Gaseous State Exercise # 3

GASEOUS STATE Exercise # 3 1. Part # I [Matrix Match Type Questions] For a fixed amount of the gas match the two column : Column-I Column-II (V3,T3) (V2,T2) (A) PV (V1,T1) (p) T1 > T2 > T3 P (P3V3) (P2V2) (B) V (P1V1) (q) P1 > P2 > P3 T (P3V3) (P2V2) (C) P (P1V1) (r) V1 > V2 > V3 T (d3T3) (d2T2) (D) d (d1T1) (s) d1 > d2 > d3 P 2. Column-I Column-II (A) P1V1 = P2V2 = P3V3 = .................. (1) Dalton’s law of partial pressures at constant temperature V1 V2 V3 (B) T = T = T = ................ at constant 1 2 3 pressure. (2) Kinetic equation of ideal gases. 1 d (D) P = P1 + P2 + P3 +............... (C) r  a   (E) (V – b)  P + 2  = RT T V   (F) R/N (G) Molar volume 1 mnc2 3 (I) Graph between P and V at constant temperature (J) Graph between V and T at constant pressure (H) PV = (3) 22.4 litre at STP (4) Isotherm (5) Isobar (6) Charles’ law (7) Graham’s law (8) Boyle’s law (9) Equation for real gases. (10). Boltzmann’s constant. 3. Column – I Column – II pb RT (A) At low pressure (p) Z = 1 + (B) At higher pressure (q) Z = 1 – V RT m (C) At low density of gas (r) gas is more compressible than ideal gas (D) For H2 and He at 0°C (s) gas is less compressible than ideal gas Part # II a [Comprehension Type Questions] Comprehension # 1 Critical constant of A gas T4 D When pressure is incerases at constant temp volume of gas P3 TC decreases P AB  gases, BC  vapour + liquid, CD  liquid P2 critical point : At this point all the physical properties of T3 (T3 >T 2) isotherm liquid phase will be same as the physical properties in vapour T2 >T1 C P1 B such as, density of liquid = density of vapour T1 TC or critical temp : Temperature above which a gas can VC Vm not be liquified PC or critical pressure : minimum pressure which must be applied at critical temp to convert the gas into liquid. VC or critical volume : volume occupied by one mole of gas at TC & PC Critical constant using vander wall equations  a  P + 2 Vm    (Vm – b) = RT T  PVm3 + aVm – PbVm2 – ab – RTVm2 = 0  ( PVm2 + a ) (Vm – b) = RT Vm2  RT  a V  ab Vm3 + Vm2  b + + – =0  P  P m P  since equation is cubic in Vm hence there will be three roots of equation at any temperature and pressure. At critical point all three roots will coincide and will give single value of Vm = Vc at critical point, Vander Waal equation will be  a ab RTC  Vm3 – Vm2  b + P  + P Vm – P = 0 C   C C ...(1) But at critical point all three roots of the equation should be equal, hence equation should be : Vm, = Vc (Vm – Vc)3 = 0 Vm3 – 3Vm2 VC + 3Vm VC2 – VC3 = 0 ...(2) comparing with equation (1) RTC b + P = 3VC C ....(i) a 2 PC = 3 VC ...(ii) ab 3 PC = VC ...(iii) 464 GASEOUS STATE From (ii) and (iii), VC = 3b a From (ii) PC = 3V 2 C a a PC = 3 (3b)2 = 27b 2 substituting RTC PC = 3 VC – b = 9b – b = 8b From (i) 8a TC = 27 Rb  At critical point, the slope of PV curve (slope of isotherm) will be zero at all other point slope will be negative zero is the maximum value of slope.  P    =0  Vm TC   P    Vm  Vm TC = 0 ...(i) ....(ii) {Mathematically such points an known as point of inflection (where first two derivatives becomes zero)} using the two TC PC and VC can be calculate by 1. RT B C A scientist proposed the following equation of state P = V – V 2 + V3 . If this equation leads to the critical m m m behaviour then critical temperature is : (A) 8B 27RC (B) B 8RC (C) B2 3RC (D) None of these , 2. If the critical constants for a hypothetical gas are VC = 150 cm3 mol–1. PC = 50 atm and TC = 300 K. Then the approximate radius of the molecule is : [Take R = 1/3  75   (A)   2πN A  3. 1/3  75   (B)   8πN A  1 Ltr atm mol–1 K–1] 12 1/3  3   (C)   πN A  Identify the wrong statement related to the above graph : (A) between 50 K and 150 K temperature and pressure ranging from 10 atm to 20 atm matter may have liquid state. (B) zero is the maximum value of the slope of P-V Curve (C) If vander wall equation of state is applicable above critical temperature then cubic equation of Vm will have one real and two imaginary roots. (D) At 100 K and pressure below 20 atm it has liquid state only 1/3   3  (D)  256πN A   P C 200K 20 atm 10 atm VC 150K 50K V 464 Comprehension # 2 MEASUREMENT OF PRESSURE Barometer A barometer is an instrument that is used for the measurement of pressure. The construction of the barometer is as follows Perfect Vaccum Hg is filled upto brim Hg P0 Thin nairow glass tube P0 P0= Patm P0 Cross sectional view of the capillary column A thin narrow calibrated capillary tube is filled to the brim, with a liquid such as mercury, and is inverted into a trough filled with the same fluid. Now depending on the external atmospheric pressure, the level of the mercury inside the tube will adjust itself, the reading of which can be monitored. When the mercury column inside the capillary comes to rest, then the net forces on the column should be balanced. Applying force balance, we get, Patm × A= m×g (‘A’ is the cross-sectional area of the capillary tube) If ‘ ’ is the density of the fluid, then m= × g × h (‘h’ is the height to which mercury has risen in the capillary) hence, Patm × A = (× g × h) × A or, Patm = gh Faulty Barometer An ideal barometer will show a correct reading only if the space above the mercury column is vacuum, but in case if some gas column is trapped in the space above the mercury column, then the barometer is classified as a faulty barometer. The reading of such a barometer will be less than the true pressure. For such a faulty barometer P ×A gas or P0A = mg + Pgas A P0 = gh + Pgas gh = P0 – Pgas Patm × A mg 1. A tube closed at one end is dipped in mercury as shown in figure-3 such that the closed surface coincides with the mercury level in the container. By how much length of the tube should be extended such that the level of Hg in the tube is 5 cm below the mercury level inside the container. (assume temperature remains constant) Patm = 70 cm of Hg 20 cm Fig.-3 (A) 18 cm 2. (B) 19 cm (C) 24 cm If above tube is placed vertically with the open end upward then the length of the air column will be (assume temperature remains constant) (A) 20 cm (C) 18 cm (B) 36 cm (D) 15 cm (D) 30 cm 24 cm Patm = 75 cm 10cm 10cm, of Hg = 20.4 gm/ml 466 GASEOUS STATE 3. A gas column is trapped between closed end of a tube and a mercury column of length (h) when this tube is placed with its open end upwards the length of gas column is (1) the length of gas column becomes (2) when open end of tube is held downwards (as shown in fig.-4). Find atmospheric pressure in terms of height of Hg column. (assume temperature remains constant) Fig.-4 h (l 1 + l 2 ) (A) (l – l ) 2 1 h (l 2 – l 1 ) (B) (l + l ) 1 2 (l 1 + l 2 ) (C) h (l – l ) 2 1 (D) (h11 + h22) Comprehension # 3 One of the important approach to the study of real gases involves the analysis of a parameter Z called the compressibility factor Z = PVm where P is pressure, Vm is molar volume, T is absolute temperature and R is the RT  Vm real universal gas constant. Such a relation can also be expressed as Z =   Vm ideal   (where Vm ideal and Vm real are the molar   volume for ideal and real gas respectively). Gas corresponding Z > 1 have repulsive tendencies among constituent particles due to their size factor, whereas those corresponding to Z < 1 have attractive forces among constituent particles. As the pressure is lowered or temperature is increased the value of Z approaches 1. (reaching the ideal behaviour) 1. Choose the conclusions which are appropriate for the observation stated. Observation Conclusion I. Z=1 I. The gas need not be showing the ideal behaviour II. Z>1 II. On applying pressure the gas will respond by increasing its volume III. Z<1 III. The gas has the ability to be liquefied. IV. Z  1 for low P IV. The gas is approaching the ideal behaviour. (A) All conclusions are true (C) Conclusions I,III & IV are true 2. (B) Conclusions I, II & IV are true (D) Conclusions III & IV are true For a real gas ‘G’ Z > 1 at STP Then for ‘G’ : Which of the following is true : (A) 1 mole of the gas occupies 22.4 L at NTP (B) 1 mole of the gas occupies 22.4 L at pressure higher than that at STP (keeping temperature constant) (C) 1 mole of the gas occupies 22.4 L at pressure lower than that at STP (keeping temperature constant) (D) None of the above 466 3. Following graph represents a pressure (P) volume (V) relationship at a fixed temperature (T) for n moles of a real gas. The graph has two regions marked (I) and (II). Which of the following options is true. (A) Z < 1 in the region (II) (B) Z = 1 in the region (II) (C) Z = 1 for the curve (D) Z approaches 1 as we move from region (II) to region (I) Region (II) P Region (I) Comprehension # 4 V The rate of change of pressure (p) of a gas at constant temperature and constant external pressure due to effusion of gas from a vessel of constant volume is related to rate of change of number of molecules present by dp kT dN = dt V dt where k = Boltzmann constant, T = temperature, V = volume of vessel & N = No. of molecules and dN –pA 0 = , where A0 = area of orifice and m = mass of molecule. dt (2πmkT)1/ 2 Time required for pressure inside vessel to reduce to 1/e of its initial value in (ln e = 1) 1/2 1.  2πm   (A)   kT  1/2  kT   (B)   2πm  V A0 V A0 1/2 2πm V (D) kT A 0  2πmkT  (C)  A  0   If the gas inside the vessel had molecular weight 9 times the gas in previous example and area of orifice was 2. doubled and temperature maintained at 4T, time required for pressure to fall to 1/e times of its initial value would be (t = answer of previous option) (A) 1.33 t (B) 4.24 t (C) 0.75 t (D) 1.125 t The incorrect statement (s) is/are. 3. [I] Pressure will not fall to zero in infinite time. [II] Time required for pressure to decrease to half its initial value is independent of initial pressure. [III] The relations given above are true for real gases also. (A) I (B) II (C) III (D) I and III Comprehension # 5 Sketch shows the plot of Z v/s P for a hypothetical gas for one mole at three distinct temperature. 2 200K 500K 1000K Z1 1 Z 500 250 P (atm.) 750 H2 0°C CO2 P Boyle's temperature is the temperature at which a gas shows ideal behaviour over a pressure range in the low a pressure region. Boyle's temperature (Tb) = . If a plot is obtained at temperatures well below Boyle's Rb GASEOUS STATE temperature then the curve will show negative deviation, in low pressure region and positive deviation in the high pressure region. Near critical temperature the curve is more likely as CO2 and the temperature well above critical temperature curve is more like H2 at 0°C as shown above. At high pressure suppose all the constant Pb temperature curve varies linearly with pressure according to the following equation Z = 1 + RT (R = 2 cal mol–1 K–1) 1. Which of the following is correct : a a < 0.4 k cal mol–1 (B) 0.4 k cal mol–1 < < 2 k cal mol–1 b b a a (C) > 0.4 k cal mol–1 (D) = 1 K cal mol–1 b b For 500 K plot value of Z changes from 2 to 2.2 if pressure is varied from 1000 atm to 1200 atm (high pressure) b then the value of will be :RT (A) 10–3 atm–1 (B) 2 × 10–3 atm–1 (C) 5 × 10–4 atm–1 (D) 10–4 atm–1 (A) 2. 3. As shown in the figure at 200 K and 500 atm value of compressibility factor is 2 (approx). Then volume of the gas at this point will be :(A) 0.01 L 4. 5. (B) 0.09 L (C) 0.065 L (D) 0.657 L Plot at Boyle's temperature for the gas will be : (A) (B) (C) (D) In very high pressure region if Z v/s P is plotted at 1200 K for the above gas then it will have greatest slope. (A) true (B) false (C) can't say (D) not related to the paragraph 468 ANSWER KEY EXERCISE - 3 : PART # I 1. A (s), B (q, s), C (r), D (p) 2. 3. A (8), B (6), C (7), D (1), E (9), (F) (10), (G) (3), (H) (2), (I) (4), (J) (5) A (q, r), B (p, s), C (q, r), D (p, r) PART # II Comprehension # 1 : 1. C 2. D 3. D Comprehension # 2 : 1. B 2. C 3. A Comprehension # 3 : 1. D 2. B 3. D Comprehension # 4 : 1. A 2. C 3. C Comprehension # 5 : 1. B 2. A 3. C 4. C 5. B 486