1. Xtra Edge February 2012 - Career Point

More documents

Recommendations

Info

4. A piston can freely move inside a horizontal cylinder closed from both ends. Initially, the piston separates the inside space of the cylinder into two equal parts each of volume V0, in which an ideal gas is contained under the same pressure p0 and at the same temperature. What work has to be performed in order to increase isothermally the volume of one part of gas η times compared to that of the other by slowly moving the piston ? Sol. Let volume of chambers changes by ∆V. According to the problem, the final volume of left chamber is η times final volume of right chamber. ∴ V0 + ∆V = η(V0 – ∆V) or ⎛ η −1 ⎞ ∆V = ⎜ ⎟V0 ⎝ η + 1⎠ P0,v0,T0 P0,v0,T0 As piston is moved slowly therefore, change in kinetic energy is zero. By work-energy theorem, we can write ext Wgas in right chamber + Wgas in left chamber + W Agent = ∆KE ext W Agent = (Wgas(R) + Wgas(L)) We know that in isothermal process, work done is given by ⎛V f ⎞ W = nRT ln ⎜ ⎟ ⎜ ⎟ ⎝ Vi ⎠ ∴ Work done by gas in left chamber (WL) ⎛V ⎞ = P0V0 ln ⎜ 0 + ∆V ⎛ 2η ⎞ ⎟ = P0V0 ln ⎜ ⎟ ⎝ V0 ⎠ ⎝ η + 1⎠ Similarly, work done by gas in right chamber (WR) ⎛V ⎞ = P0V0 ln ⎜ 0 − ∆V ⎛ 2η ⎞ ⎟ = P0V0 ln ⎜ ⎟ ⎝ V0 ⎠ ⎝ η + 1⎠ ext ⎛ 2η ⎞ W Agent = –P0V0 ln ⎜ ⎟ ⎝ η + 1⎠ 2 ⎛ η + 1⎞ = P0V0 ln ⎜ ⎟ ⎝ 4η ⎠ ⎛ 2η ⎞ – P0V0 ln ⎜ ⎟ ⎝ η + 1⎠ 5. A smooth vertical tube having two different sections is open from both ends equipped with two pistons of different areas figure. Each piston slides within a respective tube section. One mole of ideal gas is enclosed between the pistons tied with a nonstretchable thread. The cross-sectional area of the upper piston is ∆S greater than that of the lower one. The combined mass of the two pistons is equal to m. The outside air pressure is P0. By how many kelvins must the gas between the pistons be heated to shift the pistons through l. Sol. Let A1 = Cross section of upper piston A2 = Cross section of lower piston T = Tension in the string P = Gas pressure m1 = Mass of upper piston m2 = Mass of lower piston Now, consider FBD of upper piston XtraEdge for IIT-JEE 28 FEBRUARY 2012 P0 P0 P0 A1 PA1 m1g From equilibrium consideration of upper piston we get, P0A1 + T + m1g = PA1 Similarly, consider FBD of lower piston T PA2 P0 A2 m2g ∴ P0A2 + T = m2g + PA2 Eliminating T, we get ( m1 + m2) g P = P0 + A1 − A2 According to problem m = m1 + m2 and ∆S = A1 – A2 ∴ mg P = P0 + ∆ S Now, PV = RT or P∆V = R∆T P∆ V or ∆T = R But ∆V = (A1 – A2)l = ∆S. l ⎛ mg ⎞ ∴ ∆T = ⎜ P0 + ⎟ ∆S.l ⎝ ∆S ⎠ l l l
KEY CONCEPT Organic Chemistry Fundamentals Reduction of Aldehydes and Ketones by Hydride Transfer : R δ+ δ– H3B – H + C = O R´ R R – H – OH H – C – O H – C – O – H Hydride transfer Alkoxide ion Alcohol R R´ These steps are repeated until all hydrogen atoms attached to boron have been transferred. Sodium borohydride is a less powerful reducing agent than lithium aluminum hydride. Lithium aluminum hydride reduces acids, aldehydes, and ketones but sodium borohydride reduces only aldehydes and ketones : O Reduced by LiAlH4 O C < C < C < O– R OR´ R R´ R O Ease of reduction R´ Reduced by NaBH4 O C H Lithium aluminum hydride reacts violently with water, and therefore reductions with lithium aluminum hydride must be carried out in anhydrous solutions, usually in anhydrous ether. (Ethyl acetate is added cautiously after the reaction is over to decompose excess LiAlH4; then water is added to decompose the aluminum complex.) Sodium borohydride reductions, by contrast, can be carried out in water or alcohol solutions. The Addition of Ylides : The Wittig reaction : Aldehydes and ketones react with phosphorus ylides to yield alkenes and triphenylphosphine oxide. (An ylide is a neutral molecule having a negative carbon adjacent to a positive heteroatom.) Phosphorus ylides are also called phosphoranes : CARBONYL COMPOUND + .. R´´ C = O + (C6H5)3P – C R´´´ XtraEdge for IIT-JEE 29 FEBRUARY 2012 R R Aldehyde or ketone Phosphorus ylide or phosphorane R R´ C = C R´´ + O =P(C6H5)3 R´´´ Alkene [(E) and(Z) isomers] Triphenyl phosphine oxide This reaction, known as the Wittig reaction, has proved to be a valuable method for synthesizing alkenes. The Wittig reaction is applicable to a wide variety of compounds, and although a mixture of (E) and (Z) isomers may result, the Wittig reaction offers a great advantage over most other alkene syntheses in that no ambiguity exists as to the location of the double bond in the product. (This is in contrast to E1 eliminations, which may yield multiple alkene products by rearrangement to more stable carbocation intermediates, and both E1 and E2 elimination reactions, which may produce multiple products when different β hydrogens are available for removal.) Phosphorus ylides are easily prepared from triphenylphosphine and primary or secondary alkyl halides. Their preparation involves two reactions : General Reaction Reaction 1 (C6H5)3P : + CH – X → (C6H5)3P – CH X – R´´ R´´ + R´´´ R´´´ Triphenylphosphine An alkyltriphenylphosphonium halide Reaction 2 (C6H5)3P – C – H : B – ⎯→ (C6H5)3P – C : – R´´ R´´ + + + H:B R´´´ R´´´ A phosphorus ylide Specific Example Reaction 1 (C6H5)3P : + CH3Br ⎯→ (C6H5)3P – CH3Br – C6H6 + Reaction 2 Methyltriphenylphosphonium bromide (89%)
Page 3 and 4: Volume - 7 Issue - 8 February, 2012
Page 5 and 6: 25% increase in number of girls fro
Page 7 and 8: Dr Amitabha Ghosh was the only Asia
Page 9 and 10: Sol. (i) The capacitor A with diele
Page 11 and 12: Sol. (i) Let m be the mass of elect
Page 13 and 14: Step 5. 1 gram equivalent acid neut
Page 15 and 16: = 2π ∫ π 2 / 3 2 t sec dt 2 t 1
Page 17 and 18: (A) (C) A 2B0πqR 3 B 0 PR 3 2 ×
Page 19 and 20: PHYSICS 1. AB and CD are two ideal
Page 21 and 22: 4. In a Young's experiment, the upp
Page 23 and 24: Matter Waves : Planck's quantum the
Page 25 and 26: The intercept of V0 versus ν graph
Page 27 and 28: PHYSICS FUNDAMENTAL FOR IIT-JEE The
Page 29: etween total mass and number of mol
Page 33 and 34: KEY CONCEPT Inorganic Chemistry Fun
Page 35 and 36: 1. It is possible to supercool wate
Page 37 and 38: z 1 K2 = = y( 0. 48M − z) 1. 26×
Page 39 and 40: MATHEMATICAL CHALLENGES 1. as φ (a
Page 41 and 42: 9. Point P (x, 1/2) under the given
Page 43 and 44: 10. 11 ∴ n(E) = 10 + 9 + 8 + ...
Page 45 and 46: dx (ii) ∫ 2 ax + bx + c This can
Page 47 and 48: MATHS Functions with their Periods
Page 49 and 50: a PHYSICS Questions 1 to 9 are mult
Page 51 and 52: R P ⊗ B O ω0 14. Magnitude of cu
Page 53 and 54: XtraEdge for IIT-JEE 51 FEBRUARY 20
Page 57 and 58: 7. NaOH can be prepared by two meth
Page 59 and 60: (C) X = 100 ml of SO2 at 1 bar, 25
Page 61 and 62: 21. Column-I Column-II (A) If y = 2
Page 63 and 64: (A) potential energy (PE) increases
Page 65 and 66: 2. In the reaction 3 Cu + 8 HNO3
Page 67 and 68: (D) P A T B (S) Pressure is increas
Page 71 and 72: 14. Derive the expression for the c
Page 73 and 74: (a) temperature and (b) Pressure ?
Page 75 and 76: should be bought to meet the requir
Page 77 and 78: This work done is stored inside the
Page 79 and 80: Use : It can be used to accelerate
Page 81 and 82:
10. (i) (ii) Benzene H CH3 - C = O
Page 83 and 84:
NH3 NH3 NH3 Co +3 NH3 NH3 NH3 No un
Page 85 and 86:
7. f(A) = A 2 - 5A + 7 I2 ⎡ 3 1
Page 87 and 88:
y 2 = - (2 - x) 2 + 9 This is the e
Page 89 and 90:
(0, 4) ( 2 3, 2) (4, 0) ⎡ 2 4 2
Page 91 and 92:
15. X = ? Y = 20 Ω l1 = 40 cm l -
Page 93 and 94:
(i) Emission : - If radiations emit
Page 95 and 96:
OH (ii) + Br2 Phenol H2O 13. I st M
Page 97 and 98:
28. (a) Acetylene is first oxidized
Page 99 and 100:
OR Let A → total of 8 in first th
Page 101 and 102:
= x 2 ⎡ ⎢ ⎢x ⎢ ⎣ 0 1 2 1
Page 103 and 104:
28. x x y ∴ x → side of square
Page 105 and 106:
C 'XtraEdge for IIT-JEE IIT JEE bec
Page 107 and 108:
XtraEdge for IIT-JEE 105 FEBRUARY 2
show all

1. Xtra Edge February 2012 - Career Point

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?