Physics-18.11-Heat and thermodynamics

11-Heat and thermodynamics HEAT AND THERMODYNAMICS Heat Heat is that form of energy. Zeroth Law Of Thermodynamics And Temperature If a system A is in thermal equilibrium with system B and the system B is in thermal equilibrium with system C, then systems A and C are in thermal equilibrium with each other. Temperature scales Relation between Celsius, Kelvin and Fahrenheit Scale: Example 1: The electrical resistance of pure platinum increases linearly with increasing temperature. This property is used in a Platinum resistance thermometer. The relation between R (Resistance at K) and R0 (Resistance at 0K) is given as where  = temperature coefficient of resistance. Now, if a Platinum resistance thermometer reads 00 Celsius when its resistance is 80 and 100o when its resistance is 90, find the temperature at which its resistance is 86. Solution: Using the given relationship, we have . . . (i) . . . (ii) Where is the desired temperature. Taking the ratio of (i) & (ii) CALORIMETRY Principle of Calorimetry When two objects having different temperatures are brought in contact, heat flows from the hot object to the cold object. If the system is sufficiently thermally isolated from its surrounding, the heat lost by the hot object = the heat gained by the cold object. Specific heat capacity The amount of heat needed to raise the temperature of unit mass of a substance by 1 is known as its specific heat capacity. If Q amount of heat raises the temperature of m mass of a material by , then its specific heat capacity is given as: Also the amount of heat supplied per unit increase in temperature for any body is known as its heat capacity, . Specific Latent Heat In order to change the state of a substance (from solid to liquid or from liquid to gas) heat has to be supplied to it. During this process temperature remains constant. The amount of heat supplied per unit mass for such a process is known as the Specific Latent Heat of that substance for that process. Example 2: 5 gm of water at 30oC and 5 gm of ice at - 20oC are mixed together in a calorimeter. Find the final temperature of mixture and also the final masses of ice and water. Water equivalent of calorimeter is negligible, specific heat of ice = 0.5 cal/gmoC and latent heat of ice = 80 cal/gm. Solution: In this case heat is given by water and taken by ice Heat available with water to cool from 30oC to 0oC = ms = . Heat required by 5 gm ice to increase its temperature up to 0oC ms Out of 150 cal heat available, 50 cal is used for increasing temperature of ice from - 20oC to 0oC. The remaining heat 100 cal is used for melting the ice. if mass of ice melted is m gm then  Thus 1.25 gm ice out of 5 gm melts and the mixture of ice and water is at 0oC. Mechanical Equivalent of Heat 1 calorie is the quantity of heat required to raise the temperature of pure water at 1 atm pressure from 14.5C to 15.5C. When work is completely converted to heat, the quantity of heat produced (Q) is found to be proportional to the quantity of work (W) that was converted into heat W  Q Or, W = JQ, where J is known as the Mechanical Equivalent of Heat. J = 4.2  107 erg cal-1 = 4.2 J/cal Example 3: A bullet splinter of mass of 10 gm moving with a speed of 400 m/s hits an ice block of mass 990 gm kept on a frictionless floor and gets stuck in it. How much ice will melt if 50% of the lost Kinetic energy goes to ice? (Temperature of ice block = 0oC). Solution: Velocity of bullet + ice block, V = m/s V =4 m/s Loss of K.E. = = [0.01(400)21(4)2] = [1600  16]= (1584/2) J Heat generated = = 95 Cal Mass of ice melted = = 1.2 gm. Thermal Expansion An increase in the temperature of a body is generally accompanied by an increase in its size. This is known as Thermal Expansion. The value of linear coefficient of thermal expansion at temperature T can be found by taking limit . or For most of the solids the value of is small and nearly independent of T. In such cases the linear dimension of an object at a different temperature is given by: (Here is the initial length and is the change in temperature). Example 4: A clock with an iron pendulum is made so as to keep correct time at 10oC. Given . How fast or slow does the clock move per day if the temperature rises to 250 C? Given iron = 12  106 per 0C. Solution: When the pendulum keeps correct time, its period of vibration is 2 sec and so it makes Vibration /day If length of pendulum at 10oC is and at 25oC is as i.e. T  i.e. n is no. of vibrations per sec. = 43200[1  0.00009] = That is the clock makes (43200  43196.12) = 3.88 vibration loss per day. That is clock losses 3.88  2 = 7.76 sec per day. Area and Volume expansion If the temperature of a two-dimensional object (lamina) is changed, its area changes. If the coefficient of linear expansion of the material of lamina is small and constant, then its final area is given by , where is the initial area, is the change in temperature and is the area coefficient of thermal expansion. It can be shown that . Similar relation also holds for the volume of a three-dimensional object (Here is known as the coefficient of volume expansion). For most solids,  ~ 106/C Thermal Expansion in liquids: The experimental measurement of  for a liquid becomes slightly difficult due to expansion of the container, when a liquid is heated in a container. The initial level of the liquid falls due to expansion of the container. But afterwards it rises due to faster expansion of the liquid. The actual increase in the = volume of the liquid the apparent increase in the volume of liquid + The increase in the volume of the container. liquid = apparent + container , for liquids, is of the order of 104/C Anomalous expansion of water: If the temperature of water is increased from 0oC, it contracts until the temperature reaches 4oC and expands thereafter. In other words the density of water is highest at 4oC. Example 5: A glass vessel of volume is completely filled with a liquid and its temperature is raised by . What volume of the liquid will overflow? Coefficient of linear expansion of glass = and coefficient of volume expansion of the liquid = . Solution: Volume of the liquid over flown = Increase in the volume of the liquid  increase in the volume of the container = = = Thermal Stress When a rod of length L is held between two rigid supports and the temperature of rod is increased by , the rigid supports prevent the rod from expanding. This causes compressive stress in the rod. As the length of the rod remains unchanged, Thermal expansion = Mechanical compression Thermal stress , which is compressive. KINETIC THEORY OF GASES Fluids consist of molecules which are in random motion, which collide with each other and with the walls of the container. The intermolecular separation in a gas is larger by an order of magnitude than the intermolecular separation in a liquid. The intermolecular separation in a gas increases as the temperature increases, and decreases with increasing density. it is experimentally observed that most gases follow a universal equation of state pV = n RT Gas molecules are relatively free: the interactions between them are small. This means that the total energy of the gas molecules is mostly kinetic. A simple model of a gas, with point particles representing molecules and their collisions with the walls of the confining vessel causing the pressure exerted by the gas leads to some important conclusions: (a) The pressure exerted by an ideal gas is given by p =  mc , where  = number of molecules per unit volume, m = mass of each molecule, crms = the rms speed of each molecule. (b) The total internal energy of an ideal gs is U = Nf where N = number of molecules in the gas, f = number of degree of freedom, T = absolute temperature of the gas, kB = = Boltzmann’s constant. (C) The rms speed is given by crms= , where M represents the molar mass of the gas (i.e. mass of 1 mole of the gas) (d) The pressure of an ideal gas is given by p = , where Utran is the translational energy of the molecules of the gas. Example6: Find r.m.s speed of Hydrogen molecules at room temperature (=300 k). Solution: Mass of 1 mole of Hydrogen gas= 2 gm = 2 10-3 kg  Vrms = = = 1.93  103 m/s. 2. When the container contains more than one gas, total pressure exerted by all the gases on the wall is sum of pressures exerted by each gas as it would while filling the container alone. In a way, each gas behaves independent of each other. Thus we have P = P1+P2+P3+……, where P1, P2, P3 are the partial pressures of gases 1, 2 & 3 respectively This is known as Dalton’s Law of partial pressures. 3. One mole of any gas occupies a volume of 22.4 litre at standard temperature and Pressure which are 273.15 (=0°C) and 1.013105 Pa (=1atm) respectively, Example 7: 4 gm Hydrogen is mixed with 11.2 litre of He at S.T.P. in a container of volume 20 litre. If the final temperature is 300 K find the pressure. Solution: 4 gm Hydrogen = 2 moles Hydrogen 11.2  He at S.T.P. = 1/2 mole of He P = PH + PHe = (nH+nHe) = (2+½) = 3.12 105 N/m2. Internal Energy Internal energy, of any body is sum total of kinetic energies and potential energies of its constituents (at molecular level). In case of an ideal gas, as there are no intermolecular forces, except during collision the possibility of potential energy is ruled out, so it is only kinetic energy. The kinetic energy of the molecules can be of three types. (i) Translational (ii) Rotational (iii) Vibrational In a way, it means that the energy of molecules is shared in various modes. These independent modes of motions are called degrees of freedom. The table given below gives the number of degrees of freedom for various types of molecules at normal temperature. Nature of motion Atomicity Degree of Freedom (f) Translational Rotational Vibrational* Total (1) Monoatomic 3 0 0 3 (2) Diatomic 3 2 0 5 Poly Linear 3 2 0 5 Non-linear 3 3 0 6 * At room temperature the energy associated with vibrational motion is negligibly small in comparison to translational and rotational K.E. Equipartition of Energy According to the Law of equipartition of energy the average K.E. of a molecule is equally shared among different degrees of freedom. The average energy per degree of freedom of a molecule is ½ kT, where k is the Boltzmann’s constant and T is the absolute temperature. Thus, for a monoatomic ideal gas: U (the internal energy) = kT Also for one mole U = RT Example 8: Find the average kinetic energy per molecule at temperature T for an equimolar mixture of two ideal gases A and B where A is monoatomic and B is diatomic. Solution: No. of degrees of freedom per molecule for A = 3 No. of degrees of freedom per molecule for B = 5 Since the mixture is equimolar, the average kinetic energy per molecule will be given by the average of the two values i.e. kT = 4kT where k is Boltzmann’s constant. Work Done In Different Processes Work done by an enclosed gas on its surroundings is given by W = p dV For different types of processes, we have got different relations between p (Pressure) and V (Volume) and accordingly we have different expressions for work. (a) Isochoric Process Here, the volume is constant throughout and therefore the work done by the gas, irrespective of the manner in which pressure varies, is zero Wisochoric = 0 (b) Isobaric Process In this case, pressure of the gas remains constant throughout the process. Hence,  pdV = p V = n RT(n = number of moles) ( T= change in absolute temperature) (C) Isothermal process The temperature remains constant throughout the process. Using ideal gas equation, we get, p= Hence,  pdV = n RT = n RT (d) Adiabatic Process: pV = Constant = C (say)  p = CV-  W =  pdV = = = nCv(T2  T1) The work done by a gas can also be evaluated from the p-V diagram of the process. Area enclosed by the curve in a p-V diagram = work done by the gas First Law of Thermodynamics First law of thermodynamics is simply a re-statement of the principle of conservation of energy for a thermally isolated system. If Q, U & W represent the heat given to the system, change in its internal energy and the work done by the system respectively, the first law of thermodynamics states that, Q = U + W The heat transferred to the system (Q) is either utilised to do work (W) or increase the internal energy of the system (U). Example 9: 3000 J of heat is given to a gas at constant pressure of 2  105 N/m2. If its volume increases by 10 litres during the process find the change in the internal energy of the gas Solution: Q = 3000 J W = P  V = (2105 N/m2) (10 10-3m3) = 2  103 J U = Q – W = 3000  2000= 1000 J. Specific Heat Capacities of Gases S = where Q = amount of heat required for ‘T’ temperature change. m = mass of the gas. In case of gases, the concept of a molar heat capacity is useful. Molar heat capacity is the amount of heat required to raise the temperature of one mole of the gas by one degree. So, if Q amount of heat goes to change the temperature of ‘n’ moles of a gas in a particular process, molar heat capacity ‘C’ can be mathematically given by: C = In terms of differentials, C = Two special cases are:- (i) If volume is kept constant during the process then CV = This is the molar heat capacity of the gas at constant volume Note: Since U is independent of the process. U = n Cv T is true for all processes. (ii) If pressure remains constant, then Cp = This is the molar heat capacity of the gas at constant pressure Relation Between Cp and Cv Cp  Cv = R This is known as Mayer’s relation. The Values of Cp and Cv If f is the number of degrees of freedom of a gas molecule then the internal energy of n moles of that gas is given as U = f/2 n RT  U = f/2 n RT = n CvT  Cv = f/2 R From Mayer’s Relation Cp = Cv + R Cp = (f/2+1)R And the ratio of specific heats  = =  = = Example 10: Find the molar heat capacity of an ideal gas with adiabatic exponent ‘’ for the polytropic process P V  = constant. Solution: We have, from first law of thermodynamics C = Cv + (n = number of moles) We have, P V = constant From Ideal gas equation P V = n RT Taking ratio, T V1 = Constant Differentiating we get =  Putting it in the equation for ‘C’. C = Cv  = Cv  = Cv  C = Second law of thermodynamics (i) Kelvin Statement:- It is impossible to derive a continuous supply of work by cooling a body to a temperature lower than that of the coldest of its surroundings. (ii) Clausius Statement:-It is impossible for a self acting machine, unaided by an external agency to transfer heat from a body to another at higher temperature. Reversible Process: A process which can be made to proceed in the reverse direction by variations in its conditions so that all changes occurring in any part of the direct process are exactly reversed in the corresponding part of the reverse process is called a reversible processes. Irreversible Process: A process which can not be made to proceed in the reverse direction is called an irreversible process. Heat Engine: It is a device which continuously converts heat energy into the mechanical energy in a cyclic process. Efficiency of heat engine:  = = Where Q1 is the heat supplied by the source and Q2 is the heat rejected to the sink. Carnot Engine: It is an ideal heat engine which is based on Carnot's reversible cycle. It works in four steps viz. Isothermal expansion, adiabatic expansion, isothermal compression and adiabatic compression. The efficiency of a Carnot engine is given by  = 1  = 1  where T1 and T2 are the temperatures of source and sink respectively. Example 11: The efficiency of a Carnot cycle is 1/6. If on reducing the temperature of the sink by 650C, the efficiency becomes 1/3, find the initial and final temperatures between which the cycle is working. Solution: Given 1 = , 2 = If the temperatures of the source and the sink between which the cycle is working are T1 and T2, then the efficiency in the first case will be 1 = 1 - = In the second case 2 = 1  = Solving T¬1 = 390 K and T2 = 325 K. OBJECTIVE 1: Calculate the root mean square speed of smoke particles of mass kg in Brownian motion in air at NTP. Boltzmann constant (A) 1.5 cm/s (B) 2.2 cm/s (C) 2.3 cm/s (D) 4.4 cm/s Ans. (a) Solution: PV =  = where  = mass of one molecule.  = cm/s 2: During an experiment an ideal gas is found to obey an additional law = constant. The gas is initially at temp T and volume V. What will be the temperature of the gas when it expands to a volume 2V? (A) (B) (C) (D) Ans. (b) Solution: According to the given problems VP2 = constant From the gas law PV = nRT    i.e,  Q.3-5 We have two vessels of equal volume, one filled with hydrogen and the other with equal mass of Helium. The common temperature is 27oC. 3: What is the relative number of molecules in the two vessels ? (A) (B) (C) (D) Ans. (C) 4: If pressure of Hydrogen is 2 atm, what is the pressure of Helium ? (A) pHe = 2 atm. (B) pHe = 3 atm. (C) pHe = 4 atm. (D) pHe = 1 atm. Ans. (d) 5: If the temperature of Helium is kept at 27o C and that of hydrogen is changed, at what temperature will its pressure become equal to that of helium ? The molecular weights of hydrogen and helium are 2 and 4 respectively. (A) 123oC (B) 140oC (C) 160oC (D) 183oC Ans. (a) Solution 3-5: 3. The masses of hydrogen and helium gases in the vessels are equal. This means that the product of the number of molecules and the mass of a molecule must be same for H2 and He gases. Since molecular masses of H2 and He are in the ratio 1: 2, their number of molecules nH and nHe in the vessels must be in the reverse ratio, that is, 4. The equation of state for one mole of a gas is pV = RT = NkT Where N is Avogadro’s number (no. of molecules in one mole) and k is Boltzmann’s constant. If a gas has n molecules, the equation of state will be pV = nkT For a given volume and a given temperature, we have p  n. Since H2 and He have same volume and same temperature (27oC), we have Here pH = 2 atm.  pHe = 1 atm. 5. Again, we have pV = nkT H2 and He have equal volumes. For having equal pressure, we must have nHTH = nHeTHe or Here THe = 27 + 273 = 300 K  TH = THe = 150 K = 150  273 = 123oC 6: A vessel contains a mixture of 7 gm of nitrogen and 11 gm of carbon dioxide at temperature T = 290 K. If pressure of the mixture P = 1 atm, calculate its density (R = 8.31 J/mol k) (A) 2.5 kg/m3 (B) 1.5 kg/m3 (C) 4.5 kg/m3 (D) 7.5 kg/m3 Ans. (b) Solution: As molecular weight of N2 and CO2 are 28 and 44, and , So, Now, according to gas law PV = = and m = 7+11 = 18 gm = 18  10-3 kg so,  = Q.7-10.The pressure of a monoatomic gas increases linearly from N/m2 to N/m2 when its volume increases from 0.2 m3 to 0.5 m3. Calculate 7: work done by the gas (A) 2.8×105 J (B) 1.8×106 J (C) 1.8×105 J (D) 1.8×102 J Ans. (C) 8: increase in internal energy (A) (B) (C) (D) Ans. (a) 9: amount of heat supplied (A) (B) (C) (D) Ans. (C) 10: molar heat capacity of the gas [R = 8.31 J/mol k] (A) 20.1 J/molK (B) 17.14 J/molK (C) 18.14 J/molK (D) 20.14 J/molK Ans. (b) Solution 7- 10: 7. Work done by the gas, = area under P-v curve = = = 8. Change in internal energy of a gas is given by As the gas is monoatomic,  = 5/3 So  9. From 1st law of thermodynamics = 10. Molar heat capacity is defined as i.e, C = J/molK Q.11-13.Two moles of Helium gas ( = 5/3) are initially at temperature 27oC and occupy a volume of 20 litres. The gas is first expanded at constant pressure until the volume is doubled. Then it undergoes an adiabatic change until the temperature returns to its initial value. 11: What are the final volume. (A) (B) (C) (D) Ans. (A) 12: What are the final pressure of gas? (A) (B) (C) (D) Ans. (a) 13: What is the work done by the gas? (Gas constant R = 8.3 T/mole K) (A) (B) (C) (D) Ans. (d) Solution:11. From ideal gas equation PV = nRT initial pressure When volume of gas is doubled at constant pressure, its temperature is also doubled. This process is shown on P-V curve by line AB. The gas then cools to temperature T adiabatically. This is shown by curve BC. The whole process is represented by curve ABC. At point B, pressure Volume , Temperature Now from adiabatic equation = constant We have = 23/2 Final volume 12. final pressure 13. The work done by gas in isobaric process AB The work done by gas during adiabatic process BC . Net work done by gas 14-15. When 1 gm of water changes from liquid to vapour phase at constant pressure of 1 atmosphere, the volume increases from 1 cm3 to 1671 c.c. The heat of vaporization at this pressure is 540 cal/gm. Find 14: The work done (in J) in change of phase (A) Joule (B) Joule (C) 190. 78 Joule (D) Joule Ans. (d) 15: Increase in internal energy of water. (A) 2099.33 J (B) 3099.33 J (C) 4099.33 J (D) 5099.33 J Ans. (a) Solution:14. As the process is isobaric = = Joule [1 erg = 10-7J] 15. From 1st law of thermodynamics cal = 2268 J, [ 1 cal = 4.2J] so, = = 2099.33 J 16: A glass flask of volume one litre at is filled level full of mercury at this temperature. The flask and mercury are now heated to 100oC. How much mercury will spill out if coefficient of volume expansion of mercury is and linear expansion of glass is respectively? (A) 14.2 c.c. (B) 15.2 c.c. (C) 18.2 c.c. (D) 20.2 c.c.J Ans. (b) Solution: In case of thermal expansion of liquid, change in volume of liquid relative to container is given by Here V = 1 litre = 1000 c.c. = 3 glass = So, = 15.2 c.c. 17: Two cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at 300K. The piston A is free to move, while that of B is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in A is 30K, then the rise in temperature of the gas in B is. (A) 30K (B) 18K (C) 50K (D) 42K Solution: For cylinder A. For cylinder B dQ = nCPdT1 dQ = nCvdT2  nCPdT1 = nCvdT2 From (I) and (II) For diatomic gas . 18: 80 gm of water at is poured on a large block of ice at . The mass of ice that melts is (A) 30 gm (B) 80 gm (C) 150 gm (D) 1600 gm Solution: Since the block of ice at is large, the whole of ice will not melt, hence final temperature is .  = heat given up by water in cooling up to = = 2400 cal If m gm be the mass of ice melted, then = ML = Here A is correct. 19: A gas at pressure is contained in a vessel. If the masses of all the molecules are halved and their speeds doubled, the resulting pressure would be (A) 4Po (B) 2Po (C) Po (D) Solution: where m = mass of one gas molecules n = total no. of gas molecules i.e, P  m and P  Vrms Here m is halved and Vrms is doubled  pressure will be doubled Hence, (B) is correct 20: The volume V versus temperature T graphs for a certain amount of a perfect gas at two pressures P1 and P2 are shown in the figure. Here (A) P1 < P2 (C) P1 = P2 (B) P1 > P2 (D) can’t be Solution: For a perfect gas,  So, the slope of the graph is Slope  Hence P1 > P2 Hence, (C) is correct 21: At room temperature the rms speed of the molecules of a certain diatomic gas is found to be 1930 m/s. The gas is (A) (B) (C) (D) Solution: It is molecular weight of hydrogen . 22: The latent heat of vaporization of water is 2240 J. If the work done in the process of vaporization of 1 gm is 168 J, then increase in internal energy is (A) 2408 J (B) 2240 J (C) 2072 J (D) 1904 J Solution: L = 2240 J, m = 1 gm dW = 168 J dQ = mL = dU + dW or dU = 2072 J Hence, (C) is correct 23: For a gas, y = 1.286. What is the number of degrees of freedom of the moleculas of this gas ? (A) 3 (B) 5 (C) 6 (D) 7 Ans. (d) Solution: (D) 24: Which of the following temperatures is the highest? (A) 100 K (B) –13oF (C) –20oC (D) –30oC Solution: (B ) –13oF is (13+32)o below ice point on F scale. 25: An ideal gas ( = 1.5) is expanded adiabatically. How many times has the gas to be expanded to reduce the root mean square velocity of molecules 2.0 times (A) 4 times (B) 16 times (C) 8 times (D) 2 times Ans. (B) Solution:  Vrms is to reduce two times i.e, temperature of the gas will have to reduce four times or During adiabatic process or,  Hence, (B) is correct 26: A thin copper wire of length L increases in length by 1% when heated from to . If a thin copper plate of area is heated from to , the percentage increase in its area will be (A) 1% (B) 2% (C) 3% (D) 4% Ans. (b) Solution: L = Lo  = percentage increase in length =  Hence or Hence, (B) is correct 27: Gas at pressure Po is contained in a vessel. If the masses of all the molecules are doubled and their speed is halved, the resulting pressure P will be equal to (A) 2Po (B) Po/4 (C) Po (D) Po/2 Ans. (d) Solution: Po = P = where m = 2m, putting the value  = P = Po/2 28: The molar heat capacity in a process of a diatomic gas if it does a work of Q/4, when Q amount of heat is supplied to it is (A) (B) (C) (D) Ans. (C) Solution: dU = CV dT = From 1st law of thermodynamics dU = dQ – dW or dU = Now molar heat capacity Hence (C) is correct 29: For an ideal gas: (A) the change in internal energy in a constant pressure process from temperature T1 to T2 is equal to nCv (T2 - T1), where Cv is the molar specific heat at constant volume and n the number of moles of the gas. (B) the change in internal energy of the gas and the work done by the gas are equal in magnitude in an adiabatic process. (C) the internal energy does not change in an isothermal process. (D) no heat is added or removed in an adiabatic process. (A) A, B (B) A, B, C (C) A, B, C, D (D) A, C Solution: (C) Change in internal energy depends only on change in temperature since internal energy is a function of state only i.e. dU = nCv,dT. In adiabatic process, dQ = 0, Hence, dU + dW = 0  dU =  dW i.e. magnitude of change in internal energy is equal to magnitude of work done. 30: Heat is supplied to a diatomic gas at constant pressure. The ratio of is (A) 5:3:2 (B) 5:2:3 (C) 7:5:2 (D) 7:2:5 Ans. (C) Solution: , , and  Hence, C is correct 31: Two mole of argon are mixed with one mole of hydrogen, then Cp/Cv for the mixture is nearly (A) 1.2 (B) 1.3 (C) 1.4 (D) 1.5 Ans.(C) Solution: Average degree of freedom fav = mix = 1 + = 1 + = = 1.4 Hence, C is correct Answer. 32: An ideal gas is taken through the cycle A BC A as shown in figure. If the net heat supplied to the gas in the cycle is 5J, the work done by the gas in the process CA is, (A) -5J (B) -10J (C) -15J (D) -20J Solution: (A) 33: When an ideal gas at pressure P, temperature T and volume V is isothermally compressed to a V/n, its pressure becomes Pi . If the gas is compressed adiabatically to V/n, its pressure becomes Pa. The ratio Pi / Pa is (A) 1 (B) n (C) n  (D) n 1- Solution: (D) For isothermal process, PV = constant. Therefore PiVi = PV or Pi or Pi = nP ……(i) For adiabatic process, PV = constant. Therefore Pa (Va)  = PV or  or Pa = nP …..(ii) From (i) and (ii) we get 34: When an ideal monatomic gas is heated at constant pressure, the fraction of heat energy supplied which increases the internal energy of the gas is (A) 2/5 (B) 3/5 (C) 3/7 (D) 3/4 Solution: (B) [for monoatomic gas  = 5/3 ] 35: A monatomic ideal gas, initially at temperature T1 is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature T2 by releasing the piston suddenly. If L1 and L2 are the lengths of the gas column before and after expansion respectively, then T1/T2 is given by (A) (L1/L2)2/3 (B) L1/L2 (C) L2/L1 (D) (L2/L1) 2/3 Solution: (D) Initial position T1(L1A)–1 = constant Final position T2 (L2A)–1 = constant [for monoatomic gas  = 5/3] 36: An ideal mono atomic gas at 300K expands adiabatically to twice its volume. What is the fine temperature (A) 189K (B) 289K (C) 30Kj (D) Non of these Solution: (A) 37: What will be P-V graph corresponding to the P-T graph (process AB) for an ideal gas shown in figure (A) Hyperbolic (B) Circle (C) Straight line (D) Elliptical Solution: (A) 38: Starting with the same initial conditions, an ideal gas expands from volume V1 to V2 in three different ways. The work done by the gas is W1 if the process is purely isothermal, W2 if purely isobaric and W3 if purely adiabatic. Then (A) W2 > W1 >W3 (B) W2 > W3 >W1 (C) W1 > W2 >W3 (D) W1 > W3 >W2 Solution: (A) 39: One mole of argon is heated using PV5/2 = = const. By which amount of heat is obtained by the process when the temperature change by T = = -26K. (A) 100J (B) 200J (C) 108J (D) 208J Solution: (C) Here n = 1 C = C = 40: 3 moles of an ideal monoatmic gas performs a cycle as shown in the fig. The gas temperatures T1 = = 400K, T2= = 800K, T3 = = 2400K, T4 = = 1200K. What will be the net work done. (A) 20J (B) 20000J (C) 200J (D) 2000J Solution: (B) WAB = CCD = 0, because process is isochoric. WAD = nRT = 3R(TA – TB) = 3  8.31 (2400 – 800) = 39884  Total work done WAD + WBC = 39888 – 19944 = 20  103J = 19944  20  103J 41: How much heat is absorbed by the system in going through the process shown in the fig. (consider that value is taken in SI system) (A) 20.4104 J (B) 30.4104 J (C) 21.4104 J (D) 25.12104 J Solution: (D) Work done = Area of PV diagram = = = 800  100  3.14 = 8  104  3.14 = 25.12  104 42: 3000J of heat is given to a gas at constant pressure of 2105 N/m2. If its volume increases by 10 litres during the process, what will be the change in the internal energy of the gas (A) 1000J (B) 100J (C) 200J (D) 2000J Solution: (A) P = 2  105 = Vi Vf = (Vi + 10  10–3) W = Pdv W = 2  105  10  10¬-3 = 2  103 3000 = 2  103 + 43: A gas at atmospheric pressure is contained in a cylinder of volume 80 litre. When it is compressed adiabatically to 20 litre its pressure rises to 7 atm. What will be the ratio of specific heats of the gas (A) 1.33 (B) 1.4 (C) 1.67 (D) 1.5 Solution: (B) Pi = 1atm = 1  105 N/m2 Vi = 80  10 – 3m3 Vf = 20  10–3 m3 Pf = 7atm = 7  105 N/m2 PiVi = PfVf 1  105  (80  10–3)  =(7  105)(20  10–3)  44: A gas consisting of rigid diatomic molecules was initially under standard conditions. Then gas was compressed adiabatically to one fifth of its initial volume. What will be the mean kinetic energy of a rotating molecule in the final state? (A) 1.44 J (B) 4.55J (C) 787.98  10–23 (D) 757.310-23J Solution: (C) (300)V1 7/5 – 1 = T2 = 571 Mean kinetic energy of rotating molecules = KT = 1.38  10 – 23  571 KT = 787.98  10–23 45: Immediately after the explosion of an atom bomb, the ball of fire produced has a radius of 100m and a temperature 105K . What will be the approximate temperature when the ball expands adiabatically to a radius of 1000m (suppose mono atomic gas is there) (A) 1000K (B) 100K (C) 105  (10–3)2/3 (D) 200K Solution: (C) r = 100 m  Vi = Ti = 105 K after explosion r = 1000 m,  Vf = Tf = 105 = 105  (10–3)2/3 = 1.05 K 46: Which of the following is false? (A) Enthalpy is a path function. (B) Work is a path function. (C) Heat is a path function. (D) Energy is a state function Solution: (C) 47: A gas mixture consists of 32 gram of oxygen and 36 gram of Ar a temperature T. Neglecting all vibration modes, the total internal energy of the system is (A) 4RT (B) 8RT (C) 9RT (D) 11RT Solution: (D) Energy = n 48: A mono atomic gas is supplied heat Q very slowly keeping the pressure constant. The work done by the gas is (A) 2/5 Q (B) 3/5 Q (C) Q /5 (D) 2/3 Q Solution: (D) For monatomic gas From the first law of thermodynamics Q = U + W 49: Which of the following parameters does not characterise the thermodynamic state of matter (A) Work (B) Pressure (C) Temperature (D) Volume Solution: (A) Work 50: Which of the following is correct (A) For an isothermal change PV = = constant (B) For a isothermal process, the change in internal energy must be equal to the work done (C) For an adiabatic change , where  is the ratio of the two specific heats (D) In an adiabatic process external work done must be equal to the heat entering the system Solution: (A) 51: An ideal gas goes through cyclic process ABC and following (P vs T) curve is obtained. This process can be represented by (A) (B) (C) (D) Solution:(B) Process A to B isothermal. Then P  . Process B  C is isobaric and C  A adiabatic. Slope of adiabatic > slope of isothermal. 52: A container contain 0.1 mol of H2 and 0.1 mol of O2 , If the gases are in thermal equilibrium then (A) Only the average kinetic energy of the molecule of H2 and O2 is same. (B) Average speed of the molecule of H2 and O2 is same. (C) Only the specific heat at constant pressure of two gases is same. (D) (d) The specific heat at constant pressure and the kinetic energy are same for both the gases. Solution: (D) The specific heat at constant pressure (Cp) is the amount of heat required to raise the temperature of one gram through 1°C when the pressure of the gas is kept constant. Again, the mean kinetic energy per molecule (3/2)kT depends only upon temperature. Clearly both the specific heats at constant pressure and mean kinetic energy are depending on the temperature which is again same for the two gases. 53: Two systems are in thermal equilibrium. The quantity which is common for them is (A) Heat (B) Momentum (C) Temperature (D) Specific heat Solution: (C) 54: Mean molecular weight is defined as (A) the number of free particles per positron mass (B) the number of free particles per electron mass (C) the number of free particles per neutral mass (D) the number of free particles per photon mass Solution: (D) 55: Which one of the following statements is true about a gas undergoing an adiabatic change (A) The temperature of the gas remains constant (B) The pressure of the gas remains constant (C) The volume of the gas remains constant (D) The gas is completely insulated from the surroundings Solution: (D) 56: If an ideal gas is allowed to expand adiabatically, the work done is equal to (A) The loss in heat (B) The loss in internal energy (C) The gain in internal energy (D) The gain in enthalpy Solution: (B) 57: For the Boyle’s law to hold, the necessary condition is (A) Isothermal (B) Adiabatic (C) Isobaric (D) Isochoric Solution: (A) 58: Specific heat of a gas undergoing adiabatic changes is (A) zero (B) infinite (C) positive (D) negative Solution: (B) 59: The internal energy of the system remains constant when it undergoes (A) a cyclic process (B) an adiabatic process (C) an isothermal process (D) an isobaric process Solution: (C) 60: The first law of thermodynamics incorporates the concepts of (A) conservation of energy (B) conservation of heat (C) conservation of work (D) equivalence of heat and work Solution: (D)