\n<ul>\n<li><strong><u>Adiabatic Relation between Volume and Temperature:<\/u><\/strong><\/li>\n<\/ul>\n<p>Again for one mole of a ideal gas<\/p>\n<p>So ideal gas eq. becomes<\/p>\n<p>or<\/p>\n<p>or<\/p>\n<p>This is a adiabatic relation between temperature and volume of an ideal gas.<\/p>\n<p><strong><u>Work done in an Adiabatic Process:<\/u><\/strong><\/p>\n<p>Suppose \u00a0moles of an ideal gas contained in a cylinder having insulating wall and insulating and frictionless piston. Let \u00a0is the pressure of the gas. Now the amount of work done by the gas to move to small volume \u00a0is<\/p>\n<p>Now suppose that gas expends adiabatically from initial stage \u00a0to final stage . Then the total amount of work done by the gas is<\/p>\n<p>Now as we know that for a adiabatic change\u00a0\u00a0 \u00a0\u00a0 or<\/p>\n<p>Or<\/p>\n<p>Or<\/p>\n<p>But<\/p>\n<p>\u27f9<\/p>\n<p>Also<\/p>\n<p>Or<\/p>\n<p>For one mole gas<\/p>\n<p>This equation gives the amount of work done during the adiabatic process.<\/p>\n<ul>\n<li>If work is done by the gas then temperature of the gas decreases As<\/li>\n<li>If work done on the gas then temperature of the gas increases e\u00a0\u00a0 if<\/li>\n<\/ul>\n<p><strong><em><u>15 <\/u><\/em><\/strong><strong><u>Specific heat of a Gas: <\/u><\/strong>Specific heat may be divided into two types.<\/p>\n<ul>\n<li><strong><u>Specific heat at constant volume (Molar specific heat at constant volume):<\/u><\/strong><\/li>\n<\/ul>\n<p>It is defined as the amount of heat required to raise the temperature of one mole of a gas through \u00a0at constant volume and denoted by<\/p>\n<ul>\n<li><strong><u>Molar Specific heat at constant pressure:<\/u><\/strong><\/li>\n<\/ul>\n<p>It is defined as the amount of heat required to raise the temperature of one mole of gas through \u00a0at constant pressure. It is denoted by<\/p>\n<ul>\n<li>At constant volume all heat supplied to gas is used to increase the temperature of the gas but at constant pressure when heat is supplied then some heat is used in expansion so specific heat at constant pressure is greater than specific heat at constant volume.<\/li>\n<\/ul>\n<p><strong><u>\u00a0<\/u><\/strong><\/p>\n<p><strong><u>Relation between <\/u><\/strong><\/p>\n<p>As according to first law of thermo dynamics<\/p>\n<p>Now let \u00a0is the amount of heat supplied at constant volume so<\/p>\n<p>So from first law<\/p>\n<p>Where \u00a0is specific heat at constant volume using in eq. (i)<\/p>\n<p>Now when if heat is supplied at constant pressure then<\/p>\n<table width=\"100%\">\n<tbody>\n<tr>\n<td>\u00d8\u00a0 <em>\u00a0(when C<sub>P<\/sub> and C<sub>V<\/sub> are measured as work)<\/em><\/p>\n<p>\u00d8\u00a0 <em>\u00a0(when C<sub>P<\/sub> and C<sub>V<\/sub> are measured as heat) where J=4.18J cal<sup>1<\/sup><\/em><\/p>\n<p>\u00d8\u00a0 <em>The ratio of two principle specific heats is represented by <\/em> <em>\u00a0 as <\/em><\/p>\n<p>\u00d8\u00a0 <em>As R is always +ve so C<sub>P<\/sub> is always greater than C<sub>V<\/sub>.<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>So eq. (i) becomes<\/p>\n<p>Now for ideal gas \u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0 Or<\/p>\n<p>\u27f9<\/p>\n<p>\u27f9<\/p>\n<p>This relation is called <strong><em>Mayer\u2019s Formula<\/em><\/strong>.<\/p>\n<p><strong><u>Limits of specific heat of a gas:<\/u><\/strong><\/p>\n<p>(i) Suppose the gas is suddenly compressed and no heat is supplied to the gas i.e \u00a0but the temperature of the gas rises due to compression<\/p>\n<p>i.e specific heat of the gas is zero.<\/p>\n<p>(ii) Now when gas is heated and allowed to expend such that heat supplied is used in expansion then there will be no rise in temperature i.e\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0 i.e specific heat of gas is .<\/p>\n<ul>\n<li>Again when heat is supplied and gas allowed to expend as well as temperature also rises in this case<\/li>\n<\/ul>\n<p>i.e specific heat of gas is<\/p>\n<ul>\n<li>Again when gas is heated and allowed to expend at such a rate that the temperature fall more due to expansion then rises in temperature due to heat supply. In this case<\/li>\n<\/ul>\n<p>i.e specific heat of gas is<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p><strong><u>16 Reversible and Irreversible Process:<\/u><\/strong><\/p>\n<ul>\n<li><strong><u>Reversible Process:<\/u><\/strong><\/li>\n<\/ul>\n<p>Any process which can be retraced in its reverse direction exactly is called reversible process.<\/p>\n<p><strong><u>Conditions:-<\/u><\/strong><\/p>\n<ul>\n<li>The process must be quasi statics. i.e it must be infinite slow so as to seems as in rest.<\/li>\n<li>The forces such as viscosity, frictions, inelasticity etc. should be absent.<\/li>\n<\/ul>\n<p><strong>Examples:-<\/strong><\/p>\n<ul>\n<li>The process of electrolysis is reversible if the resistance of electrolyte is small.<\/li>\n<li>A Carnot cycle.<\/li>\n<li>The process of gradually expansion and extension of an elastic spring is approximately reversible.<\/li>\n<li>We cannot have a perfect reversible process becomes of frictional force, viscosity etc.<\/li>\n<li><strong><u>Irreversible Process:<\/u><\/strong><\/li>\n<\/ul>\n<p>Any process which cannot be retracted in reverse direction exactly is called irreversible process.<\/p>\n<p><strong><em>Examples:<\/em><\/strong><\/p>\n<ul>\n<li>Diffusion of gases.<\/li>\n<li>Dissolution of salt in water.<\/li>\n<li>Rusting of iron.<\/li>\n<li>Sudden compression or contraction of gas.<\/li>\n<\/ul>\n<p><strong><u>17 Heat Engine:<\/u><\/strong><\/p>\n<p>It is a device which converts heat energy into mechanical energy.<\/p>\n<p><em>\u00a0A heat engine has the following essentials parts:<\/em><\/p>\n<ul>\n<li><strong><u>Source:-<\/u><\/strong><\/li>\n<\/ul>\n<p>It is heat reservoir at higher temperature \u00a0of infinite heat capacity.<\/p>\n<ul>\n<li><strong><u>Sink:-<\/u><\/strong><\/li>\n<\/ul>\n<p>It is a heat reservoir at a lower temperature \u00a0of infinite heat capacity. Any amount of heat can be supplied to it without changing its temperature.<\/p>\n<p>&nbsp;<\/p>\n<ul>\n<li><strong><u>Working Substance:-<\/u><\/strong><\/li>\n<\/ul>\n<p>It is a material which performs mechanical work heat is supplied to it.<\/p>\n<p><strong><em>Working:-<\/em><\/strong><\/p>\n<p>In every cycle of operation the working substance absorbs a definite amount of heat \u00a0from the source at higher temperature and do some work and rejects the remaining heat to sink working substance may be a cylinder with a piston which transfer mechanical energy to wheel of a vehicle via a shaft.<\/p>\n<p>Efficiency of a heat engine:<\/p>\n<p>It is defined as the ratio of net work done by a engine in one cycle to the amount of heat absorbed by the working substance from the source.<\/p>\n<p>i.e<\/p>\n<p>i.e\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 or \ud835\udec8<\/p>\n<p>Here efficiency can be clearly that it is less than 1 or 100%. The efficiency of steam engine is about 12 to 16% and of petrol engine is 26% and of diesel engine about 40%.<\/p>\n<p><strong><u>Types of Heat Engine:<\/u><\/strong><\/p>\n<p>There are of two types of heat engine.<\/p>\n<ul>\n<li><strong><u>External combustion engine:-<\/u><\/strong><\/li>\n<\/ul>\n<p>In this engine the heat needed for working substance is produced by burning the full outside the cylinder and piston arrangement of the engine. A steam engine is an example of external combustion engine.<\/p>\n<ul>\n<li><strong><u>Internal Combustion Engine:-<\/u><\/strong><\/li>\n<\/ul>\n<p>In this engine the heat needed for the engine is produced by burning the fuel inside the main cylinder. The petrol and diesel engine are the examples of internal combustion engine.<\/p>\n<ul>\n<li>A internal combustion engine is more efficient then external combustion engine.<\/li>\n<\/ul>\n<p><strong><u>18 Limitations of the first law of thermo dynamics:<\/u><\/strong><\/p>\n<ul>\n<li>It does not indicate the direction of transfer of heat as why heat flow from hot body to cold body.<\/li>\n<li>It does not tells conditions under which heat can be converted into mechanical energy,<\/li>\n<li>It does not tell the extent to which heat energy can be converted into work.<\/li>\n<\/ul>\n<p><strong><u>19 Second Law of thermo dynamics:<\/u><\/strong><\/p>\n<p>The second law&#8217;s statements are given by a number of ways but mainly two statements are famous as given below.<\/p>\n<ul>\n<li><strong><u>Kelvin \u2013 Planck\u2019s Statements:-<\/u><\/strong><\/li>\n<\/ul>\n<p>According to this statement it is impossible to get a continuous supply of work by cooling a body to a temperature lower than that of coldest of its surrounding.<\/p>\n<ul>\n<li><strong><u>Clausius Statements:-<\/u><\/strong><\/li>\n<\/ul>\n<p>It is impossible for a self acting machine to transfer heat from a cold body to hot body without any external energy.<\/p>\n<p><strong><u>Significance statement:-<\/u><\/strong><\/p>\n<ul>\n<li>It puts significant limits to the efficiency of a heat engine.<\/li>\n<li>It tells that efficiency of refrigeration can never be infinite.<\/li>\n<\/ul>\n<p><strong><u>Limitations:<\/u><\/strong><\/p>\n<ul>\n<li>It cannot be proved directly.<\/li>\n<li>It is only applicable to a cyclic process.<\/li>\n<li>It is only applicable to certain conditions.<\/li>\n<\/ul>\n<p><strong><u>20 Carnot&#8217;s reversible Engine:-<\/u><\/strong><\/p>\n<p>Sadi Carnot introduced the concept of an ideal heat engine the main parts of this heat engine are:-<\/p>\n<ol>\n<li><strong><u>Source:-<\/u><\/strong>A hot body maintained at a fixed high temperature having infinite heat capacity and any amount of heat can be taken from it without changing its temperature is called source.<\/li>\n<li><strong><u>Sink:- <\/u><\/strong>A cold body maintained at a fixed low temperature \u00a0having infinite heat capacity and any amount of heat can be rejected to it without changing its temperature is called sink.<\/li>\n<li><strong>Working substance:-<\/strong>A cylinder with insulated sides and conducting base containing perfect gas as working substance. A perfect insulated and frictionless piston is fitted into the cylinder.<\/li>\n<li><strong><u>Insulated Pad:-<\/u><\/strong><\/li>\n<\/ol>\n<p>A perfectly non conducting platform serving as a stand for the cylinder is also provided so that working substance can undergo adiabatic operations.<\/p>\n<p><strong><em><u>21 <\/u><\/em><\/strong><strong><u>Carnot Cycle:-<\/u><\/strong><\/p>\n<p>The main parts of Carnot cycle are<\/p>\n<ul>\n<li><strong><u>Isothermal Expansion Operation:<\/u><\/strong><\/li>\n<\/ul>\n<table width=\"100%\">\n<tbody>\n<tr>\n<td>P<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table width=\"100%\">\n<tbody>\n<tr>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table width=\"100%\">\n<tbody>\n<tr>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table width=\"100%\">\n<tbody>\n<tr>\n<td><em>Isothermal Expansion<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table width=\"100%\">\n<tbody>\n<tr>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table width=\"100%\">\n<tbody>\n<tr>\n<td><em>Adiabatic Expansion<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table width=\"100%\">\n<tbody>\n<tr>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table width=\"100%\">\n<tbody>\n<tr>\n<td><em>Isothermal <\/em>compression<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table width=\"100%\">\n<tbody>\n<tr>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table width=\"100%\">\n<tbody>\n<tr>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table width=\"100%\">\n<tbody>\n<tr>\n<td><em>Adiabatic <\/em>compression<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table width=\"100%\">\n<tbody>\n<tr>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table width=\"100%\">\n<tbody>\n<tr>\n<td>L\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 M<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table width=\"100%\">\n<tbody>\n<tr>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>When cylinder is placed on source then it expansion by expends by acquiring temperature \u00a0from initial stage \u00a0final stage. The \u00a0amount of heat is absorbed by the gas and \u00a0amount of work is done by the gas as \u00a0\u00a0\u00a0 .<\/p>\n<ul>\n<li><strong><u>Adiabatic Expansion Operation:-<\/u><\/strong><\/li>\n<\/ul>\n<table width=\"100%\">\n<tbody>\n<tr>\n<td>V<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>When a gas is placed on sink at tem T<sub>2<\/sub> and allowed to expand slowly till temp falls up to T<sub>2<\/sub>. Let w<sub>2<\/sub> is the amount of work done during adiabatic expansion from stage P<sub>2<\/sub>V<sub>2<\/sub>T<sub>1<\/sub> to P<sub>3<\/sub>V<sub>3<\/sub>T<sub>2<\/sub> as<\/p>\n<ul>\n<li><strong><u>Isothermal compression<\/u><\/strong><\/li>\n<\/ul>\n<p>Now the gas is placed in thermal contact with the sink at temperature . The gas now compressed isothermally. If \u00a0is the amount of heat rejected by to the sink and \u00a0is the amount of work done by the surrounding on the system from stage \u00a0to \u00a0as<\/p>\n<p><strong>Step4: adiabatic compression.<\/strong><\/p>\n<p>Now the cylinder is again placed on the insulating pad, the gas is suddenly compressed till it return to initial stage . Then amount of work done on the gas is<\/p>\n<p>The net amount of work done per cycle can be calculated as the sum of total work done by gas and on the gas as\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Total work done by gas<\/p>\n<p>Total work done on gas<\/p>\n<p>Net work done \u00a0\u00a0\u00a0 But<\/p>\n<p>\u27f9\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0 Or<\/p>\n<p>Hence in a Carnot cycle, the mechanical work done by the gas is equal to the area under the Carnot cycle.<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<ol start=\"3\">\n<li><strong> 3. Consider a Carnot&#8217;s cycle operating between <\/strong> <strong>producing 1 kJ of mechanical work per cycle. Find the heat transferred to the engine by the reservoirs.<\/strong><\/li>\n<\/ol>\n<p>Ans.\u00a0 Given:<\/p>\n<p>Efficiency of Carnot\u2019s engine, \ud835\udec8<\/p>\n<p>But \ud835\udec8 \u00a0\u00a0\u00a0 As<\/p>\n<p><strong><u>22Efficiency of Carnot Engine:<\/u><\/strong><\/p>\n<p>It is defined as the ratio of the total work done by the Carnot cycle to the amount of heat absorbed by the working substance<\/p>\n<p>i.e.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \ud835\udec8<\/p>\n<p>Also\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u2026\u2026\u2026\u2026\u2026(1)<\/p>\n<p>And\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u2026\u2026\u2026\u2026\u2026(2)<\/p>\n<p>On dividing (i) by (ii) we get<\/p>\n<p>Or\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \ud835\udec8 \u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0 clearly<\/p>\n<p><strong><em><u>Non Practicability of Carnot Engine:<\/u><\/em><\/strong><\/p>\n<p>It is not possible to make source and sink of infinite heat capacity.<\/p>\n<ul>\n<li>There is no ideal gas.<\/li>\n<li>The cylinder not be practically provided perfectly frictionless piston.<\/li>\n<li>Practically a reversible process is not possible.<\/li>\n<\/ul>\n<p><strong><u>23 Refrigerator of heat pump: <\/u><\/strong><\/p>\n<p><strong><u>Refrigerator<\/u><\/strong>:<\/p>\n<p>A refrigerator is a Carnot&#8217;s heat engine working in the reverse direction.<\/p>\n<p>In a refrigerator the working substance absorbs heat from a cold reservoir and transfer to source at high temperature with the help of external agency motor having Freon gas .<\/p>\n<ul>\n<li>In refrigerator gas is allowed to expand suddenly (adiabatically) from high to low pressure. This cools it and converts it into vapor liquid mixture.<\/li>\n<li>The cold fluid allowed absorbing heat from reservoir.<\/li>\n<li>After which the gas is compressed till it reaches at temperature greater than atmospheric pressure and temperature.<\/li>\n<li>Finally the vapor is compressed isothermally to reject heat to surrounding and then return to initial stage.<\/li>\n<\/ul>\n<p>It is defined as the ratio of heat removed \u00a0per cycle to the mechanic work required to done on it.<\/p>\n<p>&nbsp;<\/p>\n<p>As<\/p>\n<p>\u27f9<\/p>\n<p>Or<\/p>\n<p>Neither the value of \u03b2 better is the refrigerator.<\/p>\n<p><strong><u>24 Carnot Theorem:-<\/u><\/strong><\/p>\n<p>It states that no engine working between two given temperature can have efficiency greater than that of Carnot engine working between same two temperature or the efficiency of Carnot heat engine does not depends on the nature of working substance.<\/p>\n<p><strong><u>25 Third Law of Thermo dynamic:-<\/u><\/strong><\/p>\n<p>According to third law of thermo dynamics the entropy of the universe is increasing continuously.<\/p>\n<p>&nbsp;<\/p>\n<p><strong>\u00a0<\/strong><\/p>\n<p><strong>\u00a0<\/strong><\/p>\n<p><strong>\u00a0<\/strong><\/p>\n<p><strong>\u00a0<\/strong><\/p>\n<p><strong>\u00a0<\/strong><\/p>\n<p><strong>\u00a0<\/strong><\/p>\n<p><strong>\u00a0<\/strong><\/p>\n<ol>\n<li><strong>The first law of thermodynamics is a statement of<\/strong><\/li>\n<\/ol>\n<p>(a) Conservation of heat\u00a0\u00a0\u00a0 (b) conservation of work (c) conservation of momentum\u00a0\u00a0 (d) conservation of energy\u009f<\/p>\n<ol start=\"2\">\n<li><strong> If heat is supplied to an ideal gas in an isothermal process,<\/strong><\/li>\n<\/ol>\n<p>(a) The internal energy of the gas will increase\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (b) the gas will do positive work\u009f<\/p>\n<p>(c) The gas will do negative work\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (d) the said process is not possible.<\/p>\n<ol start=\"3\">\n<li><strong> The pressure p and volume V of an ideal gas both increase in a process. (Both correct h)<\/strong><\/li>\n<\/ol>\n<p>(a) The temperature of the system must increase.\u00a0 \u00a0(b) The work done by the system is positive. \u009f<\/p>\n<p>(c) Such a process is not possible. \u009f \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(d) Heat supplied to the gas is equal to the change in internal energy.<\/p>\n<ol start=\"4\">\n<li><strong> in a process on a system, the initial pressure and volume are equal to the final pressure and volume.<\/strong><\/li>\n<\/ol>\n<p>(a) The initial temperature must be equal to the final temperature. \u009f<\/p>\n<p>(b) The initial internal energy must be equal to the final internal energy. \u009f<\/p>\n<p>(c) The net heat given to the system in the process must be zero.<\/p>\n<p>(d) The net work done by the system in the process must be zero.<\/p>\n<ol start=\"5\">\n<li><strong> The internal energy of an ideal gas decreases by the same amount as the work done by the system.<\/strong><\/li>\n<\/ol>\n<p>(a) The process must be adiabatic. \u009f\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (b) The process must be isothermal.<\/p>\n<p>(c) The process must be isobaric.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (d) The temperature must decrease. \u009f<\/p>\n<ol start=\"6\">\n<li><strong> The Zeroth law of thermodynamics allows us to de\ufb01ne :<\/strong><\/li>\n<li>work B. pressure\u00a0\u00a0 C. temperature\u009f\u00a0\u00a0 D. thermal equilibrium<\/li>\n<li><strong> Heat has the same units as:<\/strong><\/li>\n<\/ol>\n<p>(a). temperature\u00a0\u00a0\u00a0 (b). Work\u009f\u00a0\u00a0\u00a0\u00a0 (c). Energy\/time\u00a0\u00a0 (d).\u00a0\u00a0 \u00a0heat capacity<\/p>\n<ol start=\"8\">\n<li><strong> A calorie is about:<\/strong><\/li>\n<\/ol>\n<p>(a). 0.24 J\u00a0\u00a0\u00a0\u00a0 (b). 8.3J\u00a0\u00a0\u00a0\u00a0\u00a0 (c). \u00a0250 J\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (d). 4.2J\u009f<\/p>\n<ol start=\"9\">\n<li><strong> The first law of thermodynamics is concerned with the conservation of <\/strong><\/li>\n<\/ol>\n<p>(a)\u00a0 Momentum (b) Energy\u009f \u00a0\u00a0\u00a0(c) Mass (d) Temperature<\/p>\n<ol start=\"10\">\n<li><strong> Which of the following is not thermodynamically function? <\/strong><\/li>\n<\/ol>\n<p>(a)\u00a0 Enthalpy (b) Work done\u009f (c) Gibb\u2019s energy (d) Internal energy<\/p>\n<ol start=\"11\">\n<li><strong> First law thermodynamics states that <\/strong><\/li>\n<\/ol>\n<p>(a)\u00a0 System can do work (b) System has temperature (c) System has pressure (d) Heat is a form of energy\u009f<\/p>\n<ol start=\"12\">\n<li><strong> Temperature is a measurement of coldness or hotness of an object. This definition is based on <\/strong><\/li>\n<\/ol>\n<p>(a)\u00a0 Zeroth law of thermodynamics\u009f \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b) First law of thermodynamics<\/p>\n<p>(c) \u00a0Second law of thermodynamics\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (d) Newton\u2019s law of cooling<\/p>\n<ol start=\"13\">\n<li><strong> First law of thermodynamics is a special case of <\/strong><\/li>\n<\/ol>\n<p>(a)\u00a0 Newton&#8217;s law\u00a0\u00a0\u00a0 (b) \u00a0Law of conservation of energy\u009f (c)\u00a0 Charle&#8217;s \u00a0law \u00a0(d)\u00a0 Law of heat exchange<\/p>\n<ol start=\"14\">\n<li><strong> in an isothermal change, an ideal gas obeys <\/strong><\/li>\n<\/ol>\n<p>(a)\u00a0 Boyle&#8217;s law\u009f\u00a0\u00a0 (b) Charle\u2019s law\u00a0\u00a0\u00a0 (c) Gay lussac law (d) none of the above<\/p>\n<ol start=\"15\">\n<li><strong> The specific heat of a gas in an isothermal process is <\/strong><\/li>\n<\/ol>\n<p>(a)\u00a0 Infinite\u009f\u00a0\u00a0 (b) Zero \u00a0\u00a0(c) Negative (d) remains constant<\/p>\n<ol start=\"16\">\n<li><strong> A container that suits the occurrence of an isothermal process should be made of <\/strong><\/li>\n<\/ol>\n<p>(a)\u00a0 Copper\u009f\u00a0\u00a0 (b) Glass \u00a0\u00a0(c) Wood (d) Cloth<\/p>\n<ol start=\"17\">\n<li><strong> The work done in an adiabatic change in a gas depends only on <\/strong><\/li>\n<\/ol>\n<p>(a)\u00a0 Change is pressure (b) Change is volume \u00a0\u00a0(c) Change in temperature\u009f\u00a0 (d) None of the above<\/p>\n<ol start=\"18\">\n<li><strong> An adiabatic process occurs at constant <\/strong><\/li>\n<\/ol>\n<p>(a)\u00a0 Temperature\u00a0\u00a0 (b) Pressure (c) Heat\u009f (d) Temperature and pressure<\/p>\n<ol start=\"19\">\n<li><strong> A cycle tyre bursts suddenly. This represents an <\/strong><\/li>\n<\/ol>\n<p>(a)\u00a0 Isothermal process (b) Isobaric process\u00a0\u00a0 (c) Isochoric process (d) Adiabatic process\u009f<\/p>\n<ol start=\"20\">\n<li><strong> The internal energy of the gas increases In <\/strong><\/li>\n<\/ol>\n<p>(a)\u00a0 Adiabatic expansion (b) Adiabatic compression\u009f \u00a0(c) Isothermal expansion\u00a0\u00a0 (d) Isothermal compression<\/p>\n<ol start=\"21\">\n<li><strong> If the door of a refrigerator is kept open, then which of the following is true? <\/strong><\/li>\n<\/ol>\n<p>(a)\u00a0 Room is cooled \u00a0\u00a0(b) Room is heated\u009f (c) Room is either cooled or heated (d) Room is neither cooled nor heated<\/p>\n<ol start=\"22\">\n<li><strong> A measure of the degree of disorder of a system is known as <\/strong><\/li>\n<\/ol>\n<p>(a)\u00a0 Isobaric (b) Isotropy (c) Enthalpy (d) Entropy\u009f<\/p>\n<ol start=\"23\">\n<li><strong> Even Carnot engine cannot give 100% efficiency because we cannot <\/strong><\/li>\n<\/ol>\n<p>(a)\u00a0 Prevent radiation \u00a0\u00a0(b)\u00a0 Find ideal sources\u00a0\u00a0 (c)\u00a0 Reach absolute zero temperature\u009f (d)\u00a0 Eliminate friction<\/p>\n<ol start=\"24\">\n<li><strong> The work done in which of the following processes is zero <\/strong><\/li>\n<\/ol>\n<p>(a)\u00a0 Isothermal process (b) Adiabatic process \u00a0\u00a0\u00a0(c) Isochoric process\u009f\u00a0 (d) none of these<\/p>\n<ol start=\"25\">\n<li><strong> in a cyclic process, work done by the system is <\/strong><\/li>\n<\/ol>\n<p>(a)\u00a0 Zero\u00a0\u00a0\u00a0\u00a0 \u00a0(b) Equal to heat given to the system\u009f<\/p>\n<p>(C)\u00a0 More than the heat given to system \u00a0\u00a0\u00a0(d) Independent of heat given to the system<\/p>\n<p><strong>\u00a0<\/strong><\/p>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Chapter\u201312: Thermodynamics Thermal equilibrium and definition of temperature (zeroth law of thermodynamics), heat, work and internal energy. First law of thermodynamics, isothermal and adiabatic processes. Second law of thermodynamics: reversible and irreversible processes, Heat engine and refrigerator. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Unit-8 Thermodynamics 1 Thermodynamics: The name thermodynamics is consist of two words thermo and dynamics. Here thermo [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"_uag_custom_page_level_css":"","site-sidebar-layout":"default","site-content-layout":"default","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"var(--ast-global-color-4)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"class_list":["post-252","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_hostinger_reach_plugin_has_subscription_block":false,"_hostinger_reach_plugin_is_elementor":false,"uagb_featured_image_src":{"full":false,"thumbnail":false,"medium":false,"medium_large":false,"large":false,"1536x1536":false,"2048x2048":false},"uagb_author_info":{"display_name":"sandeep.soni484@gmail.com","author_link":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/author\/sandeep-soni484gmail-com\/"},"uagb_comment_info":0,"uagb_excerpt":"Chapter\u201312: Thermodynamics Thermal equilibrium and definition of temperature (zeroth law of thermodynamics), heat, work and internal energy. First law of thermodynamics, isothermal and adiabatic processes. Second law of thermodynamics: reversible and irreversible processes, Heat engine and refrigerator. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Unit-8 Thermodynamics 1 Thermodynamics: The name thermodynamics is consist of two words thermo and dynamics. Here thermo&hellip;","_links":{"self":[{"href":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/wp-json\/wp\/v2\/pages\/252","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/wp-json\/wp\/v2\/comments?post=252"}],"version-history":[{"count":0,"href":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/wp-json\/wp\/v2\/pages\/252\/revisions"}],"wp:attachment":[{"href":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/wp-json\/wp\/v2\/media?parent=252"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}