(Type of intermolecular force) dipole interaction

Transcription

1 Q. Match column-i with column-ii No. Column A (Pair of molecules) Column B (Type of intermolecular force) Two molecules of Hydrogen bonding HCI Two propane Dipole induced molecules dipole interaction 3 CH 4 and HCI Dipole-dipole molecules interaction 4 Two molecules of London dispersion NH3 force Explanation (A-R, B-S, C-Q, D-P) Column C Id of Additional Answer Dipole-dipole interaction London dispersion force Dipole induced dipole interaction Hydrogen bonding NH 3 shows hydrogen Bonding CH 4 Symmetrical molecule (non-polar molecule) London or dispersive forces exist Propane similar reason as in CH 4 Q. Which of the following exhibits the weakest intermolecular force? Option NH 3 Option HCI Option 3 He Option 4 H O Correct Answer 3 Explanation Except He, all are polar In He, only dispersive forces act which is weakest force Q.3 Match column-i with column-ii No. Column A Column B Column C Id of Additional Answer Plot of V against P for ideal gas at constant temperature and number of moles Plot of V against for ideal gas at T constant pressure and number of moles

2 3 Plot of P against T for ideal gas at constant volume and number of moles 4 Plot of V against for ideal gas at P constant temperature and number of moles Explanation (A-R, B-S, C-P, D-Q) A) at constant temperature and number of moles PV = K constant K K R P= = V V V i P P = = = V V K V Assume = y andp = x, V B) PV = nrt Assume V = y, =x T= T x p y=nr x nr y= (since n and p is constant) p x

3 C) PV = nrt nr V = T P d) PV = nrt similar to (A) Q.4 A thin tube sealed at both ends, is 00 cm long. It lies horizontally, the middle 0. m containing mercury and the two ends containing air at standard atmospheric pressure. If the tube is turned to a vertical position, by what amount will the mercury be displaced? (in cm) Correct Answer 0003 Is Integer Type Explanation In this case Boyle s law should be applicable

4 Assume cross-sectional area of then tube is a Applying Boyle s law on A ( ) ( ) a =P 45+y P = 45+y Similarly for B a =P 45-y ( ) ( ) P = 45-a Now, P and P are related by following equation : P + 0 = P = 45+y 45-y =0 45-y 45+y y 3cm y +684y-45 =0 Q.5 Air is trapped in a horizontal glass tube by 40 cm mercury column as shown below: If the tube is held vertical keeping the open end up, length of air column shrink to 9 cm. length in cm by which the mercury column shifts down is Correct Answer 000 Is Integer Type Explanation 0 cm

5 Let by y cm mercury column shift down Then, in horizontal position, length of air column = (y + 9) Applying Boyle s law on air column P V = P V ( ) ( ) ( ) 76 y+9 a= a Where a is cross- sectional area. ( ) ( ) 76 y+9 = y + 76 = 6 4y = 40 y =0cm Q.6 Calculate the number of moles in the glass bulb shown in figure at 300 K. Given: d glycerin =.7 g/ml; d mercury = 3.6g/mL Option 0.36 mol Option 0.94 mol Option mol Option mol Correct Answer Explanation P ges = P atm + f gh h gly = 5m If glycerin is replaced by Hg f gly h gly = f Hg h Hg fgly hges.7 h Hg = = 50 fhg 3.6 P ges = 76cm + 00cm 76 P= atm, v = 0lt, T = 300K 76

6 76 0 PV n= = 76 =0.94m/s RT Q.7 Which one of the following plot will be a parabola at constant temperature? Option P vs /V Option PV vs P Option 3 V vs P Option 4 None of these Correct Answer 4 Explanation At constant temperature PV = constant = k Q.9 To raise the volume of the gas by four times, which of the following methods are correct? Option T is doubled and P is also doubled Option Keeping P constant, T is raised by four times Option 3 Temperature is doubled and pressure is halved Option 4 Keeping temperature constant, pressure is reduced to /4th of its initial value. Option 5 Option 6 Explanation nrt PV=nRT V= P ( ) nr( T) a V final = =V P nr( 4T) b V final = =4V P nr( T ) c V final = =4V P nrt d V final = =4V P ( ) ( ) ( ) Q.0 Among the plots, which of them represent Charle s law, at constant pressure?

7 Option Option Option 3 Option 4 Explanation PV = nrt nr V = T P V = KT

8 logv = logt + logk y = mx + c Assume, T = y =x v= V x =ky y= x kx Q. The graph of P vs V is given at different temperatures and number of rule curves, n, n, n 3 are number of moles The correct relationship are Option T > T > T 3 Option T < T < T 3 Option 3 n < n < n 3 Option 4 n 3 < n < n Explanation

9 Q. Which of the following curves represent (s) Boyle s law? Option Option Option 3 Option 4 Explanation K PV=K P= V V = P K

10 Q.3 Assertion: The pressure of a gas is inversely proportional to the volume at constant temperature and n. Reason: The gas volume is directly proportional to n at constant temperature and pressure. Option If both assertion and reason are correct and reason is the correct explanation of assertion. Option If both assertion and reason are true but reason is not the correct explanation of assertion Option 3 If assertion is true but reason is false Option 4 If reason is true but assertion is false Correct Answer Explanation Assertion is Boyle s law Reason is Avogadro s law Q.4 Assertion: for a given amount of an ideal gas, a plot of pressure (p) versus (/V) is a straight line at a constant temperature. Reason: A given amount of an ideal gas at a constant temperature obeys Boyle s law. Option If both assertion and reason are correct and reason is the correct explanation of assertion. Option If both assertion and reason are true but reason is not the correct explanation of assertion Option 3 If assertion is true but reason is false Option 4 If reason is true but assertion is false Correct Answer Explanation If both assertion and reason are true but reason is not the correct explanation of assertion Q.5 Assertion: According to Charles law, for a given amount of an ideal gas at a given pressure, plot of gas volume versus absolute temperature is a straight line passing through origin. Reason: Gaseous volume never vanishes. Option If both assertion and reason are correct and reason is the correct explanation of assertion. Option If both assertion and reason are true but reason is not the correct explanation of assertion Option 3 If assertion is true but reason is false Option 4 If reason is true but assertion is false Correct Answer Explanation If both assertion and reason are true but reason is not the correct explanation of assertion Q.6 Match the following

11 No. Column A Column B Column C Id of Additional Answer Temperature is increasing Temperature is constant 3 Volume is constant 4 Pressure is increasing Explanation A-P, S ; B-R,P,S ; C-Q ; D-P,R,S Assume that number of moles of gas is constant PV = nrt a) If P and V both increase, T also increase b) Follows Gay lussac's law d) Follows Gay lussac's law Q.7 A flask containing air (open to atmosphere) is heated from 300 K to 500 K. The percentage of air escaped to the atmosphere is nearly Option 6.6 Option 40 Option 3 60 Option 4 0 Correct Answer Explanation Open to atmospheres means- constant press vol. is constant (Assumed as not mentioned) PV = nrt nt=const n T = n T n = 00, T = 300k n = x, T = 500k =x 500 x = 60 % escaped = = 40%

12 Q.8 At N.T.P. the volume of a gas is found to be 73 ml. What will be the volume of this gas at 600 mm Hg and 73 0 C? Option 39.8 ml Option 380 ml Option ml Option ml Correct Answer 3 Explanation PV PV = T T 760mm 73ml 600mm V = 73k 546k V =69.6ml Q.9 The vapour density of a gas is.. The volume occupied by. g of this gas at N.T.P. is Option L Option.L Option 3.4L Option 4 0L Correct Answer Explanation VD od gass =. ; MW of gas =. molar mass of gas =.4 gm moles of gas in.gm. = =.4 volume of moles of gas of NTP = =. Q.0 If the weight of 5.6 litres of a gas at NTP is gm. The gas may be Option PH 3 Option COCI Option 3 NO Option 4 N O Correct Answer 4 Explanation gives mass vol.of gasatntp number of moles= = molar mass = = molar mass.4 molar mass of gas = 44 gm Hence gas is N O Q. An open flask contains air at 7 0 C temperature and one atm pressure. The flask is heated to 7 0 C at the same pressure. What fraction of the original air will remain in the flask? Option /5th Option /3rd Option 3 /7th Option 4 3/4th Correct Answer 4

13 Explanation Open container means, P and V is const. nt =const n T =n T n 300 =n 400 n = n Q. If the pressure is tripled and temperature (in kelvins) is halved, the volume of a given mass of an ideal gas becomes Option 3/ times its original volume Option /3rd of its original volume Option 3 /6th of its original volume Option 4 6 times its original volume Correct Answer 3 Explanation PV PV = T T PV 3P V = V = V T T / 6 Q.3 How many O molecules are present in.0 L of oxygen gas at 7 0 C temperature and 3.0 atm pressure? Option Option Option Option Correct Answer Explanation PV 3 n= = RT Number of molecules of oxygen 3 3 = = Q.4 If the absolute temperature of a gas is doubled and the pressure is reduced to onehalf, the volume of the gas will Option Remain unchanged Option Be doubled Option 3 Increase four-fold Option 4 Be reduced to /4th Correct Answer 3 Explanation PV PV = T T P V P V = V =4V T T Q.5 For an ideal gas, number of moles per litre in terms of its pressure P, gas constant R and temperature T is Option PT/R Option PRT

14 Option 3 P/RT Option 4 RT/P Correct Answer 3 Explanation PV = nrt n P = V RT number of moles per litre Q.6.0 litre of N and 7/8 litre of O at the same temperature and pressure were mixed together. What is relation between the masses of the two gases in the mixture? Option m =3m N O Option m =8m N O Option 3 m =m N O Option 4 m =6m N O Correct Answer 3 Explanation Volume (v) is directly proportional to moles (n) V = kn where k is const. 7 V n= ;n = ;n = 8 K N O k k 7 w N = 8;w = O k 8k Q.7 Five grams each of the following gases at 87 0 C and 750 mm pressure are taken. Which of them will have the least volume? Option HF Option HCI Option 3 HBr Option 4 HI Correct Answer 4 Explanation V n ( Avogadro'slaw ) w V MW V MW Least vol. highest MW Q.8 The vapour density of a gas is.. The volume occupied by. gm of the gas at N.T.P. is Option litre Option. litre Option 3.4 litre Option litre Correct Answer Explanation VD od gass =. ; MW of gas =.

15 molar mass of gas =.4 gm. moles of gas in.gm = =.4 volume of moles of gas of NTP = =. Q.9 Densities of two gases having same molar mass are in the ratio : and their temperature are in the ratio :, then the ratio of their respective pressure is Option : Option : Option 3 : Option 4 4: Correct Answer Explanation PV = nrt W PV= RT ( M MWofgas) M W PM= RT V PM= ρ RT P M ρ T P M = ρ T ρ T = ; = ρ T M = M PM PM = P =P ρ T T ρ Q.30 The density of a gas is.4 g/ml at one atmosphere pressure and 7 0 C. At what pressure will the gas have density thrice this value, the temperature is kept constant? Option Same pressure Option atmosphere Option 3 3 atmosphere Option 4 4. atmosphere Correct Answer 3 Explanation PM = ρ RT P ρ P P P = = f f P =3atm Q.3 The value (pv m ) depends only on Option Temperature Option Pressure Option 3 Molar mass Option 4 Volume Correct Answer Explanation PV = nrt

16 V P =RT n PV =RT m V m molar volume Q.3 A mixture of CO and CO is found to have a density of.5 g/l at 30 0 C and 740 torr. The composition of mixture is Option CO is 35.5% Option CO is 35.7% Option 3 CO is 64.5% Option 4 CO is 64.3% Explanation PV = nrt PM= ρ RT ρ RT M= = P = 38.3 x MWofco+ ( 00-x) M Molecular wtofmix = 00 x 8+ ( 00-x) = 00 x = 35.5 % Co = 35.5 % Co = 64.5 In general Avg MW of gaseous mix nm +nm +n3m = n +n +n Q.33 A vessel containing gm of oxygen at a pressure of 0 atm and a temperature of 47 0 C. It is found that because of a leak, the pressure drops to 5/8 th of its original value and the temperature falls to 7 0 C. Find the mass of oxygen that has leaked out in grams (0 - ) Correct Answer 0003 Is Integer Type Explanation PV = nrt 0 V = 0.08 ( ) 3 V = 0.08 lt. After weak PV n = = 8 O RT = 48 w O = 3gm= gm 48 3 Mass of gas that has leaked out

17 =- = gm 3 3 = 0.33 gm = gm gm Q.34 If an ideal gas at 00 K is heated to 09 K, the pressure increases by x%. x is Correct Answer 0009 Is Integer Type Explanation To solve the problem vol. is assumed to be P P = T T P = 00atm T = 0 0 k P =?, T = 09k 00 P = P =09atm % increase in Pressure = 9% Q g of a gas occupies 4 ml at STP. The gas could be Option N Option CO Option 3 C H 4 Option 4 N O 4 Explanation Similar to problem no = MW 400 MW = 8 Hence gas can be N, Co, C t Passage Text Density of gas is inversely proportional to temperature and directly proportional to pressure. P dt d = constant T P d T d T = P P Density at a particular temperature and pressure can be calculated by using ideal gas equation mass PV = nrt PV = molar mass (M) RT mass P M = RT volume P M = d RT PM =d RT Q.36 Which of the following has maximum density? Option O at 5 0 C and atm Option O at 0 0 C and atm Option 3 O at 73 0 C and atm Option 4 O at 0 0 C and atm

18 Correct Answer Explanation PM= ρ RT ( ρ isdensity ) PM ρ = RT ρ P ρ T Q.37 The density Of CO at atm and 73 K is Option.96 g L - Option. g L - Option 3.09 g L - Option 4.0 g L - Correct Answer Explanation atmand73k STP condition Vol. of mole of CO (44gm) will be.4 lt 44 - denaity( ρ ) =.96gm lt.4 Q.38 The density of gas is 3.8 gl - at STP. The density at 7 0 C and 700 mm Hg pressure will be Option 3.85 g L - Option 3.85 g ml - Option kg L - Option kg ml - Correct Answer Explanation molar mass of gas =3.8.4gm =85. 85gm MW of gas = 85 Density of 7 0 C and 700 mm Hg = gmlt Q.39 To which of the following gaseous mixtures in Dalton s law not applicable? Option Ne + He + SO Option NH 3 + HCI + HBr Option 3 O + N + CO Option 4 N + H + O Correct Answer Explanation Dalton s law is only applicable for non-reacting gaseous mixtures. Q.40 Equal weights of ethane and hydrogen are mixed in an empty container at 5 0 C. The fraction of the total pressure exerted by hydrogen is

19 Option : Option : Option 3 :6 Option 4 5:6 Correct Answer 4 Explanation P A =PT XA P H =PT X H w PH n H =X = = H P T n H +n w w CH lets assume w = 30 gm PH 5 5 = = P 5+ 6 T Q.4 A open ended mercury manometer is used to measure the pressure exerted by a trapped gas as shown in the figure. Initially manometer shows no difference in mercury level in both columns as shown in diagram. After sparking A dissociated according to following reaction A(g) B(g) + 3C(g) If pressure of gas A decreases to 0.9 atm. Then (Assume temperature to be constant and is 300 K) Option Total pressure increased to.3 atm Option Total pressure increased by 0.3 atm Option 3 Total pressure increased by.3 cm of Hg Option 4 Difference in mercury level is 8 mm. Explanation Initially pressure of A = 76cm of Hg = atm Final pressure A = =0.9atm A( g) B( g ) +3C( g) at t=0atm t=tfinal -x x 3x -x = 0.9 x = 0.

20 P A = 0.9atm, P B = 0.atm, P C =0.atm 3=0 P T = P A + P B + P C = =.3 a =0.3 76=.8 Passage Text Consider the adjacent diagram to answer these questions : Initially, flask A contained oxygen gas at 7 0 C and 950 mm of Hg, and flask B contained neon gas at 7 0 C and 900 mm. Finally, the two flasks were joined by means of a narrow tube of negligible volume equipped with a stopcork and gases were allowed to mixupfreely. The final pressure in the combined system was found to be 90 mm of Hg. Q.43 What is the correct relationship between volumes of the two flasks? Option V B = 3V A Option V B = 4V A Option 3 V B = 5V A Option 4 V B = 4.5V A Correct Answer Explanation After opening the stopcork, total number of moles of gases remain const. In bulb A, In bulb B, 950 VA 950 V n O = ; R 300 B ne R 300 After opening the stopcork 90 ( VA + VB ) Total moles, n= R VA 90 VB 90 ( V A+VB ) + = R 300 R 300 R V A + 950V B = 90V A + 90V B 40V A = 0V B V B = 4V A Q.44 If volume of flask B was measured to be 0L, mass of oxygen gas present initially in flask A was Option.00 g Option 4.00 g Option g Option g Correct Answer Explanation V B = 0L 0 V A = =.5L 4

21 950.5 n = 760 O W O =no 3gm = 4.06 gm 4gm Q.45 How many moles of gas are present in flask A in the final condition? Option 0.55 Option 0. Option Option 4 0. Correct Answer Explanation PV = nrt 90.5 PV P VA n= = = 760 = RT RT Q.46 Helium diffuses twice as fast as another gas B. If the vapour density of helium is, the molecular weight of B is Option 4 Option 8 Option 3 6 Option 4 4 Correct Answer 3 Explanation VD of He = ; MW of He = VD=4 r M M = = r M 4 He B B B He M =6 B Q.47 The atomic weight of helium is 4 times of hydrogen. Its rate of diffusion as compared to hydrogen is Option Twice Option times Option 3 times Option 4 /4 th Correct Answer Explanation rhe = = r 4 H He r = r H Q.48 X ml of H gas effuses through a hole in a container in 5 seconds. The time taken for the effusion of the same volume of the gas specified below under identical condition is

22 Option 0 seconds : He Option 0 seconds : O Option 3 5 seconds : CO Option 4 55 seconds : CO Correct Answer Explanation Rate of diffusion = V r t t M = = = r V t M t t M = 5 If t = 0, M = 3 Volume diffused time taken Q ml of gas A diffuses through a membrance in the same times as for the diffusion of 40 ml of gas B under identical pressure - temperature conditions. If the molecular weight of a is 64 that of B would be Option 00 Option 50 Option 3 00 Option 4 80 Correct Answer Explanation V r t M ( B) = = r V M ( A) t ( ) 50 M B = M ( B) 5 = M ( B ) = Q.50 In the following questions, a statement of assertion (A) is followed by a statement of reason (R) Assertion : A lighter gas diffuses more rapidly than a heavier gas. Reason : At a given temperature, the rate of diffusion of a gas is inversely proportional to the density. Option If both Assertion & Reasons are true and the reasons is the correct explanation of the assertion Option If both Assertion & Reasons are true but the reasons is not the correct explanation of the assertion Option 3 If Assertion is true statement but Reason is false Option 4 If both Assertion and Reason are false statement Correct Answer 3 Explanation r where d is density d Q.5 If a gas expands at a constant temperature Option The pressure decreases

23 Option The kinetic energy of the molecule remains the same Option 3 The kinetic energy of the molecule decreases Option 4 The number of molecules of the gas increases Explanation At constant temperature for same gas Boyle s law is applied V P KE T If T is constant, KE is constant Q.54 At the same temperature and pressure, which of the following gases will have the highest kinetic energy per mole? Option Hydrogen Option Oxygen Option 3 Methane Option 4 All the same Correct Answer 4 Explanation KE per mole = 3 RT Independent of matter of gas Q.55 The kinetic energy of one mole of gas is given by the expression : K.E. = 3PV and K.E. = 3 RT Hence, it can be said that Option K.E. α P at constant temperature Option K.E. α T at constant pressure Option 3 K.E. is not directly proportional to volume at constant temperature Option 4 K.E. α V at constant temperature Explanation At constant P, V T T V T V KE T at constant P. At constant T, KE per mole is constant i.e 3 RT Q.56 Assertion : According to kinetic theory of gases, gas molecules occupy some space. Reason : Gases can be liquefied and solidified. Option If both Assertion & Reasons are true and the reasons is the correct explanation of the assertion Option If both Assertion & Reasons are true but the reasons is not the correct explanation of the assertion Option 3 If Assertion is true statement but Reason is false Option 4 If both Assertion and Reason are false statement Correct Answer 4 Explanation According to kinetic theory if gases, vol. of gas molecules are negligible as compared to vol. of container Real gases can be liquefied and solidified

24 Q.57 Assertion : For a real gas, the gas pressure is always less than the pressure calculated from kinetic theory of gases. Reason : In kinetic theory, both molecular volumes and intermolecular attractions were ignored. Option If both Assertion & Reasons are true and the reasons is the correct explanation of the assertion Option If both Assertion & Reasons are true but the reasons is not the correct explanation of the assertion Option 3 If Assertion is true statement but Reason is false Option 4 If both Assertion and Reason are false statement Correct Answer 4 Explanation Follow theory Q.58 The molecules of which of the following gases has the highest speed? Option O at 0 0 C Option N at C Option 3 CH 4 at 98 K Option 4 H 4 at C Correct Answer 4 Explanation 3RT rms velocity = M Q.59 The energy of given amount an ideal gas depends only on its Option Pressure Option Volume Option 3 Number of moles Option 4 Temperature Correct Answer 4 Explanation 3 KE= nrt If n is constant, KE depends on T. Q.60 According to kinetic theory of gases, for a diatomic molecule Option The pressure exerted by the gas is proportional to the mean velocity of the molecule. Option The pressure exerted by the gas is proportional to the root mean square velocity of the molecules Option 3 The root mean square velocity of the molecule is inversely proportional to the temperature. Option 4 The kinetic energy of the molecules is proportional to the absolute temperature. Correct Answer 4 Explanation 3 KE= KT Where k is Boltzmann constant 3RT V rms = = M 3p d Q.6 At what temperature will the average speed of CH 4 molecules have the same value as O has at 300 K? Option 00 K

25 Option 50 K Option K Option K Correct Answer Explanation RT R 300 = MCH 4 3 T 300 = MCH 4 3 T 300 = 6 3 T=50k Q.6 Assertion : Most probable velocity is the velocity possessed by maximum fraction of molecules at the same temperature. Reasons : On collision, more and more molecules acquire higher speed at the same temperature. Option If both Assertion & Reasons are true and the reasons is the correct explanation of the assertion Option If both Assertion & Reasons are true but the reasons is not the correct explanation of the assertion Option 3 If Assertion is true statement but Reason is false Option 4 If both Assertion and Reason are false statement Correct Answer 3 Explanation Follow definition of most probable velocity. At constant temperature, a fraction of molecules always possess same velocity Q.63 Assertion : Different gases at the same condition of temperature and pressure have same root mean square speed. Reason : Average KE of gas is directly proportional to temperature in Kelvin. Option If both Assertion & Reasons are true and the reasons is the correct explanation of the assertion Option If both Assertion & Reasons are true but the reasons is not the correct explanation of the assertion Option 3 If Assertion is true statement but Reason is false Option 4 If both Assertion and Reason are false statement Correct Answer 4 Explanation Velocity of gas depends on its molecular wt and temperature 3 Any. KE per mole = RT Q.64 Which of the following is correct? Option The kinetic energy of H at 600 K is similar to kinetic energy of SO at 37 0 C Option The density of N is less than that of CO Option 3 The rms velocity of CH 4 at 400 K is similar to rms velocity of He at 00 K Option 4 The volume of an ideal gas can never be zero Explanation 37 0 C = = 600 K. Both H and SO gas are at Same temperature Average KE per mole will also be

26 At STP density of N = 8gm.4lt At STP density of CO = 44gm.4lt rms velocity of CH 4 at 400 K = = 3R R 4 Q.65 If root mean square speed of CH 4 (methane) at 48 K is same as the most probable speed of H at TK, then T is Correct Answer 0009 Is Integer Type Explanation V rms of CH 4 at 48 K = V mp of H at T K 3 R 48 R T = 6 3 R 48 R T = 6 T=9k Q.66 The factor(s) which measure(s) the deviation from ideal behaviour of a gas are Option Collision diameter Option Compressibility factor Option 3 Van der Waals constant a Option 4 Collision frequency Explanation Follow theory Q.67 Assertion : The gases He and H are very different in their behaviour at any temperature and pressure but their compressibility factors are nearly the same at the critical point. Reason : They have nearly the same critical constant. Option If both Assertion & Reasons are true and the reasons is the correct explanation of the assertion Option If both Assertion & Reasons are true but the reasons is not the correct explanation of the assertion Option 3 If Assertion is true statement but Reason is false Option 4 If both Assertion and Reason are false statement Correct Answer 3 Explanation If Assertion is true statement but Reason is false Q.68 Which of the following gases has the highest value of the van der Waals constant a? Option CCl 4 (l) Option NH 3 (g) Option 3 CO (g) Option 4 H O(g) Correct Answer 4 Explanation due to stronger Hydrogen Bonding Q.69 The constant a in van der Waal s equation is maximum in

27 Option Helium Option Hydrogen Option 3 Oxygen Option 4 Ammonia Correct Answer 4 Explanation due to stronger Hydrogen Bonding He, H, O has dispersive forces NH 3 has hydrogen Bonding (strong attractive forces) Q.70 na P+ V (V- nb) = nrt Which of the following statements is /are correct? Option It is real gas equation Option Higher the value of a more easily the gas can be liquefied Option 3 a is expressed in atm L mol - and b in L mol - Option 4 At high temperature this equation reduced into PV = nrt Explanation a attractiveforces between molecules n a =atm V mol a =atm a=atmlt m lt liquefaction is ersie Q.7 For NH 3 gas Z <. So the volume occupied by one mole of NH 3 (V m ) at STP is Option V m =.4L Option V m >.4L Option 3 V m <.4L Option 4 V m = 0 Correct Answer 3 Explanation Vm of real g as Z = at same T and P V of idealgas m If = < where V m molar vol. (vol.of me) V m of real gas < V m of ideal gas V m of real gas <.4 it at STP Q.74 Match Column -I with Column -II No. Column A Column B Column C Id of Additional Answer Real gas at high PV = RT + Pb pressure Force of attraction PV = RT + Pb among gas molecules is negligible 3 At high temperature and low pressure PV = nrt, Z = 4 Real gas at N.T.P. an P+ V (V nb) =

28 Explanation nrt A-P; B- ; C-Q,R ; D-S Vander Waals real gas equation for made of gas a P+ ( V-b ) =RT V Case I at low press (Vol. is large) V-b V a P+ V=RT v PV a =- RT VRT Z < -ve definition due to force of attraction Case II at high press a P+ p V ( ) P V b =RT PV Pb =+ RT RT Case III At very low press. (Vol. is every large) a P+ P as 0 V V V-b V PV=RT PV = RT Z = between as an ideal For ideal gas Z = Q.75 Assertion : Compressibility factor (Z) for non - ideal gases is always greater than. Reason : Non - ideal gases always exert higher presser than expected. Option If both Assertion & Reasons are true and the reasons is the correct explanation of the assertion Option If both Assertion & Reasons are true but the reasons is not the correct explanation of the assertion Option 3 If Assertion is true statement but Reason is false Option 4 If both Assertion and Reason are false statement Correct Answer 4 Explanation If both Assertion and Reason are false statement Q.77 Match the properties in the left column with corresponding function in the right column. No. Column A Column B Column C Id of Additional Answer (A) Compression factor (Z) of ideal gas.00 Z for real gas at low P a - VRT

29 3 Z for real gas at high pressure 4 Critical temperature (T c ) Explanation A-R ; B-S ; C-Q ; D-P pb + RT 8a 7Rb 8a Critical temperature (T c ) = 7Rb Follow question number 74 solution Q.78 In the following questions, a statement of assertion (A) is followed by a statement of reason (R) Assertion : When hydrogen gas expands adiabatically from high pressure to low pressure at room temperature then heating effect is observed. Reason : Hydrogen gas at room temperature is above its inversion temperature. Option If both Assertion & Reasons are true and the reasons is the correct explanation of the assertion Option If both Assertion & Reasons are true but the reasons is not the correct explanation of the assertion Option 3 If Assertion is true statement but Reason is false Option 4 If both Assertion and Reason are false statement Correct Answer Explanation Joule Thomson Effect Q.79 NH 3 can be liquefied at ordinary temperature without the application of pressure. But O cannot, because Option Its critical temp. is very high Option Its critical temp. is low Option 3 Its critical temp. is moderate Option 4 Its critical temperature is higher than that of ammonia. Correct Answer Explanation If critical temperature is high, liquidity is easier Q.80 The van der Waals constant of a gas are a = 0.75 dm 6 atm mol - b = 0.06 dm 3 mol - Hence, Option V c = dm 3 mol - Option V c = dm 3 mol - Option 3 P c = 54.5 atm Option 4 T c = 0 K Explanation 8a a T c =,Vc=3b,P c = 7Rb 7t Q.8 Match the following No. Column A Column B Column C Id of Additional Answer Attractive tendency Z < dominates At the Boyle s temperature in the high pressure region Z >

30 3 For a gas at very very low pressure and very very high temperature Z = 4 At the critical point Z = 3/8 Explanation A-Q ; B-R ; C-Q, D-P At critical point (P c, V c, T c ) a 3b P V c c Z = = 7b c RTc 8a R 7Rb = 3 8 If Z < attractive forces dominate Z > repulsive forces dominate Q.8 Assertion : Above critical temperature gaseous state cannot be observed. Reason : Above critical temperature, the average kinetic energy of a molecule is always greater than the maximum kinetic energy a molecule can possess in liquid state. Option If both Assertion & Reasons are true and the reasons is the correct explanation of the assertion Option If both Assertion & Reasons are true but the reasons is not the correct explanation of the assertion Option 3 If Assertion is true statement but Reason is false Option 4 If both Assertion and Reason are false statement Correct Answer Explanation A-P ; B-P ; C-R, D-S Boyle s Temperature (T B ) = a Rb Inversion Temperature (T B ) = a Critical Temperature (T C ) = Critical pressure (P C ) = a 7b Rb 8a 7Rb