Chemistry-82.Two Year CRP-Solutions

CHEMISTRY SOLUTIONS SECTION - A Part - I 1. B 2. A 3. D 4. A 5. C 6. B 7. B 8. D 9. C 10. D Part - II 1. Given conditions Final conditions V1 = 103 ml V2 = ? ml P1 = 763 mm P2 = 721 mm Applying Boyels law, V2 comes out to be 109 ml 2. Given conditions New conditions V1 = Vml V2 = V ml P1 = 760 mm = 1 atm P2 = 3 atm T1 = 300 K T2 = ? By applying gas equation T2 = 300  3 = 900 K Thus the temperature above which the tube will burst = 900 – 273 = 627C 3. Volume of the container = Volume of the gas  V = 5 litres Molecular mass of O2 = 32  n = moles T = 300 K R = 0.0821 Using PV = nRT We will get P = 1.54 atm SECTION - B Part - I 1. A 2. C 3. D 4. B 5. B Part - II Solution: = = = 5.45 1017 s–1 ½ mv2 = h–ho = (6.6310–27 ergs–sec) (5.451017s–1) –2.62 10–9 ergs K.E. = 9.93 10–10 ergs 3

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